\| \cdot \|_{\Omega,m} : \overline{\mathscr{C}}^{m} (\Omega) \to \mathbb{R}_{\geq 0}, \quad
f \mapsto
Definition 4.6.1
Assume to be given a locally convex topological vector space \mathrm{V} over the field \mathbb{K} of real or complex numbers.
Let (v_i)_{i \in I} be a family of elements of \mathrm{V}. Let \mathscr{F} (I) be the
set of finite subsets of I and note that it is filtered by set-theoretic inclusion.
The family (v_i)_{i \in I} then gives rise to the net
\Big( \sum_{i \in J} v_i \Big)_{J \in \mathscr{F}(I)}. One calls the family (v_i)_{i \in I}
summable to an element v \in \mathrm{V} if the net
\Big( \sum_{i \in J} v_i \Big)_{J \in \mathscr{F}(I)} converges to v. In other words this means that
for every convex zero neighborhood U \subset \mathrm{V} and \varepsilon > 0 there exists an element
J_{U,\varepsilon} \in \mathscr{F} (I) such that for all finite sets J with J_{U,\varepsilon} \subset J \subset I
p_U \left( v - \sum_{i \in J} v_i \right) < \varepsilon .
As before, p_U denotes here the gauge of U.
If \mathrm{V} is Hausdorff, the limit v of a summable family (v_i)_{i \in I} is
uniquely determined, and one writes in this situation
v = \sum_{i \in I} v_i .
We denote the space of summable families in \mathrm{V} over the given index set I by
\ell^1 (I ,\mathrm{V}). For E=\mathbb{C} we just write \ell^1 (I ) instead of \ell^1 (I ,\mathbb{C}). If in addition the index set
coincides with \mathbb{N}, we briefly denote \ell^1 (\mathbb{N} ) by \ell^1.
Proposition 4.6.2[Cauchy criterion for summability]
Let \mathrm{V} be a complete locally convex topological vector space.
A family (v_i)_{i \in I} of elements of \mathrm{V} then is summable
to some v\in \mathrm{V} if and only if it satisfies the following Cauchy condition:
Proof.
By completeness of \mathrm{V} it suffices to verify that the net \Big( \sum_{i \in J} v_i \Big)_{J \in \mathscr{F}(I)} is a Cauchy net
if and only if condition (C) is satisfied. Recall that one calls
\Big( \sum_{i \in J} v_i \Big)_{J \in \mathscr{F}(I)} a Cauchy net if for every convex zero neighborhood U \subset \mathrm{V}
all \varepsilon > 0 there exists an element J_{U,\varepsilon} \in \mathscr{F} (I) such that for all J,J' \in \mathscr{F} (I)
containing J_{U,\varepsilon} as a subset the relation
p_U \left( \sum_{i \in J} v_i - \sum_{i\in J'} v_i \right) < \varepsilon
holds true.
But that is clearly equivalent to condition (C).
4.6.3 Several other notions of summability have been introduced in the analysis and functional analysis literature.
These are mainly either used to establish summability criteria or are used in the study of
topological tensor products and nuclearity of locally convex topological vector spaces, see
Definition 4.6.4
A family (v_i)_{i \in I} in \mathrm{V} is called weakly summable to v \in \mathrm{V} if for
every continuous linear form \alpha : \mathrm{V} \to \mathbb{K} the net
\Big( \sum_{i \in J} \alpha( v_i ) \Big)_{J \in \mathscr{F}(I)} converges in \mathbb{K} to \alpha ( v ).
In other words this means that for every \alpha \in \mathrm{V}' and \varepsilon > 0 there exists a finite set
J_{\alpha,\varepsilon} \subset I such that for all finite sets J with J_{\alpha,\varepsilon} \subset J \subset I
\left| \alpha (v) - \sum_{j\in J} \alpha( v_i ) \right| < \varepsilon .
The set of all weakly summable families in \mathrm{V} with index set I is denoted \ell^1 [I, \mathrm{V}].
Definition 4.6.5
A family (v_i)_{i \in I} in \mathrm{V} is called absolutely summable if for every circled convex
zero neighborhood U \subset \mathrm{V} there exists some C\geq 0 such that
\sum_{i \in J} p_U \left( v_i\right) \leq C \quad \text{for all } J \in \mathscr{F} (I) .
We denote the set of all absolutely summable families in \mathrm{V} by \ell^1 \{ I, \mathrm{V} \}.
Proposition 4.6.6
A family (v_i)_{i \in I} \subset \mathrm{V} is absolutely summable if and only if for every element U
of a basis of circled convex zero neighborhoods there exists a C\geq 0 such that
\sum_{i \in J} p_U \left( v_i\right) \leq C \quad \text{for all } J \in \mathscr{F} (I) .
Proof.
Definition 4.6.7
A family (v_i)_{i \in I} in \mathrm{V} is called totally summable if there exists a bounded
absolutely convex subset B\subset \mathrm{V} and a C\geq 0 such that
\sum_{i \in J} p_B \left( v_i\right) \leq C \quad \text{for all } J \in \mathscr{F} (I) .
We write \ell^1 \langle I, \mathrm{V} \rangle for the set of all totally summable families in \mathrm{V}.
\addcontentslinetocsubsectionSummable families of complex numbers
Lemma 4.6.8[cf.~
\left|\sum_{i\in J} z_i \right| \leq C \quad \text{for all } J \in \mathscr{F} (I) .
Then one has the estimate
\sum_{i\in J}\left| z_i \right| \leq 4 C \quad \text{for all } J \in \mathscr{F} (I) .
Proof.
We assume first that all z_i are real. Then let I^+ the set of all indices i\in I such that z_i \geq 0,
and I^- the set of all i\in I such that z_i < 0. Then, for all finite J\subset I
\sum_{i\in J} \left| z_i \right| = \sum_{i\in J\cap I^+} \left| z_i \right| + \sum_{i\in J\cap I^-} \left| z_i \right|
= \left| \sum_{i\in J\cap I^+} z_i \right| + \left| \sum_{i\in J\cap I^-} z_i \right| \leq 2 C .
In the general case decompose z_i into real and imaginary parts x_i = \Re z_i and y_i = \Im z_i.
By the triangle inequality one obtains for all finite J\subset I
\sum_{i\in J} \left| z_i \right| \leq \sum_{i\in J} \left| x_i \right| + \sum_{i\in J} \left| y_i \right| \leq 4 C .
Proposition 4.6.9
For a family (z_i)_{i\in I} of complex numbers the following are equivalent.
In case that one hence all of the conditions are fulfilled, the estimate
\left| \sum_{i\in I} z_i \right| \leq \sum_{i\in I}\left| z_i \right|
holds true.
Proof.
Assume that (z_i)_{i\in I} is absolutely summable. Since \mathbb{C} is normed with norm given by the absolut value
this just means that there exists some C > 0 such that
\sum_{i\in J}\left| z_i \right| \leq C for all J\in \mathscr{F} (I). Hence the supremum
c = \sup \left\{ \sum_{i\in J}\left| z_i \right| \mid J \in \mathscr{F} (I)\right\} exists and is \leq C.
For given \varepsilon > 0 choose J_\varepsilon \in \mathscr{F} (I) such that
c- \varepsilon \leq \sum_{i\in J_\varepsilon}\left| z_i \right| \leq c .
Then one has for all K\in \mathscr{F} (I) with K\cap J_\varepsilon = \emptyset
\left| \sum_{i\in K} z_i \right| \leq \sum_{i\in K} \left| z_i \right| \leq \varepsilon .
Hence \left( \sum_{i\in J} z_i \right)_{J \in \mathscr{F} (I)} is a Cauchy net, so has to
converges by completeness of \mathbb{C}. This proves summability of (z_i)_{i\in I}.
Vice versa, assume now that (z_i)_{i\in I} is summable. Then
\left( \sum_{i\in J} z_i \right)_{J \in \mathscr{F} (I)} is a Cauchy net. Hence there exists an element
J_1 \in \mathscr{F} (I) such that for all K\in \mathscr{F} (I) with K \cap J_1 = \emptyset
the inequality
\left| \sum_{i\in K} z_i \right| < 1
holds true. Let C = \sum_{i\in J_1} \left| z_i\right|. Then one has for all J\in \mathscr{F} (I)
\left| \sum_{i\in J} z_i \right| \leq \left| \sum_{i\in J\setminus J_1} z_i \right| +
\left| \sum_{i\in J\cap J_1} z_i \right| \leq 1 + C .
By the preceding lemma the set of partial sums \sum_{i\in J} \left| z_i \right|, where J runs through the
finite subsets of I, is then bounded by 4 + 4C, hence (z_i)_{i\in I} is absolutely summable.
\addcontentslinetocsubsectionSummability in Banach spaces
Proposition 4.6.10
Let \mathrm{V} be a normed vector space.
For a family (v_i)_{i\in I} of elements in \mathrm{V} the following are equivalent:
If \mathrm{V} is even a Banach space, these conditions are all equivalent to
Proof.
(II) and (III)
are equivalent by \Crefthm:summability-criteria-family-complex-numbers
Assume now that (I) holds true.
to do: Carl Neumann series
\addcontentslinetocsubsectionProperties of and relations between the various summability types
Theorem 4.6.11
Let I be a non-empty index set. Then the spaces \ell^1 (I, \mathrm{V} ) of summable families,
\ell^1 [I, \mathrm{V} ] of weakly summable families, \ell^1 \{I, \mathrm{V} \} of absolutely summable families and
\ell^1 \langle I, \mathrm{V} \rangle of totally summable families in E are all subvector spaces of
the product vector space E^I = \Pi_{i\in I} E. Furthermore one has the following chain of inclusions:
\ell^1 \langle I, \mathrm{V} \rangle\subset\ell^1 \{I, \mathrm{V} \} \quad\text{and} \quad \ell^1 (I, \mathrm{V}) \subset \ell^1 [I, \mathrm{V} ] .
If E is complete, then one even has
\ell^1 \{I, \mathrm{V} \} \subset \ell^1 (I, \mathrm{V})
Proof.
Now let (v_i) be a summable family and \alpha : \mathrm{V} \to \mathbb{K} a continuous linear form.
 Then U absorbes B, so there exists r > 0 such that
B \subset rU. Hence
Definition 4.7.1[cf.~
The locally convex vector topology \tau is called strongly compatible with the tensor product structure, a strongly admissible tensor product topology or briefly strongly admissible if it satisfies:
4.7.2
The admissible respectively strongly admissible vector topologies on \mathrm{V} \otimes \mathrm{W} are obviously partially ordered
by set-theoretic inclusion. Therefore, the following definition makes sense.
Definition 4.7.3
6.1.1
Let us first remind the reader that as before \mathbb{K} stands for the field
of real or of complex numbers. We will keep this notational agreement
throughout the whole chapter.
Definition 6.1.2
By a sesquilinear form on a \mathbb{K}-vector space \mathrm{V} one understands a map
\langle \cdot , \cdot \rangle : \mathrm{V} \times \mathrm{V} \to \mathbb{K}
with the following two properties:
A hermitian form is a sesquilinear form \langle \cdot , \cdot \rangle on \mathrm{V} with the following additional property:
A sesquilinear form \langle \cdot , \cdot \rangle is called weakly-nondegenerate if it satisfies axiom
Finally, one calls a hermitian form \langle \cdot , \cdot \rangle on \mathrm{V} positive semidefinite if
Remark 6.1.3
Recall that a map \langle \cdot , \cdot \rangle : \mathrm{V} \times \mathrm{V} \to \mathbb{K} is called bilinear,
if it satisfies ( SF1) and
If the underlying ground field \mathbb{K} coincides with the field of real numbers,
a sesquilinear form is by definition the same as a bilinear form,
and a hermitian form the same as a symmetric bilinear form.
6.1.4
Given a positive semidefinite hermitian form \langle \cdot , \cdot \rangle on a
\mathbb{K}-vector space \mathrm{V}, one calls two vectors v,w \in \mathrm{V} orthogonal
if \langle v , w \rangle = 0. Since the hermitian form
\langle \cdot , \cdot \rangle is assumed to be positive semidefinite, the map
\| \cdot \| : \mathrm{V} \to \mathbb{R}_{\geq 0} , \: v \mapsto \|v \| = \sqrt{\langle v , v \rangle}
is well-defined. We will later see that \| \cdot \| is a seminorm on \mathrm{V}
and therefore call the map \| \cdot \| the seminorm associated to
\langle \cdot , \cdot \rangle. The following formulas are immediate consequences of the
properties defining a positive semidefinite hermitian form and the definition
of the associated seminorm:
\tag{6.1.1}
\| v+w \|^2 = \| v\|^2 + 2\, \Re \, \langle v , w \rangle + \| w\|^2 \quad \text{for all }
v,w \in \mathrm{V} ,
\| v+w \|^2 = \| v\|^2 + \| w\|^2 \quad \text{for all orthogonal } v,w \in \mathrm{V} ,
\| v+w \|^2 + \|v-w\|^2 = 2 \big( \| v\|^2 + \| w\|^2 \big) \quad \text{for all } v,w \in \mathrm{V},
\| rv \| = \sqrt{ |r|^2 \langle v , v \rangle } = |r| \| v \|
\quad \text{for all } v,w \in \mathrm{V} \text{ and } r\in \mathbb{K} .
Formula \eqrefeq:pythagorean-theorem is an abstract version of the pythagorean theorem,
\Crefeq:parallelogram-identity is called the parallelogram identity.
The triangle inequality for the map \| \cdot \| will turn out to be a consequence of the
following result.
Proposition 6.1.5[Cauchy--Schwarz inequality]
Given a positive semidefinite hermitian form \langle \cdot , \cdot \rangle on a \mathbb{K}-vector space
\mathrm{V} the following inequality holds true:
\tag{6.1.2}
|\langle v, w \rangle| \leq \|v\|\|w\| \quad \text{for all $v,w \in \mathrm{V}$}.
Equality holds if and only if v and w are linearly dependant.
Proof.
If v or w is 0, the proof is trivial, hence we can assume both to be nonzero. Put
c = - \frac{\langle v, w \rangle}{\|v\|^2}
and compute
\tag{6.1.3}
0 \leq \|c v + w\|^2 = \langle c v + w, c v + w \rangle =
|c|^2 \langle v, v \rangle + c \langle v, w \rangle +
\overline{c}\langle w, v \rangle + \langle w, w \rangle =
= |c|^2\|v\|^2 + 2 \, \Re \big( c\langle v, w \rangle \big) + \|w\|^2 =
\frac{|\langle v, w \rangle|^2}{\|v\|^2} - 2\frac{|\langle v, w \rangle|^2}{\|v\|^2} + \|w\|^2 =
= \|w\|^2 - \frac{|\langle v, w \rangle|^2}{\|v\|^2} .
Hence
|\langle v, w \rangle|^2 \leq \|v\|^2\|w\|^2
which entails the Cauchy--Schwarz inequality.
If v,w are linearly dependant nonzero elements of \mathrm{V}, then there exists a nonzero scalar
a\in \mathbb{K} such that v = a w. Hence
|\langle v , w \rangle| = |a| \, \| w \|^2
= \| v \| \| w \| \, .
Vice versa, if equality holds, then \Crefeq:expanding-norm-of-sum-in-inner-product-space
entails that c v +w = 0, which means that v and w are linearly dependant.
Lemma 6.1.6
A positive semidefinite hermitian form \langle \cdot , \cdot \rangle on a \mathbb{K}-vector space
\mathrm{V} is weakly-nondegenerate if and only if it is positive definite that is if and only if
Proof.
A positive definite real bilinear or complex hermitian form \langle \cdot , \cdot \rangle is
non-degenerate since for every v \in \mathrm{V} \setminus \{ 0 \} the linear form
\langle - ,v \rangle : \mathrm{V} \to \mathbb{K} then is nonzero because \langle v,v \rangle > 0.
Conversely, if \langle - ,v \rangle : \mathrm{V} \to \mathbb{K} is nonzero for all
v \in \mathrm{V} \setminus \{ 0 \}, then there exists an element w \in \mathrm{V} such that
\langle w,v \rangle \neq 0. The Cauchy--Schwarz inequality entails
0 < |\langle w,v \rangle |^2 \leq \langle w,w \rangle \, \langle v,v \rangle ,
which implies \langle v,v \rangle > 0. Hence \langle \cdot , \cdot \rangle is positive definite.
Proposition 6.1.7
The map
\| \cdot \| :V \to \mathbb{R}_{\geq 0} , \: v \mapsto \|v \| = \sqrt{\langle v , v \rangle}
associated to a positive semidefinite hermitian form
\langle \cdot , \cdot \rangle on a \mathbb{K}-vector space \mathrm{V} is a seminorm.
If the hermitian form is positive definite, then \| \cdot \| is even a norm.
Proof.
Absolute homogeneity (N2) is given by Eq.~\eqrefeq:absolute-homogeneity.
The triangle inequality is a consequence of the Cauchy--Schwarz inequality:
\| v + w \|^2 = \| v\|^2 + 2 \, \Re \, \langle v , w \rangle + \| w\|^2 \leq
\| v\|^2 + 2 \, \| v \| \, \| w \| + \| w\|^2 = \big(\| v\|+\| w\| \big)^2 .
Finally, if \langle \cdot , \cdot \rangle is positive definite, then
\| v \| = \sqrt{\langle v , v \rangle} > 0 for all v \in \mathrm{V} \setminus \{ 0 \},
so \| \cdot \| is a norm.
Hilbert space definition
Definition 6.1.8
By an inner product or a scalar product on a \mathbb{K}-vector space \mathfrak{H} one
understands a positive definite hermitian form on \mathfrak{H}. A \mathbb{K}-vector space \mathfrak{H}
together with an inner product \langle \cdot , \cdot \rangle : \mathfrak{H} \times \mathfrak{H} \to \mathbb{K} is
called an inner product space or a pre-Hilbert space.
A hermitian form on a \mathbb{K}-vector space \mathfrak{H} which is only positive semidefinite is called a semi-inner product or a semi-scalar product.
A Hilbert space is an inner product space (\mathfrak{H} , \langle \cdot , \cdot \rangle) which is
complete as a normed vector space. In other words, a Hilbert space is Banach space where the
norm on the space is induced by an inner product.
Examples 6.1.9
Theorem giving condition to relate inner product with norm
Theorem 6.1.10
Let \mathrm{V} be a normed \mathbb{K}-vector space. Then the norm \|\cdot\| : \mathrm{V} \to \mathbb{R}_{\geq 0}
is associated to an inner product \langle \cdot , \cdot \rangle : \mathrm{V} \times \mathrm{V} \to \mathbb{K}
if and only if the parallelogram identity
\| v + w \|^2 + \| v - w\|^2 = 2\|v\|^2 + 2\|w\|^2
holds true for all v,w \in \mathrm{V}. In this case, the inner product of two elements v,w\in \mathrm{V}
can be expressed by the polarization identity for \mathbb{K} = \mathbb{R}
\tag{6.1.5}
\langle v , w \rangle = \frac 14 \left( \| v + w \|^2 - \| v - w \|^2 \right)
= \frac 12 \left( \| v + w \|^2 - \| v \|^2 - \| w \|^2 \right)
respectively by the polarization identity for \mathbb{K} = \mathbb{C}
\tag{6.1.6}
\langle v , w \rangle = \frac 14 \sum_{k=1}^4 \mathfrak{i}^k \, \| v + \mathfrak{i}^k \, w \|^2 .
Proof.
The forward direction is a consequence of
6.1, Eq.~(6.1.1).
To show the backward direction we consider two cases \mathbb{K} = \mathbb{R} and
\mathbb{K} = \mathbb{C} separately.
\textit1.~Case. Given the norm \|\cdot\| define \langle \cdot , \cdot \rangle : \mathrm{V} \times \mathrm{V} \to \mathbb{R} by real polarization
\langle v , w \rangle = \frac 14 \left( \| v + w \|^2 - \| v - w \|^2 \right), \quad \text{where } v,w\in \mathrm{V} .
Note that the parallelogram identity entails
\frac 14 \left( \| v + w \|^2 - \| v - w \|^2 \right) =
\frac 12 \left( \| v + w \|^2 - \| v \|^2 - \| w \|^2 \right) .
Observe that by definition \langle v , w \rangle = \langle w , v \rangle and \| v \| = \sqrt{ \langle v , v \rangle}.
Let us show additivity in the first variable. Let v_1,v_2,w \in \mathrm{V} and compute using the parallelogram identity
\| v_1 + v_2 + w \|^2 = 2 \| v_1 + w \|^2 + 2 \| v_2\|^2 - \| v_1 + w - v_2\|^2 ,
\| v_1 + v_2 + w \|^2 = 2 \| v_2 + w \|^2 + 2 \| v_1\|^2 - \| v_2 + w - v_1\|^2 .
Hence
\| v_1 + v_2 \pm w \|^2 =
\| v_1 \pm w \|^2 + \| v_2 \pm w \|^2 + \| v_1\|^2 + \| v_2\|^2 - \| v_1 \pm w - v_2\|^2 - \| v_2 \pm w - v_1\|^2 .
Subtracting the - version from the + version of this equation entails
\langle v_1 + v_2 , w \rangle \, = \frac 14 \left( \| v_1 + v_2 + w \|^2 - \| v_1 + v_2 - w \|^2 \right)=
= \frac 14 \left( \| v_1 + w \|^2 + \| v_2 + w \|^2 - \| v_1 - w \|^2 - \| v_2 - w \|^2 \right) =
\langle v_1 , w \rangle + \langle v_2 , w \rangle ,
so additivity in the first variable is proved. By induction one derives from this that
for all natural n
\tag{6.1.7}
\langle nv , w \rangle = n \langle v , w \rangle \quad \text{for all } v,w \in \mathrm{V} .
Since then \langle - nv , w \rangle - n \langle v , w \rangle = \langle -nv + nv , w \rangle = 0 for all n\in \mathbb{N},
Eq.~\eqrefeq:natural-number-homogeneity also holds for n\in \mathbb{Z}.
Now let p\in \mathbb{Z} and q \in \mathbb{N}_{ > 0}.
Since addition, multiplication by scalars and the norm are continuous, the function
\mathbb{R} \to \mathbb{R}, \:r \mapsto \langle rv , w \rangle - r \langle v , w \rangle = \frac 14 \left( \| r v + w \|^2 + r \| v - w \|^2 - \| rv - w \|^2 - r \| v + w \|^2 \right)
is continuous. Since it vanishes over \mathbb{Q}, it has to coincide with the zero map. Therefore,
Eq.~\eqrefeq:rational-number-homogeneity holds for all r\in \mathbb{R}. So \langle \cdot , \cdot \rangle is linear in
the first coordinate. By symmetry, it is so too in the second coordinate. Hence \langle \cdot , \cdot \rangle is a
symmetric bilinear form inducing \| \cdot \|.
\textit2.~Case. In the case \mathbb{K} = \mathbb{C} use complex polarization and put
\langle v , w \rangle = \frac 14 \sum_{k=1}^4 \mathfrak{i}^k \, \| v + \mathfrak{i}^k \, w \|^2
\quad \text{for all } v,w\in \mathrm{V} .
Then \langle \cdot , \cdot \rangle is conjugate symmetric, since
\overline{\langle v , w \rangle} = \frac 14 \sum_{k=1}^4 (- \mathfrak{i})^k \, \| v + \mathfrak{i}^k \, w \|^2
= \frac 14 \sum_{k=1}^4 (-\mathfrak{i})^k \, \| (-\mathfrak{i})^k \, v + w \|^2 =
\langle w , v \rangle .
Next compute
\Re \langle v , w \rangle = \frac 14 \left( \| v + w \|^2 - \| v - w \|^2 \right)
and
\Im \langle v , w \rangle = \frac 14 \left( \| v + \mathfrak{i} w \|^2 - \| v - \mathfrak{i} w \|^2 \right) .
By the first case one concludes that \Re \langle \cdot , \cdot \rangle and \Im \langle \cdot , \cdot \rangle
are both \mathbb{R}-linear in the first and the second coordinate. Moreover,
\Re \langle \mathfrak{i} v , w \rangle = \frac 14 \left( \| \mathfrak{i} v + w \|^2 - \| \mathfrak{i} v - w \|^2 \right)
= \frac 14 \left( \| v - \mathfrak{i} w \|^2 - \| v + \mathfrak{i} w \|^2 \right) =
- \Im \langle v , w \rangle
and
\Im \langle \mathfrak{i} \, v , w \rangle =
\frac 14 \left( \| \mathfrak{i} v + \mathfrak{i} w \|^2 - \| \mathfrak{i} v - \mathfrak{i} w \|^2 \right) =
\Re \langle v , w \rangle ,
hence \langle \cdot , \cdot \rangle is complex linear in the first coordinate.

This finishes the proof that \langle \cdot , \cdot \rangle is a complex inner product inducing the
norm \| \cdot \|.
6.1.11
Next we will turn Hilbert spaces into a category. To this end one needs to know what morphisms in this
category should be. There are two options each giving rise to a category of Hilbert spaces. These categories
just differ by their morphism classes. The first one is to
have as morphisms linear maps A:\mathfrak{H}_1\to\mathfrak{H}_2 preserving the inner products which means that
they fulfill
\langle Av_1,Av_2 \rangle = \langle v_1,v_2 \rangle \quad \text{for all } v_1,v_2 \in \mathfrak{H}_1 .
By \Crefthm:parallegram-identity-guarantees-that-norm-comes-from-some-inner-product this property is
equivalent to
\| Av \| = \| v \| \quad \text{for all } v \in \mathfrak{H}_1 ,
that is to A being norm preserving or isometric. Obviously, the identity map between two
Hilbert spaces is isometric and the composition of two composable isometric linear maps between Hilbert
spaces is again isometric and linear. Hence Hilbert spaces together with norm preserving linear maps between
them form a category which we denote by \mathsf{Hilb_{np}}. The isomorphisms in this category
are the surjective and inner product preserving linear maps between Hilbert spaces. Such maps
are called unitary. The condition of a linear map being norm preserving is pretty restrictive, so
the category \mathsf{Hilb_{np}} contains only few morphisms. This can be healed by allowing all
bounded linear maps between Hilbert spaces to be morphisms that is of all
linear A:\mathfrak{H}_1\to\mathfrak{H}_2 for which there exists a C\geq 0 such that
\| Av \| \leq C \| v \| \quad \text{for all } v \in \mathfrak{H}_1 .
The smallest such C is called the operator norm of A and is denoted \| A\|.
Equivalently, the operator norm is given by
\| A\| = \sup \big\{ \| Av \| \bigm\vert v\in \mathfrak{H}_1, \, \| v \| \leq 1 \big\}
= \sup \big\{ \| Av \| \bigm\vert v\in \mathfrak{H}_1, \, \| v \| = 1 \big\} .
Every norm preserving linear map is bounded with operator norm 1.
In particular the identity map on a Hilbert space is bounded. Moreover, if
A: \mathfrak{H}_1\to\mathfrak{H}_2 and B: \mathfrak{H}_2\to\mathfrak{H}_3 are bounded linear operators
between Hilbert spaces, then the composition BA : \mathfrak{H}_1\to\mathfrak{H}_3 is bounded
with operator norm \leq \| B\|\, \| A\| since for all v \in \mathfrak{H}_1 with \| v\|\leq 1
\| BAv \| \leq \| B \| \, \|Av\| \leq \| B\|\, \| A\| .
Hence Hilbert spaces as objects together with bounded linear maps as morphisms form a category which we
denote by \mathsf{Hilb} and call the category of Hilbert spaces. Note that the morphisms
in this category appear to ``forget'' the inner product and just preserve the linear and the topological
structure. John Baez
6.1.12
Last in this section we will introduce bounded bilinear and sesquilinear maps. We define them for normed
vector spaces. Their main application lies in the operator theory on Hilbert spaces, so we introduce
them here.
Definition 6.1.13
Let \mathrm{V} be a vector space over \mathbb{K} with norm \| \cdot \| : \mathrm{V} \to \mathbb{R}_{\geq 0}.
A bilinear or sesquilinear form b : \mathrm{V} \times \mathrm{V} \to \mathbb{K} is called bounded if
there exists a C > 0 such that
| b(v,w)| \leq C \, \| v\| \, \| w \| \quad \text{for all } v,w \in \mathrm{V} .
In this case,
\| b \| := \sup \big\{ | b(v,w) | \bigm\vert v,w \in \mathrm{V} \: \ \: \| v\| = \| w \| = 1 \big\}
exists and is called the norm of the form b.
Example 6.1.14
The inner product on a (pre-) Hilbert space is bounded by the Cauchy--Schwarz inequality and has norm 1.
Proposition 6.1.15
A bounded bilinear or sesquilinear form b : \mathrm{V} \times \mathrm{V} \to \mathbb{K} on a normed vector space \mathrm{V} over
\mathbb{K} is continuous. Vice versa, if \mathrm{V} is complete, then continuity of b : \mathrm{V} \times \mathrm{V} \to \mathbb{K}
implies boundedness.
Proof.
If b is bounded, then
\big\vert b(v,w) - b(v',w') \big\vert \,
\leq \big\vert b(v,w) - b(v',w) \big\vert + \big\vert b(v',w) - b(v',w') \big\vert \leq
\leq \| b \| \, \left( \| w \| \, \| v-v'\| + \| v' \| \, \| w-w'\| \right)
for all v,v',w,w' \in \mathrm{V}. Hence b is locally Lipschitz continuous, so in particular continuous.
Now assume that \mathrm{V} is a Banach space and that b is continuous. Then one can find \delta > 0 such that
for all v,w \in V of norm less than \delta the relation |b(v,w)| < 1 holds true. But that entails for all
non-zero v,w
|b(v,w)| = \frac{4 \, \|v\| \, \|w\|}{\delta^2} \cdot b\left( \delta \frac{v}{2 \|v\|}, \delta \frac{w}{2 \|w\|}\right)
\leq \frac{4}{ \delta^2} \|v\| \, \|w\| .
Hence b is bounded.
6.2.1
One of the issues with infinite-dimensional analysis is that a closed subspace of an infinite dimensional Banach space might not have a
closed complement. Fortunately, the situation in Hilbert space theory is not so grim because
every closed subspace of a Hilbert space admits an orthogonal complement. This is one of the four
crucial properties which distinguish Hilbert spaces from Banach spaces and which are stated in the following.
In this section \mathfrak{H} will always denote a Hilbert space over the field \mathbb{K}=\mathbb{R} or \mathbb{K}=\mathbb{C}.
The symbol \langle \cdot , \cdot\rangle will stand for the inner product of \mathfrak{H}.
Theorem 6.2.2[Best approximation theorem]
Every closed convex nonempty subset C of a Hilbert space
\mathfrak{H} has a unique element of minimal norm.
Proof.
Let d = \inf\{ \|v\| \mid v \in C \} which is a non-negative real number. We claim there exists a unique
v_0 \in C with \|v_0\| =d.
For uniqueness, consider two vectors v_0, v_1 satisfying the desired property, and let v = \frac{1}{2}(v_0 + v_1)
be their midpoint. Then
\|v\| = \frac{1}{2}\|v_0 + v_1\| \leq \frac{1}{2}(\|v_0\| + \|v_1\|) = d
By minimality of d this entails \|v\| =d. By the parallelogram identity
\left\|\frac{1}{2} (v_0 + v_1)\right\|^2 + \left\|\frac{1}{2}(v_0 - v_1) \right\|^2 =
2\left\|\frac{v_0}{2}\right\|^2 + 2\left\|\frac{v_1}{2}\right\|^2 = d^2 ,
hence
\left\| \frac{1}{2} (v_0 - v_1) \right\|^2 \leq d^2 - \|v\|^2 = 0 ,
proving v_0 = v_1.
For the proof of existence observe that by definition of d there exists a sequence (v_n)_{n \in \mathbb{N}} \subset C such
that \lim_{n \to \infty}\|v_n \| = d. By convexity
\frac{1}{2}(v_n + v_m) \in C
for all n,m \in \mathbb{N}, hence \frac{1}{4}\|v_n + v_m \|^2 \geq d^2. The parallelogram equality entails
0 \leq \| v_n - v_m \|^2 = 2\|v_n \|^2 + 2\|v_m \|^2 - \| v_n + v_m \|^2 \leq 2\|v_n \|^2 + 2\|v_m \|^2 - 4d^2 .
Since \lim_{n \to \infty}\|v_n \| = d there exists for given \varepsilon > 0 an N\in \mathbb{N} such that
\|v_n \|^2 -d^2 \leq \frac 14 \varepsilon^2 for all n \geq N. Hence, for n,m \geq N
0 \leq \| v_n - v_m \| \leq \varepsilon ,
and (v_n)_{n \in \mathbb{N}} is a Cauchy-sequence, so convergent by completeness of \mathfrak{H}. Put v_0:= \lim_{n \to \infty}v_n.
Then v_0 \in C since C is closed and \|v_0 \|=\lim_{n \to \infty}\|v_n \| = d. The existence claim follows
and the proof is finished.
Theorem and Definition 6.2.3[Orthogonal decomposition theorem]
Let \mathrm{V} \subset \mathfrak{H} be a closed subspace of the Hilbert space \mathfrak{H}. Then the
orthogonal complement
\mathrm{V}^\bot = \big\{ w \in \mathfrak{H} \bigm\vert \langle v,w \rangle = 0 \text{ for each } v \in \mathrm{V} \big\}
is a closed subspace of \mathfrak{H} and \mathfrak{H} = \mathrm{V} \oplus \mathrm{V}^\bot.
The map \operatorname{pr}_\mathrm{V} : \mathfrak{H} \to \mathrm{V} which maps w \in \mathfrak{H} to the unique w_1\in \mathrm{V}
such that w - w_1 \in \mathrm{V}^\bot is called the orthogonal projection onto \mathrm{V}.
It satisfies \left\| w- \operatorname{pr}_\mathrm{V} (w)\right\| = d(w,\mathrm{V}) := \inf \big\{ \| v-w \| \bigm\vert v\in \mathrm{V} \big\} that is \operatorname{pr}_\mathrm{V} (w) is the unique element of \mathrm{V} having shortest distance from w.
Proof.
For v \in \mathfrak{H} define v^\flat :\mathfrak{H} \to \mathbb{R} by v^\flat (w) = \langle w,v \rangle. Recall that this map is
continuous and linear. Hence the kernel (v^\flat)^{-1}(0) is a closed linear subspace of \mathfrak{H} and
\mathrm{V}^{\bot} = \bigcap_{v \in \mathrm{V}} (v^\flat)^{-1}(0)
is a closed linear subspace. To show \mathrm{V} \cap \mathrm{V}^\bot = \{0\}, consider
v \in \mathrm{V} \cap \mathrm{V}^\bot. Then \|v \|^2 = \langle v,v \rangle = 0.
Now, given some w \in \mathfrak{H}, it can be written as w = w_1 + w_2 with
w_1 \in \mathrm{V} and w_2 \in \mathrm{V}^\bot. To see this put C = w - \mathrm{V}. Then C is closed and convex.
By the best approximation theorem there exists a unique element w_2 \in C of minimal norm. Let w_1
be the unique element of \mathrm{V} such that w_2 = w -w_1. It remains to show w_2 \in \mathrm{V}^\bot.
Since w_2 has minimal norm among the elements of w-\mathrm{V} the following inequality holds for all
vectors v \in \mathrm{V}:
\|w_2 \|^2 \leq \| w_2 + v \|^2 = \|w_2 \|^2 + 2\, \Re\langle w_2,v\rangle + \| v \|^2 .
Hence
0 \leq 2\, \Re\langle w_2,v\rangle + \| v \|^2 \quad \text{for all } v \in \mathrm{V} .
Now assume that \|v\|=1 and choose \varphi \in \mathbb{R} such that e^{i\varphi}\langle w_2, v \rangle \in \mathbb{R}.
Setting v' = e^{-i\varphi}v, one obtains for all \lambda \in \mathbb{R} by the last inequality
0 \leq 2 \langle w_2,\lambda v'\rangle + \| \lambda v' \|^2 =
2\lambda\langle w_2, x'\rangle + \lambda^2.
For \lambda = -\langle w_2, v' \rangle this entails the estimate
\| \langle w_2, v' \rangle \|^2 =
- \left( - 2\|\langle w_2,v \rangle \|^2 + \| \langle w_2, v' \rangle \|^2 \right) = - \left( 2\lambda\langle w_2, x'\rangle + \lambda^2 \right) \leq 0 .
Hence \langle w_2, v \rangle = 0 for all unit vectors v \in \mathrm{V}, therefore w_2 \in \mathrm{V}^\bot.
The remainder of the claim is now a consequence of the construction of w_1 from the given w and the
observation that \operatorname{pr}_\mathrm{V} (w) = w_1.
Corollary 6.2.4
For every closed subspace \mathrm{V} \subset \mathfrak{H} of a Hilbert space \mathfrak{H} the relation
V = (V^\perp)^\perp
holds true.
Proof.
One has V \subset (V^\perp)^\perp by definition of the orthogonal complement.
Since
\mathfrak{H} = V \oplus V^\perp = (V^\perp)^\perp \oplus V^\perp
by the preceding theorem, the claim follows.
Theorem 6.2.5[Riesz representation theorem for Hilbert spaces]
Let \mathfrak{H} be a Hilbert space and \mathfrak{H}' its topological dual. Then the musical map
{}^\flat: \mathfrak{H} \to \mathfrak{H}',\quad
v \mapsto v^\flat = \left( \mathfrak{H} \ni w \mapsto \langle w,v\rangle \in \mathbb{K}\right)
is an isometric isomorphism which is linear in the real case and conjugate-linear in the complex case.
Proof.
Obviously, {}^\flat is linear if the ground field \mathbb{K} equals \mathbb{R} and conjugate-linear if \mathbb{K}=\mathbb{C}.
Now observe that for all v \in \mathfrak{H} by the Cauchy--Schwarz inequality
\| v^\flat \| = \sup\big\{ |\langle w,v \rangle | \bigm\vert w \in \mathfrak{H} \: \ \: \|w\| =1 \big\} =
\| v \| ,
hence {}^\flat is an isometry, so in particular injective.
It remains to show surjectivity. So assume that \alpha : \mathfrak{H} \to \mathbb{K} is a nontrivial
continuous linear form.
Let \mathrm{V} be its kernel. Then \mathrm{V} is a closed linear subspace of \mathfrak{H}.
Since \alpha is nontrivial, the orthogonal complement \mathrm{V}^\bot is nontrivial, too. Hence
\mathrm{V}^\bot \cong \mathfrak{H}/\mathrm{V} is isomorphic to \operatorname{im} \alpha = \mathbb{K}
and there exists a vector v \in \mathrm{V}^\bot \setminus \{ 0 \} such that
\alpha (v) = 1. Since v spans \mathrm{V}^\bot there exists for every w\in \mathfrak{H} a
unique \lambda_w \in \mathbb{K} such that w = \operatorname{pr}_V (w) + \lambda_w v. Then compute
\alpha (w) = \alpha (\lambda_w v ) = \lambda_w \quad \text{and} \quad
\left( \frac{v}{\|v\|^2}\right)^\flat (w) =
\frac{1}{\|v\|^2} \langle w, v \rangle = \frac{ \lambda_w}{\|v\|^2} \langle v , v \rangle
= \lambda_w .
This entails \alpha = \left( \frac{v}{\|v\|^2}\right)^\flat, and {}^\flat is surjective.
Remark 6.2.6
Sometimes, and we will follow that convention, the inverse of the musical isomorphism
{}^\flat: \mathfrak{H} \to \mathfrak{H}' is denoted {}^\sharp: \mathfrak{H}' \to \mathfrak{H}.
Corollary 6.2.7
Every Hilbert space \mathfrak{H} is reflexive that is the canonical map
H \to H'' , \: v \mapsto \left( H' \ni \lambda \mapsto \lambda(v) \in \mathbb{K} \right)
is an isometric isomorphism.
Proof.
By the Riesz Representation Theorem, the dual \mathfrak{H}' is a Hilbert space
with inner product
\langle\!\langle \cdot, \cdot \rangle\!\rangle : \mathfrak{H}' \times \mathfrak{H}' \to \mathbb{K}, \:
(\lambda,\mu) \mapsto \langle\!\langle \lambda, \mu \rangle\!\rangle = \langle \mu^\sharp,\lambda^\sharp \rangle .
Hence, by applying the Riesz Representation Theorem twice,
the map {}^\flat \circ {}^\flat : \mathfrak{H} \to \mathfrak{H}'' is an isometric linear isomorphism.
Now compute for v\in \mathfrak{H} and \lambda \in \mathfrak{H}'
(v^\flat)^\flat (\mu) = \langle\!\langle \lambda, v^\flat \rangle\!\rangle =
\langle v, \lambda^\sharp \rangle = \lambda (v) .
Hence {}^\flat \circ {}^\flat coincides with the canonical map above and the claim
follows.
Corollary 6.2.8
Let b:\mathfrak{H} \times \mathfrak{H}\to \mathbb{K} be a bounded sesquilinear form on a Hilbert space \mathfrak{H}. Then there exists
unique bounded linear map A : \mathfrak{H} \to \mathfrak{H} such that
b(v,w) = \langle Av,w \rangle \quad \text{for all } v,w\in \mathfrak{H} .
Moreover, the operator norm \| A \| coincides with \| b\|.
Proof.
First let us show uniqueness. So let A,B : \mathfrak{H} \to \mathfrak{H} be bounded and linear so that
b(v,w) = \langle Av,w \rangle = \langle Bv,w \rangle \quad \text{for all } v,w\in \mathfrak{H} .
Then \|(A-B)v \|^2 = \langle Av-Bv,(A-B)v \rangle = b(v,(A-B)v) - b(v,(A-B)v)= 0 for all v\in \mathfrak{H}
which entails equality of A and B.
To prove existence observe that for every v\in \mathfrak{H} the map \mathfrak{H} \to \mathbb{K}, w \mapsto \overline{b(v,w)}
is bounded an linear, so by the Riesz representation theorem there exists for every v an element Av \in \mathfrak{H}
such that \langle w,Av \rangle = \overline{b(v,w)} for all w\in \mathfrak{H}.
Let us show that the map A is linear.
For v_1,v_2\in \mathfrak{H} check that
\langle w,A(v_1+v_2) \rangle = \overline{b(v_1+v_2,w)} = \overline{b(v_1,w)}+ \overline{b(v_2,w)} =
= \langle w,Av_1 \rangle +\langle w,Av_2 \rangle = \langle w,Av_1+Av_2 \rangle \quad \text{for all } w\in \mathfrak{H} .
But that implies A(v_1+v_2) = Av_1+Av_2. Given r\in \mathbb{K} and v\in\mathfrak{H} one verifies
\langle w,A(rv) \rangle = \overline{b(rv,w)} = \overline{rb(v,w)} = \overline{r} \, \overline{b(v,w)}
= \overline{r} \langle w,Av \rangle = \langle w,rAv \rangle \quad \text{for all } w\in \mathfrak{H} .
Hence A(rv) = r Av and linearity of A is proved.
For the operator norm compute
\| A \| = \sup \big\{ \left|\langle Av,w \rangle \right| \bigm\vert v,w \in \mathfrak{H} \: \ \: \|v\| = \|w\| =1 \big\} =
= \sup \big\{ \left| b(v,w) \right| \bigm\vert v,w \in \mathfrak{H} \: \ \: \|v\| = \|w\| =1 \big\} = \| b\| .
6.2.9
Last in this section we will examine the Hilbert direct sum or just Hilbert sum of a family
(\mathfrak{H}_i)_{i\in I} of Hilbert spaces. It is defined by
\widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i = \left\{ (v_i)_{i\in I} \in \prod_{i\in I}\mathfrak{H}_i
\Bigm\vert \left( \|v_i\|^2 \right)_{i\in I} \text{ is summable} \right\} =
= \left\{ (v_i)_{i\in I} \in \prod_{i\in I}\mathfrak{H}_i
\Bigm\vert \exists C \geq 0 \, \forall J \in \mathscr{F} (I): \: \sum_{i\in J} \|v_i\|^2 \leq C \right\} ,
where, as usual, \mathscr{F} (I)\subset \mathscr{P}(I) denotes the set of all finite subsets of I.
Proposition 6.2.10
Let (\mathfrak{H}_i)_{i\in I} be a family of Hilbert spaces. Then the Hilbert direct sum
\widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i is a Hilbert space with inner product given by
\langle -,- \rangle : \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i \times
\widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i \to \mathbb{K}, \quad
\left( (v_i)_{i\in I} , (w_i)_{i\in I} \right) \mapsto \sum_{i\in I} \langle v_i,w_i \rangle .
Proof.
We show first that \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i is a subvector space of
the direct product \prod_{i\in I} \mathfrak{H}_i.
Let z\in \mathbb{K} and (v_i)_{i\in I}, (w_i)_{i\in I}\in \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i.
Choose C,D \geq 0 such that
\sum_{i\in J} \|v_i\|^2 \leq C \quad\text{and}\quad
\sum_{i\in J} \|w_i\|^2 \leq D \quad\text{for all} J \in \mathscr{I} .
Then
\tag{6.2.1}
\sum_{i\in J} \|z v_i\|^2 = |z| \, \sum_{i\in J} \| v_i\|^2\leq |z| \, C \quad
\text{for all } J\in \mathscr{I} ,
so (zv_i)_{i\in I} \in \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i.
Moreover, by Minkowski's inequality for finite sums,
\tag{6.2.2}
\sum_{i\in J} \|v_i + w_i\|^2 \leq
\left( \sqrt{\sum_{i\in J} \|v_i\|^2} + \sqrt{\sum_{i\in J} \| w_i\|^2} \right)^2
\leq \left( \sqrt{C} + \sqrt{D} \right)^2 \quad\text{for all } J\in \mathscr{I} .
Hence the family \left( \| v_i + w_i\|^2 \right)_{i\in I} is summable and
\left( v_i + w_i \right)_{i\in I} \in \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i.
Next observe that the map
\big\| - \big\| : \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i \to \mathbb{K},
\: (v_i)_{i\in I} \mapsto \big\| (v_i)_{i\in I} \big\| = \sqrt{\sum_{i\in I} \| v_i\|^2}
is well-defined by definition of the Hilbert direct sum. It is even a norm
by \eqrefeq:estimate-finite-sum-square-norms-multiple and
\eqrefeq:estimate-finite-sum-square-norms-sum.
Now we need to show that the inner product on \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i
is well-defined which means that the family \left( \langle v_i,w_i \rangle \right)_{i\in I} is summable
for all (v_i)_{i\in I} , (w_i)_{i\in I} \in \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i.
To this end let J\subset I be a finite subset. Then, by the triangle inequality,
the Cauchy--Schwarz inequality on the Hilbert spaces \mathfrak{H}_i and the
Cauchy--Schwarz inequality for finite sums,
\left| \sum_{i\in J} \langle v_i,w_i \rangle \right| \leq
\sum_{i\in J} \left| \langle v_i,w_i \rangle \right| \leq
\sum_{i\in J} \| v_i\| \, \| w_i\| \leq
\sqrt{\sum_{i\in J} \| v_i\|^2} \cdot \sqrt{\sum_{i\in J} \| w_i\|^2}
\leq \big\| (v_i)_{i\in I} \big\| \, \big\| (w_i)_{i\in I} \big\| .
Hence the family \left( \langle v_i,w_i \rangle \right)_{i\in I} is absolutely summable, so in particular
summable, and the inner product is well-defined.
By definition and since all the inner products on the Hilbert spaces \mathfrak{H}_i are conjugate
symmetric and positive definite, the map \langle -,- \rangle on
\widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i has to be conjugate symmetric and positive definite
as well. It remains to show linearity in the first argument.
Denote for (v_i)_{i\in I}, (w_i)_{i\in I} \in \prod_{i\in I} \mathfrak{H}_i
and J\in \mathscr{F} (I) by \langle (v_i)_{i\in I} ,(w_i)_{i\in I} \rangle_J the finite sum
\sum_{i\in J} \langle v_i,w_i \rangle. Observe that the net
\big( \langle (v_i)_{i\in I} ,(w_i)_{i\in I} \rangle_J \big)_{J\in \mathscr{F}(I)} converges to
\langle (v_i)_{i\in I} ,(w_i)_{i\in I} \rangle in case both (v_i)_{i\in I} and (w_i)_{i\in I}
are in \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i. Now let z\in \mathbb{K} and
(v_i)_{i\in I}, (v_i^\prime)_{i\in I}, (w_i)_{i\in I} \in \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i.
Then
\langle (v_i)_{i\in I} + (v_i^\prime)_{i\in I},(w_i)_{i\in I} \rangle_J =
\langle (v_i)_{i\in I} ,(w_i)_{i\in I} \rangle_J + \langle (v_i^\prime)_{i\in I},(w_i)_{i\in I} \rangle_J
\quad\text{and}
\langle z (v_i)_{i\in I},(w_i)_{i\in I} \rangle_J = z \langle (v_i)_{i\in I} ,(w_i)_{i\in I} \rangle_J .
By convergence of all the nets \big( \langle (v_i)_{i\in I} ,(w_i)_{i\in I} \rangle_J \big)_{J\in \mathscr{F}(I)},
linearity in the first argument follows.
By construction, the norm associated to the inner product \langle -,- \rangle on
\widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i coincides with the above defined norm \big\|-\big\|.
It remains to show that \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i equipped with the norm
\big\|-\big\| is complete. To this end observe that for every finite J\subset I the map
\big\| - \big\|_J : \prod_{i\in I} \mathfrak{H}_i \to \mathbb{R}_{\geq 0}, \:
(v_i)_{i\in I} \mapsto \sqrt{\langle (v_i)_{i\in I},(v_i)_{i\in I} \rangle_J}
= \sqrt{\sum_{i\in J} \| v_i \|^2}
is a seminorm and that (v_i)_{i\in I} \in \prod_{i\in I} \mathfrak{H}_i lies in
the Hilbert direct sum \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i if and only if the
family \left(\big\| (v_i)_{i\in I} \big\|_J\right)_{J\in \mathscr{F}(I)} is bounded.
Now let \left((v_i^n)_{i\in I} \right)_{n\in \mathbb{N}} be a Cauchy sequence.
Let \varepsilon > 0 and choose N_\varepsilon \in \mathbb{N} such that
\tag{6.2.3}
\big\| (v_i^m)_{i\in I} - (v_i^n)_{i\in I} \big\| < \varepsilon \quad
\text{for all } n,m\geq N_\varepsilon .
Hence
\tag{6.2.4}
\big\| (v_i^m)_{i\in I} - (v_i^n)_{i\in I} \big\|_J < \varepsilon \quad
\text{for all } J\in \mathscr{F}(I) \text{ and } n,m\geq N_\varepsilon .
Taking J =\{ j\} for j\in I this implies that the sequence
(v_j^n)_{n\in \mathbb{N}} is a Cauchy sequence in the Hilbert space \mathfrak{H}_j.
Let v_j \in \mathfrak{H}_j be its limit. The family (v_i)_{i\in I} then is
an element of \widehat{\bigoplus\limits_{i\in I}} \mathfrak{H}_i. To verify this
put N=N_1 and observe that by \eqrefeq:cauchy-criterion-sequence-finite-cutoff
for all finite J\subset I
\big\| (v_i)_{i\in I} \big\|_J \leq
\big\| (v_i^N)_{i\in I} \big\|_J + \big\| (v_i)_{i\in I} - (v_i^N)_{i\in I} \big\|_J =
= \big\| (v_i^N)_{i\in I} \big\|_J + \lim_{m\to \infty} \big\| (v_i^m)_{i\in I} - (v_i^N)_{i\in I} \big\|_J
\leq \big\| (v_i^N)_{i\in I} \big\| + 1.
Hence the family \left(\big\| (v_i)_{i\in I} \big\|_J\right)_{J\in \mathscr{F}(I)}
is bounded and (v_i)_{i\in I} lies in the Hilbert direct sum of the spaces \mathfrak{H}_i, i\in I.
Moreover, \eqrefeq:cauchy-criterion-sequence-finite-cutoff entails that
\big\| (v_i)_{i\in I} - (v_i^n)_{i\in I} \big\|_J =
\lim_{m\to \infty} \big\| (v_i^m)_{i\in I} - (v_i^n)_{i\in I} \big\|_J \leq \varepsilon \quad
\text{for all } J\in \mathscr{F}(I) \text{ and } n\geq N_\varepsilon .
Since \big\| (v_i)_{i\in I} - (v_i^n)_{i\in I} \big\| is the limit of the
net \left(\big\| (v_i)_{i\in I} - (v_i^n)_{i\in I} \big\|_J\right)_{J\in\mathscr{F}(I)},
the estimate
\big\| (v_i)_{i\in I} - (v_i^n)_{i\in I} \big\| \leq \varepsilon \quad
\text{for all } n\geq N_\varepsilon
follows and the sequence \left((v_i^n)_{i\in I} \right)_{n\in \mathbb{N}} convergence to
(v_i)_{i\in I}. This finishes the proof.
Definition 6.3.1
A (possibly empty) subset S of a Hilbert space \mathfrak{H} is called
an orthogonal system or just orthogonal if for any two different elements v,w \in S the relation
\langle v,w \rangle = 0 holds true. If in addition \left\|v\right\|=1 for all elements v \in S,
then the set is called orthonormal or an orthonormal system.
A family (v_i)_{i\in I} of vectors in \mathfrak{H} is called orthogonal if
\langle v_i,v_j \rangle = 0 for all i,j\in I with i\neq j and orthonormal
if in addition \|v_i\| =1 for all i\in I.
6.3.2
Obviously, the set of orthonormal subsets of a Hilbert space is ordered by set-theoretic inclusion.
Therefore, the following definition makes sense.
Definition 6.3.3
A maximal orthonormal set in a Hilbert space \mathfrak{H} is called an orthonormal basis
or a Hilbert basis of \mathfrak{H}.
Proposition 6.3.4
Every Hilbert space \mathfrak{H} has an orthonormal basis.
Proof.
Wothout loss of generality we can assume that \mathfrak{H} \neq \{ 0\}, because
\emptyset is a Hilbert basis for \{ 0 \}.
Let \mathscr{O} denote the set of orthonormal subsets of \mathfrak{H}. As mentioned before, \mathscr{O}
is ordered by set-theoretic inclusion. Let \mathscr{C} \subset \mathscr{O} be a non-empty chain.
Put U = \bigcup_{S\in \mathscr{C}} S. Then U is an upper bound of \mathscr{C}.
So by Zorn's lemma \mathscr{O} has a maximal element.
Remark 6.3.5
Proposition 6.3.6[Pythagorean theorem for orthogonal families]\hspace1mm
An orthogonal family (v_i)_{i\in I} in a Hilbert space \mathfrak{H} is summable if and only if
the family of norms \left(\|v_i\| \right)_{i\in I} is square summable. In this case one has
\left\|\sum_{i\in I} v_i\right\|^2 = \sum_{i\in I} \|v_i\|^2 .
Proof.
Assume that \left(\|v_i\| \right)_{i\in I} is square summable or in other words that the net of partial sums
\left( \sum_{i\in J} \|v_i\|^2 \right)_{J\in\mathscr{F} (I)} converges to some s \in \mathbb{R}.
For \varepsilon > 0 choose a finite J_\varepsilon \subset I such that for all finite J with
J_\varepsilon \subset J\subset I the relation
\left| s - \sum_{i\in J} \|v_i\|^2 \right| < \frac{\varepsilon^2}{2}
holds true. For finite K\subset I with K\cap J_\varepsilon =\emptyset one then obtains by the pythagorean theorem for
finite orthogonal families, Eq.~((6.1.1)),
\left\| \sum_{i\in K} v_i\right\|^2 = \sum_{i\in K} \left\| v_i\right\|^2 \leq
\left| s - \sum_{i\in K \cup J_\varepsilon} \|v_i\|^2 \right| + \left| s - \sum_{i\in J_\varepsilon} \|v_i\|^2 \right|
< \varepsilon^2 .
Hence \left( \sum_{i\in J} v_i \right)_{J\in\mathscr{F} (I)} is a Cauchy net in \mathfrak{H}, so convergent.
Now let (v_i)_{i\in I} be summable to v\in \mathfrak{H}.
Then there exists a J_1 \in \mathscr{F} (I) such that for all finite J \subset I
containing J_1
\left\| v - \sum_{i\in J} v_i \right\| \leq 1 .
This implies by the pythagorean theorem for finite orthogonal families
\sum_{i\in J} \left\| v_i \right\|^2 = \left\| \sum_{i\in J} v_i \right\|^2 \leq
\left( \left\| v - \sum_{i\in J} v_i \right\| + \left\| v \right\| \right)^2 \leq ( 1 + \| v\| )^2 .
Hence the net of partial sums \left( \sum_{i\in J} \|v_i\|^2 \right)_{J\in\mathscr{F} (I)} is bounded, so convergent
since each term \|v_i\|^2 is \geq 0.
By continuity of the inner product and pairwise orthogonality of the v_i we finally obtain in the convergent case
\left\| \sum_{i\in I} v_i \right\|^2 = \langle \sum_{i\in I} v_i, \sum_{j\in I} v_j \rangle =
\sum_{i\in I} \langle v_i, \sum_{j\in I} v_j \rangle = \sum_{i\in I} \sum_{j\in I} \langle v_i, v_j \rangle = \sum_{i\in I} \left\| v_i \right\|^2 .
Proposition 6.3.7
Let (v_i)_{i\in I} be an orthonormal family in a Hilbert space \mathfrak{H}. Then for every v \in \mathfrak{H}
the family \left( \langle v,v_i \rangle \right)_{i\in I} is square summable and
Bessel's inequality holds true that is
\sum_{i\in I} \left| \langle v,v_i \rangle \right|^2 \leq \| v \|^2 .
Proof.
Theorem 6.3.8
Let B be an orthonormal system in a Hilbert space \mathfrak{H}. Then the following are equivalent:
Proof.
(1) \Rightarrow (2):
If v \neq 0, then \frac{v}{\left\|v\right\|} is a unit vector orthogonal to each v_i. Hence \{v\} \cup B
is an orthonormal system which is strictly larger than B, contradicting (1).
(2) \Rightarrow (3).
First note that by the pythagorean theorem for infinite families, \Crefthm:pythagorean-theorem-infinite-families,
the canonical map \iota: \widehat{\bigoplus}_{b \in B} H_b \rightarrow H is well-defined and an isometry.
Hence \iota is injective. It remains to show that \iota is surjective.
To this end observe that \operatorname{im} \iota is closed in \mathfrak{H} since \iota is an isometry (the image is complete).
If \iota is not surjective, then \operatorname{im} \iota^\perp is not the zero vector space.
Choose v \in \operatorname{im} \iota^\perp \setminus \{ 0\}.
Then v is orthogonal to each element of B, but v \neq 0. This contradicts
(2), so \operatorname{im} \iota = \mathfrak{H}.
(3) \Rightarrow (5):
We can represent any v \in \mathfrak{H} in the form v = \iota \left( (v_b)_{b \in B} \right) =
\sum_{b\in B} v_b with \left( v_b \right)_{b\in B} \in \widehat{\bigoplus}_{b \in B} H_b.
Write v_b = r_b \, b for every b\in B, where r_b \in \mathbb{K} is uniquely determined by v_b.
Then compute using continuity of the inner product
\langle v, b \rangle = \langle \sum_{c \in B} v_c, b \rangle = \sum_{c \in B} r_c \langle c, b \rangle = r_b .
Therefore,
v = \sum_{b \in B} r_b \, b = \sum_{b \in B} \langle v, b \rangle b .
(5) \Rightarrow (6):
Fourier expansion of v, w \in H gives v = \sum\limits_{b \in B} \langle v, b \rangle b and
w = \sum\limits_{b \in B} \langle w, b \rangle b. Then, by continuity of the inner product,
\langle v, w \rangle = \sum\limits_{b \in B} \langle v, b \rangle\langle b, w \rangle .
(5) \Rightarrow (4):
Let v\in \mathfrak{H}. Then \sum\limits_{b\in J}\langle v, b \rangle b \in \operatorname{Span} (B) for all finite J \subset B.
But by Fourier expansion v is the limit of the net
\left( \sum\limits_{b\in J}\langle v, b \rangle b \right)_{J\in \mathscr{F} (B)} , so v lies in the closure
\overline{\operatorname{Span}} (B).
(4) \Rightarrow (2):
Assume that \langle v, b \rangle = 0 for all b \in B. By (4), v can be written as a limit
v = \lim\limits_{n\to\infty} v_n, where v_n \in \operatorname{Span} (B) for all n\in \mathbb{N}.
Then \langle v, v_n \rangle = 0 for all n\in \mathbb{N} by assumption.
By continuity of the inner product this implies
\left\|v\right\|^2 = \lim\limits_{n\to\infty} \langle v, v_n \rangle = 0 ,
so v=0.
(6) \Rightarrow (7):
Put v = w. Then, by assumption,
\left\|v\right\|^2 = \langle v, v \rangle = \sum\limits_{b\in B} \langle v, b \rangle \langle b, v \rangle = \sum\limits_{b \in B} {\left|\langle v, b \rangle\right|}^2 .
(7) \Rightarrow (1):
Assume (7) and that (1) is not true.
Then there exists v \in H with \left\|v\right\| = 1 and \langle v, b \rangle = 0 for all b \in B. But then
\left\|v\right\|^2 = \sum_{b \in B} {\left|\langle v, b \rangle\right|}^2 = 0,
which is a contradiction.
6.4.1
Let \mathbb{K} be the field of real or complex numbers.
Hilbert spaces over \mathbb{K} together with bounded \mathbb{K}-linear maps
between them form a category denoted by \mathbb{K}\text{-}\mathsf{Hilb}
or just \mathsf{Hilb} if no confusion can arise. This can be seen
immediately by observing that the identity map \mathbb{1}_\mathfrak{H} on a Hilbert space
is a bounded linear operator and that the composition B \circ A : \mathfrak{H}_1 \to\mathfrak{H}_3
of two bounded linear operators between Hilbert spaces
A:\mathfrak{H}_1 \to \mathfrak{H}_2 and B :\mathfrak{H}_2 \to \mathfrak{H}_3 is again a
bounded linear operator.
We want to endow the category \mathsf{Hilb} with a bifunctor
{\,\widehat{\otimes}\,} : \mathsf{Hilb} \times \mathsf{Hilb} \to \mathsf{Hilb}
so that it becomes a monoidal category. The (bi)functor {\,\widehat{\otimes}\,} will be
called the Hilbert tensor product.
Unless mentioned differently, Hilbert spaces, vector spaces and the algebraic tensor product \otimes
in this section are assumed to be taken over the ground field \mathbb{K}.
Proposition 6.4.2
Let \mathfrak{H}_1 and \mathfrak{H}_2 be two Hilbert spaces.
Then there exists a unique inner product
\langle \cdot,\cdot \rangle: (\mathfrak{H}_1 \otimes \mathfrak{H}_2) \times (\mathfrak{H}_1 \otimes \mathfrak{H}_2)
\to \mathbb{K} on the algebraic tensor product \mathfrak{H}_1\otimes \mathfrak{H}_2 such that
\tag{6.4.1}
\langle v_1 \otimes v_2 , w_1 \otimes w_2 \rangle = \langle v_1,w_1 \rangle \cdot \langle v_2,w_2 \rangle
\quad\text{for all } v_1,w_1 \in \mathfrak{H}_1, \: v_2,w_2 \in \mathfrak{H}_2 .
Proof.
Let us first provide some preliminary constructions.
Recall that for every pair of vector spaces \mathrm{V}_1 and \mathrm{V}_2
the bilinear map
\tau: \operatorname{Hom}\nolimits (\mathrm{V}_1 ,\mathbb{K}) \times \operatorname{Hom}\nolimits (\mathrm{V}_2,\mathbb{K}) \to
\operatorname{Hom}\nolimits (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K}),
(\lambda_1,\lambda_2) \mapsto \big( \mathrm{V}_1 \otimes \mathrm{V}_2 \to \mathbb{K}, \:
v_1 \otimes v_2 \mapsto \lambda_1 (v_1) \cdot\lambda_2 (v_2) \big)
induces a linear map
\widehat{\tau}: \operatorname{Hom}\nolimits (\mathrm{V}_1 ,\mathbb{K}) \otimes \operatorname{Hom}\nolimits (\mathrm{V}_2,\mathbb{K}) \to
\operatorname{Hom}\nolimits (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K})
by the universal property of the tensor product.
This map is an isomorphism. To see this choose a basis (v_{1i})_{i\in I} of V_1 and
a basis (v_{2j})_{j\in J} of V_2. Let (v^\prime_{1i})_{i\in I} and (v^\prime_{2j})_{j\in J}
denote the respective dual bases of V_1^\prime and V_2^\prime. Then
\left( v^\prime_{1i} \otimes v^\prime_{2j}\right)_{(i,j)\in I \times J} is a basis of
\operatorname{Hom}\nolimits (\mathrm{V}_1 ,\mathbb{K}) \otimes \operatorname{Hom}\nolimits (\mathrm{V}_2,\mathbb{K}) which under \widehat{\tau}
is mapped bijectively to the basis \left( (v_{1i} \otimes v_{2j})^\prime \right)_{(i,j)\in I \times J}
of \operatorname{Hom}\nolimits (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K}) dual to the basis
\left( v_{1i} \otimes v_{2j}\right)_{(i,j)\in I \times J} of \mathrm{V}_1 \otimes \mathrm{V}_2.
Hence \widehat{\tau} is a linear isomorphism as claimed, and we can identify
the tensor product \lambda_1\otimes \lambda_2 of two linear functionals
\lambda_i: \mathrm{V}_i\to\mathbb{K}, i=1,2
with its image in \operatorname{Hom}\nolimits (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K}).
Now observe that for two conjugate-linear maps \mu_1 :\mathrm{V}_1 \to \mathbb{K}
and \mu_2 :\mathrm{V}_2 \to \mathbb{K} the map
\tau^* (\mu_1,\mu_2) = \overline{\overline{\mu_1}\otimes \overline{\mu_2}} :
\mathrm{V}_1 \otimes \mathrm{V}_2 \to \mathbb{K}
is conjugate-linear and satisfies
\tag{6.4.2}
\tau^* (\mu_1,\mu_2) \, (v_1\otimes v_2) = \mu_1(v_1) \cdot \mu_2(v_2) \quad
\text{for all } v_1\in\mathrm{V}_1 ,\: v_2 \in\mathrm{V}_2 .
One obtains a map
\tau^*: \operatorname{Hom}\nolimits^* (\mathrm{V}_1,\mathbb{K}) \times \operatorname{Hom}\nolimits^* (\mathrm{V}_2,\mathbb{K})
\to \operatorname{Hom}\nolimits^* (\mathrm{V}_1 \otimes \mathrm{V}_2 ,\mathbb{K}) ,
where here the symbol \operatorname{Hom}\nolimits^* (\mathrm{V},\mathbb{K}) denotes the space of all conjugate linear
functionals on a vector space \mathrm{V}. Since \tau^* is biadditive and
since \tau^* (z\mu_1,\mu_2) = \tau^* (\mu_1,z\mu_2) for all \mu_1\in \operatorname{Hom}\nolimits^*(\mathrm{V}_1 ,\mathbb{K}),
\mu_2\in \operatorname{Hom}\nolimits^*(\mathrm{V}_2 ,\mathbb{K}), and z\in \mathbb{K}, the map \tau^* factors through a linear map
\widehat{\tau^*} :\operatorname{Hom}\nolimits^* (\mathrm{V}_1 ,\mathbb{K}) \otimes \operatorname{Hom}\nolimits^* (\mathrm{V}_2,\mathbb{K}) \to
\operatorname{Hom}\nolimits^* (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K}) .
Using the above bases (v_{1i})_{i\in I} and (v_{2j})_{j\in J} of V_1 and V_2 respectively,
one observes that \widehat{\tau^*} is an isomorphism since it maps the basis
\left( \overline{v^\prime_{1i}} \otimes \overline{v^\prime_{2j}}\right)_{(i,j)\in I \times J} of
\operatorname{Hom}\nolimits^* (\mathrm{V}_1 ,\mathbb{K}) \otimes \operatorname{Hom}\nolimits^* (\mathrm{V}_2,\mathbb{K})
bijectively to the basis \left( \overline{(v_{1i} \otimes v_{2j})^\prime} \right)_{(i,j)\in I \times J}
of the space \operatorname{Hom}\nolimits^* (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K}). So \widehat{\tau^*} is also
a linear isomorphism, which allows us to identify the tensor product \mu_1 \otimes \mu_2
of two conjugate linear functionals \mu_i: \mathrm{V}_i\to\mathbb{K}, i=1,2 with its image in
\operatorname{Hom}\nolimits^* (\mathrm{V}_1 \otimes \mathrm{V}_2,\mathbb{K}).
After these preliminary considerations we consider the map
\mathfrak{H}_1\times \mathfrak{H}_2 \to \operatorname{Hom}\nolimits^* (\mathfrak{H}_1\otimes \mathfrak{H}_2,\mathbb{K}) ,
\: (v_1,v_2) \mapsto \overline{v_1^\flat} \otimes \overline{v_2^\flat} =
\tau^* \left( \overline{v_1^\flat} , \overline{v_2^\flat} \right)=
\widehat{\tau^*} \left( \overline{v_1^\flat} \otimes \overline{v_2^\flat} \right),
which is well-defined and bilinear since the musical isomorphisms
{}^\flat: \mathfrak{H}_l \to \mathfrak{H}_l', v \mapsto \langle - ,v \rangle , l=1,2,
are conjugate-linear and since \tau^* is bilinear. Hence it factors through a linear map
\beta: \mathfrak{H}_1\otimes \mathfrak{H}_2 \to \operatorname{Hom}\nolimits^* (\mathfrak{H}_1\otimes \mathfrak{H}_2,\mathbb{K})
such that
\tag{6.4.3}
\beta (v_1\otimes v_2) (w_1\otimes w_2) = \langle v_1,w_1 \rangle \cdot \langle v_2,w_2 \rangle
\quad \text{for all } v_1,w_1\in \mathfrak{H}_1 , \: v_2,w_2\in \mathfrak{H}_2 .
Now put
\langle \cdot,\cdot \rangle :
(\mathfrak{H}_1 \otimes \mathfrak{H}_2) \times (\mathfrak{H}_1 \otimes \mathfrak{H}_2)
\to \mathbb{K}, \: (v,w) \mapsto \beta (v) (w) .
Then \langle \cdot,\cdot \rangle is sesquilinear by construction, and
\eqrefeq:defining-relation-inner-product-tensor-product holds true by
\eqrefeq:defining-relation-map-beta.
Let us show that \langle \cdot,\cdot \rangle is positive definite. Let
v = \sum_{k=1}^n v_{1k} \otimes v_{2k} \in \mathfrak{H}_1 \otimes \mathfrak{H}_2.
Choose an orthonormal basis e_1, \ldots , e_m of the linear subspace
spanned by v_{21}, \ldots , v_{2n}. Expand
v_{2k} = \sum_{i=1}^m c_{ki} e_i with c_{k1}, \ldots , c_{km} \in \mathbb{K}.
Then
\tag{6.4.4}
v = \sum_{k=1}^n v_{1k} \otimes v_{2k} = \sum_{k=1}^n \sum_{i=1}^m v_{1k} \otimes (c_{ki} e_i) =
\sum_{i=1}^m \left( \sum_{k=1}^n c_{ki} v_{1k} \right) \otimes e_i =
\sum_{i=1}^m w_{1i} \otimes e_i ,
where w_{1i} = \sum_{k=1}^n c_{ki} v_{1k} . Hence
\tag{6.4.5}
\langle v,v \rangle = \langle \sum_{i=1}^m w_{1i} \otimes e_i,\sum_{j=1}^m w_{1j} \otimes e_j \rangle
= \sum_{i=1}^m \sum_{j=1}^m \langle w_{1i}, w_{1j} \rangle \, \langle e_i,e_j \rangle =
\sum_{i=1}^m \| w_{1i} \|^2 \geq 0 .
Moreover, if \langle v,v \rangle = 0, then w_{1i}=0 for i=1,\ldots , m, which implies
v= \sum_{i=1}^m w_{1i} \otimes e_i =0.
So \langle \cdot,\cdot \rangle is an inner product on \mathfrak{H}_1\otimes \mathfrak{H}_2 satisfying
\eqrefeq:defining-relation-inner-product-tensor-product. It is uniquely determined by this
condition since the vectors v_1\otimes v_2 with v_1\in \mathfrak{H}_1 and v_2\in \mathfrak{H}_2 span
\mathfrak{H}_1\otimes \mathfrak{H}_2.
Definition 6.4.3
Let \mathfrak{H}_1 and \mathfrak{H}_2 be Hilbert spaces.
The Hilbert completion of the algebraic tensor product \mathfrak{H}_1 \otimes \mathfrak{H}_2
equipped with the unique inner product \langle \cdot,\cdot \rangle fulfilling
\eqrefeq:defining-relation-inner-product-tensor-product
will be denoted \mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2, its inner product again by \langle \cdot,\cdot \rangle. One calls the Hilbert space
\big(\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2 ,\langle \cdot,\cdot \rangle\big)
the Hilbert tensor product of \mathfrak{H}_1 and \mathfrak{H}_2 or just the
tensor product of \mathfrak{H}_1 and \mathfrak{H}_2 if no confusion can arise.
Proposition 6.4.4
Let \mathfrak{H}_1 and \mathfrak{H}_2 be Hilbert spaces.
Proof.
Proposition 6.4.5
Assigning to each pair of Hilbert spaces \mathfrak{H}_1 and \mathfrak{H}_2 the Hilbert tensor product
\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2
and to each pair of bounded linear operators A_1:\mathfrak{H}_1 \to \mathfrak{H}_3 and A_2:\mathfrak{H}_2 \to \mathfrak{H}_4
between Hilbert spaces the unique bounded extension
A_1 {\,\widehat{\otimes}\,} A_2 : \mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2 \to \mathfrak{H}_3 {\,\widehat{\otimes}\,} \mathfrak{H}_4 of the operator
A_1 \otimes A_2 : \mathfrak{H}_1 \otimes \mathfrak{H}_2 \to \mathfrak{H}_3 {\,\widehat{\otimes}\,} \mathfrak{H}_4,
v_1 \otimes v_2 \mapsto A_1(v_1) \otimes A_2(v_2)
comprises a (covariant) bifunctor
{\,\widehat{\otimes}\,} : \mathsf{Hilb} \times \mathsf{Hilb} \to \mathsf{Hilb} .
Moreover, {\,\widehat{\otimes}\,} is isometric in the sense that
\tag{6.4.6}
\| v_1 \otimes v_2\| = \|v_1\| \, \|v_2\| \quad \text{for all } v_1\in \mathfrak{H}_1, \: v_2 \in \mathfrak{H}_1 \text{ and }
\| A_1 {\,\widehat{\otimes}\,} A_2\| = \| A_1\| \, \|A_2\| \quad \text{for all }
A_1\in \mathfrak{B} ( \mathfrak{H}_1, \mathfrak{H}_3 ), \: A_2\in \mathfrak{B} ( \mathfrak{H}_2, \mathfrak{H}_4 ) .
Proof.
We first show that A_1 \otimes A_2 is a bounded operator. To this end observe that
A_1 \otimes A_2 can be written as the composition of the two operators A_1\otimes \mathbb{1}_{\mathfrak{H}_2}
and \mathbb{1}_{\mathfrak{H}_3} \otimes A_2. Hence it suffices to show that each of these linear maps is bounded.
Let v = \sum_{k=1}^n v_{1k} \otimes v_{2k} \in \mathfrak{H}_1 \otimes \mathfrak{H}_2 be of norm 1.
As in the proof of \Crefthm:construction-tensor-inner-product expand
v_{2k} = \sum_{i=1}^m c_{ki} e_i, k=1,\ldots ,n, where e_1, \ldots , e_m is an orthonormal basis of
\operatorname{Span} \{ v_{21}, \ldots , v_{2n}\} and c_{k1}, \ldots , c_{km} \in \mathbb{K}.
Equations \eqrefeq:expansion-tensor-product-vector-orthonormal-basis-second-component
and \eqrefeq:positivity-tensor-inner-product then entail that
v = \sum_{i=1}^m w_{1i} \otimes e_i \quad\text{and}\quad
1 = \langle v,v \rangle = \sum_{i=1}^m \| w_{1i} \|^2
for w_{1i} = \sum_{k=1}^n c_{ki} v_{1k} , i=1,\ldots , m.
Hence
\|( A_1 \otimes \mathbb{1}_{\mathfrak{H}_2})v \|^2 = \left\| \sum_{i=1}^m A_1 (w_{1i}) \otimes e_i\right\|^2
= \sum_{i=1}^m \| A_1 (w_{1i}) \|^2 \leq \|A_1\|^2 \sum_{i=1}^m \| w_{1i} \|^2 = \|A_1\|^2 ,
so A_1\otimes \mathbb{1}_{\mathfrak{H}_2} is bounded with norm \leq \|A_1\|.
By symmetry, \mathbb{1}_{\mathfrak{H}_3} \otimes A_2 is bounded with norm \leq \|A_2\|. Hence
A_1 \otimes A_2 = (\mathbb{1}_{\mathfrak{H}_3} \otimes A_2 )\circ (A_1\otimes \mathbb{1}_{\mathfrak{H}_2})
is bounded and
\| A_1 \otimes A_2 \| \leq \|A_1\| \, \|A_2\| .
Let us show the converse inequality. For given \varepsilon > 0 there exist
unit vectors v_i \in \mathfrak{H}_i, i=1,2 such that
\|A_iv_i \|\geq \| A_i \| - \frac{\varepsilon}{2(\| A_1\| +\|A_2\| +1)}. Then
\| (A_1 {\,\widehat{\otimes}\,} A_2) (v_1\otimes v_2) \| = \| A_1v_1\| \, \| A_2v_2\|
\geq \| A_1 \| \, \| A_2\| - \varepsilon .
This implies
\| A_1 \otimes A_2 \| \geq \|A_1\| \, \|A_2\|
and \eqrefeq:completed-tensor-product-isometry-equation-bounded-linear-maps follows.
Equality \eqrefeq:completed-tensor-product-isometry-equation-vectors is clear by
construction of the Hilbert tensor product.
Next observe that
\mathbb{1}_{ \mathfrak{H}_1} {\,\widehat{\otimes}\,} \mathbb{1}_{\mathfrak{H}_2} = \mathbb{1}_{\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2}
by definition. Given Hilbert spaces \mathfrak{H}_1,\ldots,\mathfrak{H}_6
and bounded linear operators A_i: \mathfrak{H}_{i}\to \mathfrak{H}_{i+2} and
B_i : \mathfrak{H}_{i+2}\to \mathfrak{H}_{i+4} for i=1,2, the composition
(B_1 \otimes B_2 ) \circ (A_1 \otimes A_2) coincides with
(B_1\circ A_1) \otimes (B_2\circ A_2) by functoriality of the algebraic tensor product.
By continuity of the operators A_1 {\,\widehat{\otimes}\,} A_2 and B_1 {\,\widehat{\otimes}\,} B_2 and
by density of \mathfrak{H}_1\otimes \mathfrak{H}_2 in \mathfrak{H}_1{\,\widehat{\otimes}\,} \mathfrak{H}_2
the equality
(B_1 {\,\widehat{\otimes}\,} B_2 ) \circ (A_1 {\,\widehat{\otimes}\,} A_2) =
(B_1\circ A_1) {\,\widehat{\otimes}\,} (B_2\circ A_2)
follows. Hence {\,\widehat{\otimes}\,} is a bifunctor as claimed.
Proposition 6.4.6
For every Hilbert space \mathfrak{H} one has two natural isomorphisms
\widehat{u}_\mathfrak{H} : \mathbb{K} {\,\widehat{\otimes}\,} \mathfrak{H} \to \mathfrak{H} ,\: z\otimes v \to z v \quad \text{and} \quad
{}_\mathfrak{H} \widehat{u} : \mathfrak{H} {\,\widehat{\otimes}\,} \mathbb{K} \to \mathfrak{H} ,\: v \otimes z \to z v
called the left and the right unit, respectively. Furthermore, for every triple of Hilbert spaces
\mathfrak{H}_1,\mathfrak{H}_2,\mathfrak{H}_3 there is a natural isomorphism, called associator
\widehat{a}_{\mathfrak{H}_1,\mathfrak{H}_2,\mathfrak{H}_3} : (\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2) {\,\widehat{\otimes}\,} \mathfrak{H}_3 \to
\mathfrak{H}_1 {\,\widehat{\otimes}\,} ( \mathfrak{H}_2 {\,\widehat{\otimes}\,} \mathfrak{H}_3 ),\:
(v_1\otimes v_2)\otimes v_3 \mapsto v_1\otimes (v_2 \otimes v_3 ).
These data fulfill the so-called coherence conditions that is the pentagon diagram
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\node (P2) at (90+2*72:2.5cm) {\mathllap{\mathfrak{H}_1{\,\widehat{\otimes}\,} ((\mathfrak{H}_2{\,\widehat{\otimes}\,} \mathfrak{H}_3)}{\,\widehat{\otimes}\,} \mathfrak{H}_4)};
\node (P3) at (90+3*72:2.5cm) {\mathfrak{H}_1{\,\widehat{\otimes}\,} (\mathfrak{H}_2\mathrlap{{\,\widehat{\otimes}\,} (\mathfrak{H}_3{\,\widehat{\otimes}\,} \mathfrak{H}_4))}};
\node (P4) at (90+4*72:2.5cm) {(\mathfrak{H}_1{\,\widehat{\otimes}\,} \mathfrak{H}_2){\,\widehat{\otimes}\,} (\mathfrak{H}_3{\,\widehat{\otimes}\,} \mathfrak{H}_4)};
\draw
(P0) edge[- > , > =angle 90] node[left] {\widehat{a}_{\mathfrak{H}_1,\mathfrak{H}_2,\mathfrak{H}_3} {\,\widehat{\otimes}\,} \, \mathbb{1}_{\mathfrak{H}_4}\hspace{2mm}} (P1)
(P1) edge[- > , > =angle 90] node[left] {\widehat{a}_{\mathfrak{H}_1,\mathfrak{H}_2{\,\widehat{\otimes}\,} \mathfrak{H}_3 , \mathfrak{H}_4}} (P2)
(P2) edge[- > , > =angle 90] node[below] {\hspace{5mm}\mathbb{1}_{\mathfrak{H}_1}{\,\widehat{\otimes}\,} \, \widehat{a}_{\mathfrak{H}_2,\mathfrak{H}_3,\mathfrak{H}_4}} (P3)
(P4) edge[- > , > =angle 90] node[right] { \widehat{a}_{\mathfrak{H}_1 , \mathfrak{H}_2,\mathfrak{H}_3 {\,\widehat{\otimes}\,} \mathfrak{H}_4}} (P3)
(P0) edge[- > , > =angle 90] node[right] { \widehat{a}_{\mathfrak{H}_1{\,\widehat{\otimes}\,} \mathfrak{H}_2, \mathfrak{H}_3 , \mathfrak{H}_4}} (P4);
\end{tikzpicture}
and the triangle diagram
\begin{tikzcd}
( \mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathbb{K}) {\,\widehat{\otimes}\,} \mathfrak{H}_2
\ar[rrrr,"\widehat{a}_{\mathfrak{H}_1,\mathbb{K},\mathfrak{H}_3}"]
\ar[drr,"{}_{\mathfrak{H}_1}\!\widehat{u} \, {\,\widehat{\otimes}\,} \,\mathbb{1}_{\mathfrak{H}_2}",swap]
\mathfrak{H}_1 {\,\widehat{\otimes}\,} (\mathbb{K} {\,\widehat{\otimes}\,} \mathfrak{H}_2 )
\ar[dll,"\mathbb{1}_{\mathfrak{H}_1} {\,\widehat{\otimes}\,} \, \widehat{u}_{\mathfrak{H}_2}"]
\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2
\end{tikzcd}
commute for all Hilbert spaces \mathfrak{H}_1,\mathfrak{H}_2,\mathfrak{H}_3,\mathfrak{H}_4. In other words,
the category \mathsf{Hilb} endowed with the Hilbert tensor product {\,\widehat{\otimes}\,} is a monoidal category.
Proof.
The category of \mathbb{K}-vector spaces with the usual tensor product as tensor functor is monoidal.
Denote the corresponding unit isomorphisms and associator by
_{-}u, u_{-}, and a_{-,-,-}, respectively. Then observe that by construction
\mathbb{K} {\,\widehat{\otimes}\,} \mathfrak{H} = \mathbb{K} \otimes \mathfrak{H} and \mathfrak{H} {\,\widehat{\otimes}\,} \mathbb{K} = \mathfrak{H} \otimes \mathbb{K}
for every Hilbert space \mathfrak{H}. In particular this means that putting
\widehat{u}_\mathfrak{H} = u_\mathfrak{H} and {}_{\mathfrak{H}}\widehat{u} = {}_{\mathfrak{H}}u gives the desired units.
Next recall that \mathfrak{H}_1 \otimes \mathfrak{H}_2 is dense in \mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2
which by \Crefthm:totality-tensor-product-total-sets implies density of (\mathfrak{H}_1 \otimes \mathfrak{H}_2) \otimes \mathfrak{H}_3
and \mathfrak{H}_1 \otimes (\mathfrak{H}_2 \otimes \mathfrak{H}_3) in (\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2) {\,\widehat{\otimes}\,} \mathfrak{H}_3
and \mathfrak{H}_1 {\,\widehat{\otimes}\,} (\mathfrak{H}_2 {\,\widehat{\otimes}\,} \mathfrak{H}_3), respectively.
Similarly one argues that \mathfrak{H}_1 \otimes (\mathfrak{H}_2 \otimes (\mathfrak{H}_3\otimes \mathfrak{H}_4))
is dense in \mathfrak{H}_1 {\,\widehat{\otimes}\,} (\mathfrak{H}_2 {\,\widehat{\otimes}\,} (\mathfrak{H}_3{\,\widehat{\otimes}\,} \mathfrak{H}_4)),
and so on. Since the associator map
a_{\mathfrak{H}_1,\mathfrak{H}_2,\mathfrak{H}_3} : (\mathfrak{H}_1 \otimes \mathfrak{H}_2) \otimes \mathfrak{H}_3 \to
\mathfrak{H}_1 \otimes ( \mathfrak{H}_2 \otimes \mathfrak{H}_3 ) is bounded, it extends in a unique way to a linear
bounded map \widehat{a}_{\mathfrak{H}_1,\mathfrak{H}_2,\mathfrak{H}_3} : (\mathfrak{H}_1 {\,\widehat{\otimes}\,} \mathfrak{H}_2) {\,\widehat{\otimes}\,} \mathfrak{H}_3 \to
\mathfrak{H}_1 {\,\widehat{\otimes}\,} ( \mathfrak{H}_2 {\,\widehat{\otimes}\,} \mathfrak{H}_3 ). Using density, continuity, and commutativity
of the pentagon and triangle diagrams for the tensor product functor one concludes that the coherence conditions for
{\,\widehat{\otimes}\,} with the unit and associator maps _{-}\widehat{u}, \widehat{u}_{-}, and \widehat{a}_{-,-,-} are satisfied.
6.5.1
Throughout this section, \mathfrak{H} stands for a Hilbert space over the
field \mathbb{K} of real or complex numbers.
Let A \in \mathfrak{B} (\mathfrak{H}) that is let
A:\mathfrak{H} \rightarrow \mathfrak{H} be linear and bounded. Then the map
b_A : \mathfrak{H} \times \mathfrak{H} \to \mathbb{K}, \: (v,w) \mapsto \langle v, Aw \rangle
is sesquilinear and bounded with norm
\left\|b_A\right\| = \sup \big\{ \left| b_A(v,w)\right| \bigm\vert
v,w\in \mathfrak{H} \:\\: \left\|w\right\|=\left\|v\right\|=1 \big\} =
\left\|A\right\| .
By \Crefthm:correspondence-bounded-sesquilinear-forms-bounded-operators
to the Riesz representation theorem there exists a unique element
A^* \in \mathfrak{B}(\mathfrak{H}) such that
b_A (v,w) = \langle A^* v, w \rangle \quad
\text{for all } v,w \in \mathfrak{H} .
This operator satisfies \left\|A^*\right\|= \left\|b_A\right\| = \left\|A\right\|.
Definition 6.5.2
The unique operator A^*\in \mathfrak{B} (\mathfrak{H}) associated to some
A\in \mathfrak{B}(\mathfrak{H}) such that
\langle v,Aw \rangle = \langle A^* v, w \rangle \quad
\text{for all } v,w \in \mathfrak{H}
is called the adjoint of A.
Proposition 6.5.3
If A \in \mathfrak{B} (\mathfrak{H}), then so is its adjoint A^* \in \mathfrak{B} (\mathfrak{H}).
Proof.
First we show that A^* is linear. Given v, w, w' \in H, we compute
\langle v, A^*(w + w') \rangle = \mu_{A, w+w'}(v) = \langle Av, w + w' \rangle = \langle Av, w \rangle + \langle Av, w' \rangle
= \mu_{A, w}(v) + \mu_{A, w'}(v) = \langle v, A^*w \rangle + \langle v, A^*w' \rangle = \langle v, A^*w + A^*w' \rangle
Since this is true for all v \in H, this implies that A^*(w+w') = A^*w'. Furthermore, given \lambda \in \mathbb{K}, we have
\langle v, A^*(\lambda w) \rangle = \mu_{A, \lambda w}(v) = \langle Av, \lambda w \rangle = \overline \lambda \langle Av, w \rangle
= \overline \lambda \mu_{A, w}(v) = \overline \lambda \langle v, A^* w \rangle = \langle v, \lambda A^*w \rangle.
Again, since this is true for all v \in H, we know A^*(\lambda w) = \lambda A^*w. This proves that A^* is linear.
It remains to show that A^* is bounded. We know
\left\|A^*\right\| = \sup_{\left\|v\right\| = \left\|w\right\| = 1} {\left|\langle v, A^* w \rangle\right|} = \sup_{\left\|v\right\| = \left\|w\right\| = 1} {\left|\langle w, Av \rangle\right|} = \left\|A\right\| < \infty,
which is what we wanted to show. Note that \left\|A^*\right\| = \left\|A\right\|.
We leave it as an exercise to show that
\left\|A\right\| = \sup_{\left\|v\right\| = \left\|w\right\| = 1} {\left|\langle v, Aw \rangle\right|}
for all A \in \mathfrak{L} (\mathfrak{H}), as was used in the above proof.
Definition 6.5.4
An operator A \in \mathfrak{L} (\mathfrak{H}) is called self-adjoint if A = A^*, unitary if A^* = A^{-1}, and normal if [A, A^*] = AA^* - A^*A = 0.
We note that self-adjoint and unitary operators are always normal, but normal operators do not have to be self-adjoint or unitary. In the remainder of these notes, we gather several results on self-adjoint and normal operators.
Lemma 6.5.5
An operator A \in \mathfrak{L} (\mathfrak{H}) is self-adjoint if and only if \langle Av, v \rangle \in \mathbb{R} for all v \in H.
Proof.
\Rightarrow) If A is self-adjoint, then
\langle Av, v \rangle = \mu_{A,v}(v) = \langle v, A^*v \rangle = \langle v, Av \rangle = \overline{\langle Av,v \rangle},
which implies that \langle Av,v \rangle \in \mathbb{R}.
\Leftarrow) Suppose that \langle Av, v \rangle \in \mathbb{R} for all v \in H.  This implies that \langle Av, w \rangle + \langle Aw, v \rangle is real as well, so
\Im \langle Av, w \rangle = -\Im \langle Aw, v \rangle= \Im \langle v, Aw \rangle.
Since this holds for all w \in H, it holds for iw as well. Thus,
\Re \langle Av, w \rangle = \Im \langle Av, -iw \rangle = \Im \langle v, A(-iw) \rangle = \Im i\langle v, Aw \rangle = \Re \langle v, Aw \rangle.
Combining the above two lines yields \langle Av, w \rangle = \langle v, Aw \rangle for all v, w \in H. Since the adjoint satisfies \langle Av, w \rangle = \langle v, A^*w \rangle, this implies that A = A^*.
Proposition 6.5.6
If A \in \mathfrak{L} (\mathfrak{H}) and \langle Av, v \rangle = 0 for all v \in H, then A = 0.
Proof.
Since \langle Av, v \rangle = 0 for all v \in H, equation (*) from Lemma 4 reduces to
\langle Av, w \rangle = -\langle Aw, v \rangle = -\langle w, Av \rangle = -\overline{\langle Av, w \rangle} \quad \text{for all } v, w \in H ,
i.e.\langle Av,w \rangle has no real part for all v, w \in H. But then fixing v and setting w = Av implies \left\|Av\right\|^2 = 0 for all v \in H, so A = 0.
Proposition 6.5.7
If A \in \mathfrak{L} (\mathfrak{H}) is self-adjoint, then
\left\|A\right\| = \sup_{\left\|v\right\| = 1} {\left|\langle Av, v \rangle\right|}.
Proof.
We know
\left\|A\right\| = \sup_{\left\|v\right\| = \left\|w\right\| = 1} {\left|\langle Av, w \rangle\right|},
so we clearly have
\sup_{\left\|v\right\| = 1} {\left|\langle Av, v \rangle\right|} \leq \left\|A\right\|.
Proposition 6.5.8
If A \in \mathfrak{L} (\mathfrak{H}), then A^*A is self-adjoint and \left\|A^*A\right\| = \left\|A\right\|^2.
Proof.
For arbitrary v \in H, we have
\langle A^*Av, v \rangle = \langle Av, Av \rangle = \left\|Av\right\|^2 \in \mathbb{R},
so A^*A is self-adjoint by Lemma 4. By Proposition 6,
\left\|A^*A\right\| = \sup_{\left\|v\right\| = 1} {\left|\langle A^*Av, v \rangle\right|} = \sup_{\left\|v\right\| = 1} \left\|Av\right\|^2 = \left\|A\right\|^2.
Proposition 6.5.9
If A \in \mathfrak{L} (\mathfrak{H}), then there exist B,C \in \mathfrak{L} (\mathfrak{H}) self-adjoint such that A = B+iC. Furthermore, A is normal if and only if [B,C] = 0.
Proof.
We define
B = \frac{1}{2}(A + A^*) \quad \text{ and } \quad C = \frac{i}{2}(A^* - A).
Clearly A = B + iC. Note also that A^* = B - iC.
Finally, we compute
[A, A^*] = [B + iC, B-iC] = -i[B, C] + i[C,B] = -2i[B,C],
Clearly A is normal if and only if [B,C] = 0.
Proposition 6.5.10
If A is normal, then \left\|Av\right\| = \left\|A^*v\right\| for all v \in H.
Proof.
Using the fact that A^*A = AA^*, we compute
\left\|Av\right\|^2 = \langle Av, Av \rangle = \langle v, A^*Av \rangle = \langle v, AA^*v \rangle = \langle A^*v, A^*v \rangle = \left\|A^*v\right\|^2.
Taking a square root yields the desired result.
6.6.1
In this section \mathfrak{H} will always denote a fixed complex Hilbert space.
Definition 6.6.2
By a projection-valued measure or a spectral measure on a
measurable space (\Omega,\mathscr{A}) one understands a map
E: \mathscr{A} \to \mathfrak{B}(\mathfrak{H}) having the following properties:
Remark 6.6.3
Recall that convergence of a sequence of operators
(A_n)_{n\in \mathbb{N}} \subset \mathfrak{B}(\mathfrak{H}) in the strong operator topology to some A means
that for every v\in \mathfrak{H} the sequence (A_nv)_{n\in\mathbb{N}} converges in \mathfrak{H} to Av.
One denotes this by A ={s\,-\!\!}\lim\limits_{n\to\infty} A_n.
Likewise, B = {s\,-\!\!}\sum\limits_{n=0}^\infty A_n means that the sequence of partial sums
\left( \sum\limits_{k=0}^n A_n \right)_{n\in \mathbb{N}} converges in the strong operator topology
to some B \in \mathfrak{B} (\mathfrak{H}).
Proposition 6.6.4
A spectral measure E : \mathscr{A} \to \mathfrak{B} (\mathfrak{H}) has the following properties
in addition to the defining axioms:
Proof.
\itshape ad \hyperref[ite:spectral-measure-empty-set] (SM1').
\itshape ad \hyperref[ite:spectral-measure-finite-additivity] (SM2').
\itshape ad (SM1).
6.7.1
We now apply the foundations of Hilbert space theory built in the previous sections to
spectral theory. For the moment we will sacrifice generality and work only with bounded
linear operators. The spectral theory of unbounded linear operators will be treated later.
Let us a recall that a linear map A: \mathfrak{H}_1 \rightarrow \mathfrak{H}_2 between Hilbert spaces
is continuous if and only if it is bounded, i.e.has finite operator norm, and that
\mathfrak{B} (\mathfrak{H}_1,\mathfrak{H}_2) is a Banach space with the operator norm.
For the rest of this section, \mathfrak{H}, \mathfrak{H}_1, \mathfrak{H}_2, \ldots will always
denote complex Hilbert spaces and A, B bounded linear operators.
We will also now fix the base field to be complex, i.e.\mathbb{K} = \mathbb{C}.
Last we agree on writing I_\mathfrak{H} or just I for the identity operator on a Hilbert space
\mathfrak{H}.
\addcontentslinetocsubsectionSpectrum and Resolvent
Definition 6.7.2
Let A:\mathfrak{H} \to \mathfrak{H} be a bounded linear operator. A complex number \lambda
is then called an eigenvalue of A if there exists a nonzero v \in H such that
Av = \lambda v. For every \lambda \in \mathbb{C} one defines the \lambda-eigenspace of
A as
\operatorname{Eig}_\lambda (A) = \big\{ v \in H \bigm\vert Av = \lambda v \big\} \subset \mathfrak{H},
which is clearly a linear subspace of \mathfrak{H}.
6.7.3
By definition it is immediately clear that
\operatorname{Eig}_\lambda (A) = \ker(A - \lambda),
where the \lambda on the right stands for the operator \lambda I.
In other words this means that
\lambda\in \mathbb{C} is an eigenvalue of A if and only if A - \lambda is not injective.
Definition 6.7.4
Let A \in \mathfrak{B}(\mathfrak{H}). We make the following definitions.
6.7.1
Evidently, \sigma(A) = \mathbb{C} \setminus \varrho(A) and \sigma_{p}(A) \subset \sigma_{ap}(A) \subset \sigma(A),
and these may all be strict inclusions. Note that A - \lambda is bounded for any \lambda \in \mathbb{C},
so the open mapping theorem ref_error implies that (A - \lambda)^{-1} \in \mathfrak{B}(\mathfrak{H}) when \lambda \in \varrho(A).
We call the map
R_\bullet (A ):\varrho(A) \rightarrow \mathfrak{B}(\mathfrak{H}), \quad R_\lambda (A) = (A-\lambda)^{-1}
the resolvent of A, not to be confused with the resolvent set \varrho(A).
To keep the notation clean, we often briefly write R_\lambda for R_\lambda(A) and leave implicit that R_\lambda depends on A.
First, we prove some topological properties of the spectrum and resolvent. Recall the following lemma, which generalizes the geometric series.
Lemma 6.7.5[Carl Neumann] Let A \in \mathfrak{B}(\mathfrak{H}). If \left\|A\right\| < 1, then I - A is invertible,
(I - A)^{-1} = \sum_{n=0}^\infty A^n,
and
\left\|(I-A)^{-1}\right\| \leq \frac{1}{1 - \left\|A\right\|}.
Proof.
Since \left\|A\right\| < 1 and \left\|A^n\right\| \leq \left\|A\right\|^n by submultiplicativity of the operator norm, we know
\sum_{n=0}^\infty \left\|A^n\right\| < \infty. This implies that the family (A^n)_{n\in\mathbb{N}} is absolutely summable,
so \sum_{n=0}^\infty A^n exists. Furthermore, for every N \in \mathbb{N} we have
(I-A)\sum_{n=0}^N A^n = \left(\sum_{n=0}^N A^n \right)(I-A)= \sum_{n=0}^N A^n - \sum_{n=1}^{N+1}A^n = I - A^{N+1},
which implies that
\lim_{N \rightarrow \infty}(I-A)\sum_{n=0}^N A^n = \lim_{N \rightarrow \infty} \left(\sum_{n=0}^N A^n\right) (I-A) = I.
By continuity of multiplication in \mathfrak{B}(\mathfrak{H}) one gets
(I-A) \sum_{n=0}^\infty A^n = \left(\sum_{n=0}^\infty A^n\right) (I-A) = I,
which proves that I-A is invertible and (I-A)^{-1} = \sum_{n=0}^\infty A^n.
Finally, one concludes by the triangle inequality and submultiplicativity of the operator norm
\left\|(I - A)^{-1}\right\| \leq \sum_{n=0}^\infty \left\|A^n\right\| \leq \sum_{n=0}^\infty \left\|A\right\|^n = \frac{1}{1 - \left\|A\right\|}.
Proposition 6.7.6
Let A \in \mathfrak{B}(\mathfrak{H}).
Proof.
Next, we prove some algebraic properties of the resolvent. Hereby, [A,B] = AB - BA denotes the commutator of two operators,
as usual.
Proposition 6.7.7
Let A,B \in \mathfrak{B}(\mathfrak{H}). Then the following holds true.
Proof.
The resolvent R_\bullet (A) also has some nice analytic properties which we are going to prove next.
Proposition 6.7.8
The resolvent R_\bullet (A) :\varrho(A) \rightarrow \mathfrak{B}(\mathfrak{H}), \lambda \mapsto R_\lambda is continuous and complex
differentiable with derivative given by
R_\bullet(A)' :\: \varrho(A) \rightarrow \mathfrak{B}(\mathfrak{H}), \: \lambda \mapsto \lim_{\mu \rightarrow \lambda} \frac{R_\mu - R_\lambda}{\mu - \lambda} = R_\lambda^2
Proof.
Fix \lambda \in \varrho(A) and \varepsilon > 0. Let 0 < {\left|\mu - \lambda\right|} < \delta, where
\delta = \min\left( \frac{\varepsilon}{2\left\|R_\lambda\right\|^2},\, \frac{1}{2\left\|R_\lambda\right\|} \right) .
Note that \mu \in \varrho(A) by Proposition 6.7.6. Moreover, \left\|(\mu - \lambda)R_\lambda\right\| < 1, so I - (\mu - \lambda)R_\lambda is invertible with norm less than (1 - \left\|(\mu - \lambda) R_\lambda\right\|)^{-1} by Lemma 6.7.5.
Now observe that the first resolvent identity can be rearranged to
R_\mu = R_\lambda[I - (\mu - \lambda)R_\lambda]^{-1} .
Hence
\left\|R_\mu - R_\lambda\right\| \leq {\left|\mu - \lambda\right|}\left\|R_\mu\right\|\left\|R_\lambda\right\|
\leq {\left|\mu - \lambda\right|} \left\|R_\lambda\right\|^2 \left\|(I - (\mu - \lambda)R_\lambda)^{-1}\right\|
\leq \frac{{\left|\mu - \lambda\right|} \left\|R_\lambda\right\|^2}{1 - \left\|(\mu - \lambda)R_\lambda\right\|}
< \frac{\varepsilon/2}{1-1/2} = \varepsilon .
This proves that \lambda \mapsto R_\lambda is continuous.
As for complex differentiability, we simply use the first resolvent identity and continuity to conclude
\lim_{\mu \rightarrow \lambda} \frac{R_\mu - R_\lambda}{\mu - \lambda} = \lim_{\mu \rightarrow \lambda} R_\mu R_\lambda = R_\lambda^2.
Proposition 6.7.9
Let A \in \mathfrak{B}(\mathfrak{H}). Then \lambda R_\lambda \rightarrow -I as {\left|\lambda\right|} \rightarrow \infty. In particular, R_\lambda \rightarrow 0 as {\left|\lambda\right|} \rightarrow \infty.
Proof.
Fix \varepsilon > 0. For {\left|\lambda\right|} > \left\|A\right\|, we have by \Crefthm:resolvent-topological-properties (III)
\lambda R_\lambda = -I - \sum_{n=1}^\infty \lambda^{-n}A^n.
Since
\left\|\sum_{n=1}^\infty \lambda^{-n}A^n\right\| \leq \frac{\left\|A\right\|}{{\left|\lambda\right|}-\left\|A\right\|},
one sees that \lambda R_\lambda \rightarrow - I as {\left|\lambda\right|} \rightarrow \infty. Similarly, for {\left|\lambda\right|} > \left\|A\right\| one has
\left\|R_\lambda\right\| \leq \frac{1}{{\left|\lambda\right|}} + \frac{1}{{\left|\lambda\right|}}\sum_{n=1}^\infty \left\|\lambda^{-n}A^n\right\| \leq \frac{1}{{\left|\lambda\right|}} + \frac{1}{{\left|\lambda\right|}} \frac{\left\|A\right\|}{{\left|\lambda\right|} - \left\|A\right\|},
which shows that R_\lambda \rightarrow 0 as {\left|\lambda\right|} \rightarrow \infty.
Proposition 6.7.10
For all v, w \in \mathfrak{H}, the map
\langle R_\bullet (A) v, w \rangle : \: \varrho(A) \rightarrow \mathbb{C} , \: \lambda \mapsto \langle R_\lambda v, w \rangle
is holomorphic with derivative
\langle R_\bullet (A) v, w \rangle' : \: \varrho(A) \rightarrow \mathbb{C} , \: \lambda \mapsto \langle R_\lambda^2 v, w \rangle.
Proof.
Given \lambda \in \varrho(A), we compute
\lim_{\mu \rightarrow \lambda} \frac{\langle R_\mu v, w \rangle - \langle R_\lambda v, w \rangle}{\mu - \lambda} =
\lim_{\mu \rightarrow \lambda} \frac{\langle (\mu - \lambda)R_\mu R_\lambda v, w \rangle}{\mu - \lambda} =
\lim_{\mu \rightarrow \lambda} \langle R_\mu R_\lambda v, w \rangle = \langle R_\lambda^2v , w \rangle,
where we have used the first resolvent identity in the first step and continuity of the inner product in the last.
Proposition 6.7.11
The spectrum of an operator A \in \mathfrak{B}(\mathfrak{H}) is nonempty.
Proof.
Suppose \sigma(A) = \emptyset, hence \varrho(A) = \mathbb{C}. The map
\mathbb{C} \to \mathbb{C} , \: \lambda \mapsto \langle R_\lambda v, w \rangle
then is entire for every v, w \in \mathfrak{H}. Furthermore, one has for \left\|v\right\|, \left\|w\right\| \leq 1
{\left|\langle R_\lambda v, w \rangle\right|} \leq \left\|R_\lambda\right\| \left\|v\right\| \left\|w\right\| \leq \left\|R_\lambda\right\| .
Since \lambda \mapsto \left\|R_\lambda\right\| is continuous and \left\|R_\lambda\right\| \rightarrow 0 as {\left|\lambda\right|} \rightarrow \infty, one sees
that \left\|R_\lambda\right\| is bounded. Hence \langle R_\bullet v, w \rangle is a bounded entire function, which by Liouville's theorem implies
that it is zero for every pair v, w \in \mathfrak{H} with \left\|v\right\| = \left\|w\right\| = 1.
This entails that R_\lambda = 0 for every \lambda \in \mathbb{C}, which
is a contradiction to R_\lambda being invertible. Hence \sigma(A) \neq \emptyset.
6.8.1
In this section let \mathrm{V},\mathrm{W} always denote Banach spaces over the field
\mathbb{K} =\mathbb{R} or \mathbb{K}=\mathbb{C}. The symbols \mathfrak{H}, \mathfrak{H}_1, ... will always stand for Hilbert spaces over \mathbb{K}.
Definition 6.8.2
By an unbounded \mathbb{K}-linear operator or shortly by an
unbounded operator from \mathrm{V} to \mathrm{W} we understand a linear map
A: \operatorname{Dom} (A) \to \mathrm{W} defined on a \mathbb{K}-linear subspace \operatorname{Dom} (A)\subset \mathrm{V}.
As usual, \operatorname{Dom} (A) is called the domain of the operator A.
The space of unbounded \mathbb{K}-linear operators from V to W will be
denoted \mathfrak{L}_\mathbb{K} (V,W) or just \mathfrak{L} (V,W).
Remark 6.8.3
In this work, the term ``unbounded'' is meant in the sense of
``not necessarily bounded''. Sometimes we just say
linear operator or even only operator instead of
``unbounded linear operator''.
6.8.4 Observe that besides the domain \operatorname{Dom} (A) of an unbounded operator
A \in \mathfrak{L} (\mathrm{V},\mathrm{W}) the
kernel
\operatorname{Ker}(A)=\big\{ v \in \mathrm{V} \bigm\vert Av = 0 \big\}\subset\mathrm{V} ,
the image
\operatorname{Im}(A)=\big\{ w \in \mathrm{W} \bigm\vert \exists v\in \operatorname{Dom} (A): w = Av \big\}
\subset \mathrm{W} ,
and the graph
\operatorname{Gr}(A)=\big\{ (v,w) \in \operatorname{Dom} (A) \times \mathrm{W} \bigm\vert
w = Av \big\} \subset \mathrm{V} \times \mathrm{W}
of A are all linear subspaces. We will frequently make use of this.
Definition 6.8.5
An unbounded operator A\in \mathfrak{L} (\mathrm{V},\mathrm{W}) is called
densely defined if \operatorname{Dom} (A) is dense in \mathrm{V},
and closed if the graph \operatorname{Gr} (A)
is closed in \mathrm{V} \times \mathrm{W}.
The operator A \in \mathfrak{L} (V,W) is called closable if the closure
\overline{\operatorname{Gr}(A)} is the graph of an unbounded operator
from \mathrm{V} to \mathrm{W}.
An operator A \in \mathfrak{L} (V,W) is called an extension of
B \in \mathfrak{L} (V,W) if \operatorname{Gr} (B) \subset \operatorname{Gr} (A). One writes in this
situation B \subset A.
7.1.1
Infinite tensor products of Hilbert spaces were introduced by
7.1.2
Let R be a commutative ring and (M_i)_{i\in I} a possibly infinite family of R-modules.
Consider \prod_{i\in I} M_i, the product of the family (M_i)_{i\in I} within the category
of R-modules. For each j\in I let \pi_j : \prod_{i\in I} M_i \to M_j denote the natural
projection onto the j-th factor and \iota_j :M _j \hookrightarrow \prod_{i\in I}M_i the
uniquely determined natural embedding such that
\pi_j\circ \iota_i =
\begin{cases}
\mathrm{id}_{M_i} & \text{for}\enspace i=j \enspace\text{and} \\
0 & \text{else} .
\end{cases}
Given an R-module N one then understands by a multilinear map from \prod_{i\in I}M_i
to N a map f: \prod_{i\in I}M_i \to N
such that for each j\in I and x\in \prod_{i\in I} M_i with \pi_j (x)=0 the map M_j\to N, m\mapsto f(\iota_j(m)+x) is linear.
The set of multilinear maps from \prod_{i\in I} M_i to N will be denoted by
\mathfrak{Mlin} \big(\prod_{i\in I} M_i,N\big). It carries a natural structure of an R-module
given by pointwise addition of multilinear maps and pointwise action of a scalar on a
multilinear map that is by
f+ g = \left( \prod_{i\in I} M_i \ni x \mapsto f(x) + g(x) \in N \right) \quad\text{and}\quad
r f = \left( \prod_{i\in I} M_i \ni x \mapsto rf(x)\in N \right)
for all f,g \in \mathfrak{Mlin} \big(\prod_{i\in I} M_i,N\big) and r\in R.
Since for j\in I and x\in \prod_{i\in I} M_i with \pi_j(x)= 0 the maps
M_j\to N, m \mapsto (f+g) (\iota_j(m) + x) = f (\iota_j(m) + x) + g (\iota_j(m) + x)
and M_j\to N, m \mapsto rf (\iota_j(m) + x) are linear by assumption on f and g,
the maps f+g and rf are multilinear again, so \mathfrak{Mlin}\big(\prod_{i\in I} M_i,N\big) is an
R-module indeed with zero element the constant function mapping to 0\in N.
Remarks 7.1.3
Before proceeding further let us make several explanations concerning the notation used.
Lemma 7.1.4[cf.~
Proof.
Lemma 7.1.5
Assume to be given a non-empty family of R-modules (M_i)_{i\in I} and a partition
(I_a)_{a \in A} of the index set I. Then there exists a natural ismorphism
\kappa_{I,A}: \prod_{i \in I} M_i \to \prod_{a \in A} \prod_{i \in I_a} M_i
uniquely determined by the condition that \pi_a \circ \kappa_{I,A} = \pi_{I_a} for all
a\in A.
Proof.
By the universal property of the product the R-module map
\kappa = \kappa_{I,A}: \prod_{i \in I} M_i \to \prod_{a \in A} \prod_{i \in I_a} M_i
exists and is uniquely determined by the requirement that \pi_a \circ \kappa_{I,A} = \pi_{I_a}
for all a\in A. Naturality also follows from the universal property of the product.
It remains to show that \kappa is an isomorphism. By construction,
\pi_i(x) = \pi_i \pi_a \kappa (x) =0 for all i\in I and a(i)\in A such that i\in I_{a(i)},
hence x=0.
So \kappa is injective. It is also surjective. To see this pick x_a \in \prod_{i \in I_a} M_i
for each a\in A. With a(i) for i\in I defined as before put
x = \big( \pi_i (x_{a(i)} )\big)_{i\in I}. Then, by construction,
\pi_i \pi_{a} \kappa (x) = \pi_i \pi_{a} (x) = \pi_i (x) = \pi_i (x_{a}) for all a\in A and
i\in I_a,
hence \big(\pi_a \kappa (x) \big)_{a \in A} = (x_a)_{a \in A}
and \kappa is surjective.
Proposition 7.1.6[Exponential law for multilinear maps]
Let (M_i)_{i\in I} be a family of R-modules over a commutative ring R,
N an R-module,
and assume that J \subset I is a non-empty subset such that the complement
K = I\setminus J is also non-empty. Then the map
\eta_{I,J}: \:
\mathfrak{Mlin} \left( \prod_{j\in J} M_j , \mathfrak{Mlin} \left(\prod_{k\in K} M_k ,N \right)\right)
\rightarrow \mathfrak{Mlin} \left( \prod_{i\in I} M_i, N \right),
\hspace{1em} f \mapsto \left( \prod_{i\in I} M_i\ni (x_i)_{i\in I} \mapsto
f\big( (x_j)_{j\in J} \big)\left( (x_k)_{k\in K} \right) \in N \right)
is an isomorphism which is natural in (M_i)_{i\in I} and N.
Proof.
We first show that \eta= \eta_{I,J} is linear. To this end let
f,g \in \mathfrak{Mlin} \left( \prod_{j\in J} M_j , \mathfrak{Mlin} \left(\prod_{k\in K} M_k ,N \right)\right) and r\in R.

Hence \eta is an R-module map.
Next we show that \eta is an isomorphism by constructing an inverse.
Given f \in \mathfrak{Mlin} \big( \prod_{i\in I} M_i, N \big) we define
f^\sharp : \mathfrak{Mlin} \big( \prod_{j\in J} M_j \big) \to \mathfrak{Mlin} \big(\prod_{k\in K} M_k ,N \big) by the requirement that
f^\sharp (y) (z) = f (x_{y,z}) \quad \text{for all}\enspace y = (y_j)_{j\in J}\enspace \text{and} \enspace z = (z_k)_{k \in K} ,
where x_{y,z} is the element of \prod_{i\in I}M_i uniquely determined by
\pi_i (x_{y,z}) =
\begin{cases}
y_i & \text{for}\enspace i \in J , \\
z_i & \text{for}\enspace i \in K .
\end{cases}
One thus obtains an R-module map
(-)^\sharp_{I,J} :
\mathfrak{Mlin} \left( \prod_{i\in I} M_i, N \right) \to
\mathfrak{Mlin} \left( \prod_{j\in J} M_j , \mathfrak{Mlin} \left(\prod_{k\in K} M_k ,N \right)\right),\quad f \mapsto f^\sharp
which by construction is inverse to \eta_{I,J}.
Naturality of \eta_{I,J} in (M_j)_{j\in J} and N is clear by definition.
Definition 7.1.7
Let (M_i)_{i\in I} be a family of R-modules over a commutative ring R. By a
tensor product of (M_i)_{i\in I} one understands an R-module \bigotimes_{i\in I}M_i
together with a multilinear map
\tau : \prod_{i\in I}M_i \to \bigotimes_{i\in I}M_i such that the following universal property is fulfilled:
The linear map \overline{f} making the diagram comute will sometimes be called the linearization
of the multilinear map f.
Given a tensor product
\big( \bigotimes_{i\in I}M_i,\tau\big), we will usually denote the image of an element
(x_i)_{i\in I} \in \prod_{i\in I}M_i
under the map \tau by \otimes_{i\in I} x_i.
Remarks 7.1.8
Theorem 7.1.9
Let (M_i)_{i\in I} be a family of R-modules over a commutative ring R. Then the following
holds true.
Proof.
Lemma 7.1.10
Assume that (M_i)_{i\in I} is a finite family of R-modules such that for every i\in I a
generating set S_i of the R-module M_i has been given. Then the set
S = \tau \left( \prod_{i\in I}S_i \right) is a generating set of the tensor product
\bigotimes_{i\in I} M_i.
Proof.
By construction of the tensor product in the proof of
\Crefthm:construction-fundamental-properties-infinite-tensor-product it is clear that
a generating set of \bigotimes_{i\in I} M_i is given by the set
of elements of the form \otimes_{i\in I}x_i where (x_i)_{i\in I}\in \prod_{i\in I}M_i.
Each of the x_i can now be represented in the form
x_i = \sum_{k=1}^{n_i} r_{i,k} s_{i,k} \quad\text{with}\enspace r_{i,1},\ldots ,r_{i,n_i}\in R,\enspace
s_{i,1},\ldots ,s_{i,n_i}\in S_i .
Hence, by multilinearity of \tau and with I=\{ i_1,\ldots ,i_d\},
\otimes_{i\in I}x_i = \tau \left( (x_i)_{i\in I} \right) =
\sum_{k_{i_1} =1}^{n_{i_1}}\cdots \sum_{k_{i_d} =1}^{n_{i_d}} r_{i_1,k_{i_1}} \cdot \ldots \cdot r_{i_d,k_{i_d}}
\cdot \tau \left( (s_{i,k_i})_{i\in I} \right) ,
so \otimes_{i\in I}x_i is a linear combination of elements of S and the claim is proved.
Lemma 7.1.11
Let (M_i)_{i\in I} be a family of R-modules, (I_a)_{a\in A} a finite partition
of the index set I, and N an R-module. For a\in A put
N_a = \bigotimes_{i\in I_a} M_i and let \tau_a: \prod_{i\in I_a} M_i \to N_a
denote the canonical map. Assume that
f : \prod_{a\in A}\prod_{i\in I_a} M_i \to N is a map which is
componentwise multilinear in the following sense.
\beginitemize
\item[(\mathsf{ CM})\hspace-1mm]
Let b\in A and y=(y_a)_{a\in A} \in \prod_{a\in A}\prod_{i\in I_a} M_i
a family with y_b =0. If for all j\in I_b and families
x=(x_i)_{i\in I_b}\in \prod_{i\in I_b} M_i with x_j =0 the map
M_j \to N , \enspace m \mapsto f( \iota_b (\iota_j (m) + x) + y)
is linear, then f factors through
(\tau_a)_{a\in A}: \prod_{a\in A}\prod_{i\in I_a} M_i \to
\prod_{a\in A}N_a . More precisely, there exists
a unique multilinear map \overline{f} : \prod_{a\in A}N_a \to N such that
f = \overline{f} \circ (\tau_a)_{a\in A} .
\enditemize
Proof.
We prove the claim by induction on the cardinality of A.
If A is a singleton, then \prod_{a \in A} \prod_{i\in I_a} M_i canonically
coincides with \prod_{i\in I} M_i and f: \prod_{i\in I_a} M_i \to N
is multilinear, hence by the universal property of the tensor product there
exists a unique linear map \overline{f}: N_a \to N such that
f = \overline{f} \circ \tau_a.
Now assume that the claim holds whenever the cardinality of the index set A is
\leq n for some n \in \mathbb{N}^*.
Assume to be given initial data (M_i)_{i\in I} and §N, a partition
(I_a)_{a\in A} of A with |A| = n+1 and componentwise multilinear map
f: \prod_{a\in A}\prod_{i\in I_a} M_i \to N. Fix a \in A and put
B =A \setminus \{ a\}. Let
x = (x_i)_{i\in I_a} \in \prod_{i\in I_a}M_i
and \widetilde{x} be the element of \prod_{d\in A}\prod_{i\in I_d} M_i such that
\pi_d (\widetilde{x}) =
\begin{cases}
x &\text{for} \enspace d = a , \\
0 &\text{else} .
\end{cases}
The map
f_x: \prod_{b\in B}\prod_{i\in I_b} M_i \to N , \enspace
y \mapsto f (\iota_B (y) + \widetilde{x})
then is componentwise multilinear. Hence by inductive assumption there exists
a unique multilinear map \overline{f_x} : \prod_{b\in B}N_b \to N such that
f_x = \overline{f_x}\circ (\tau_b)_{b\in B}. By assumption on f
the map
\prod_{i\in I_a} M_i \to \mathfrak{Map} \left( \prod_{b\in B} \prod_{i\in I_b} M_i,N\right) ,
x \mapsto f_x is multilinear which implies multilinearity of
\overline{f_\bullet} :
\prod_{i\in I_a} M_i \to \mathfrak{Mlin} \left( \prod_{b\in B} N_b ,N \right),
\enspace x \mapsto \overline{f_x} .
Let F: N_a \to \mathfrak{Mlin} \left( \prod_{b\in B} N_b,N \right) be its
linearization. Application of the exponential law for multilinear maps,
\Crefthm:exponential-law-multilinear-maps, now gives a
multilinear map \eta (F) : \prod_{d \in A} N_d \to N which we denote
by \overline{f}. Given a family (x_d)_{d\in A} of families
x_d =(x_i)_{i\in I_d} one checks
\overline{f} \left( \big(\tau_d (x_d)\big)_{d\in A} \right)
= F \big( \tau_a(x_a) \big) \left( \big(\tau_b (x_b)\big)_{b\in B} \right)
= \overline{f}_{x_a} \left( \big(\tau_b (x_b)\big)_{b\in B} \right)
= f_{x_a} \left( (x_b)_{b\in B} \right) = f \left( (x_d)_{d\in A} \right) .
Hence \overline{f} \circ (\tau_d)_{d\in A} =f.
To finish the induction step it remains to prove uniqueness.
So let \overline{g} : \prod_{d \in A} N_d \to N be another multilinear map
such that \overline{g} \circ (\tau_d)_{d\in A} =f and consider
the induced linear map
\overline{g}^\sharp =\eta^{-1} (\overline{g}) : N_a \mapsto \mathfrak{Mlin} (\prod_{b\in B}N_b, N).
Then for every x\in \prod_{i\in I_a}M_i the relation
\overline{g}^\sharp (\tau_a(x)) \circ (\tau_b)_{b\in B} = f_x =
\overline{f}_x \circ (\tau_b)_{b\in B}
is satisfied.
Hence \overline{g}^\sharp (\tau(x)) = \overline{f}_x for all x\in \prod_{i\in I_a}M_i which entails
that \overline{g}^\sharp coincides with F. By
\Crefthm:exponential-law-multilinear-maps one obtains
\overline{g} = \overline{f}. This finishes the induction step and the lemma
is proved.
Proposition 7.1.12
Let (M_i)_{i\in I} be a family of R-modules and (I_a)_{a\in A} a finite partition of the index set I.
Then there exists a natural isomorphism
\alpha_{I,A}: \bigotimes_{i \in I} M_i \to \bigotimes_{a \in A} \bigotimes_{i \in I_a} M_i .
Proof.
Put N_a = \bigotimes_{i \in I_a} M_i for a \in A and let
\tau_a : \prod_{i \in I_a} M_i \to N_a be the canonical map to the tensor product.
Let \tau_A :\prod_{a \in A}N_a \to \bigotimes_{a \in A} N_a be the canonical map to
the tensor product of the modules N_a.
Define \tau_{I,A}: \prod_{i \in I} M_i\to \prod_{a \in A} N_a as the unique map so that
\pi_a \circ \tau_{I,A} = \tau_a \circ \pi_{I_a} for all a\in A.
By construction \tau_{I,A} = (\tau_a)_{a\in A} \circ \kappa_{I,A},
where \kappa_{I,A} : \prod_{i\in I} M_i \to \prod_{a \in A} \prod_{i \in I_a} M_i is the natural isomorphism from
\Crefthm:associator-cartesian-product.
The composition \tau_A \circ \tau_{I,A} then is multilinear by \Crefthm:construction-multilinear-maps-composition
(III), hence factors through a linear map
\alpha_{I,A} : \bigotimes_{i\in I} M_i \to \bigotimes_{a \in A} N_a
that is
\tag{7.1.1}
\tau_A \circ (\tau_a)_{a\in A} \circ \kappa_{I,A} =
\alpha_{I,A} \circ \tau_I .
Naturality of \alpha_{I,A} in (M_i)_{i\in I} is clear by definition so it remains to
construct an inverse to \alpha_{I,A}. Consider the composition
\tau_I \circ \kappa^{-1}: \prod_{a \in A} \prod_{i \in I_a} M_i
\to \bigotimes_{i\in I} M_i. Assume that a \in A and
(y_b)_{b\in A\setminus\{a\}} \in \prod_{b \in A\setminus\{a\}} \prod_{i \in I_b} M_i
have been chosen. Let y_a\in \prod_{i \in I_a} M_i be 0, put
\widetilde{y} =(y_d)_{d\in A} \in \prod_{d \in A} \prod_{i \in I_d} M_i, and let
y\in \prod_{i \in I} M_i be the family such that \pi_i(y) = \pi_i (y_{a(i)}) for
all i\in I, where a(i) denotes the unique element of A such that
i\in I_{a(i)}. In other words let y =\kappa^{-1} (\widetilde{y}).
For every j\in I_a and x= (x_i)_{i\in I_a}\in \prod_{i\in I_a}M_i
with \pi_j (x)=0 the map
M_j \to \bigotimes_{i\in I} M_i,\enspace m \mapsto \tau_I \circ \kappa^{-1} \left( \iota_a (\iota_j(m)+x)+\widetilde{y}\right)
= \tau_I \left( \iota_j (m) + \iota_{I_a}(x) + y \right)
then is multilinear since \tau_I is multilinear and \pi_j(\iota_{I_a}(x) + y)=\pi_j(x) +\pi_j(y_a) = 0.
Hence \tau_I \circ \kappa^{-1} is componentwise multilinear and therefore,
by \Crefthm:componentwise-multilinear-maps-factorization, factors
through the map
(\tau_a)_{a\in A} : \prod_{a\in A} \prod_{i\in I_a} M_i\to \prod_{a\in A} N_a
which means that
\tag{7.1.2}
\tau_I \circ \kappa^{-1} = \lambda_{I,A}\circ (\tau_a)_{a\in A}
for some uniquely defined multilinear map
\lambda_{I,A} : \prod_{a\in A} N_a\to\bigotimes_{i\in I} M_i.
Let
\overline{\lambda}_{I,A} : \bigotimes_{a\in A} N_a\to\bigotimes_{i\in I} M_i
be the linearization of \lambda_{I,A}.
We claim that \overline{\lambda_{I,A}} is inverse to \alpha_{I,A}.
By definition of \overline{\lambda_{I,A}} and
Eqs.~\eqrefeq:defining-equation-associator-map-tensor-product and
\eqrefeq:defining-equation-inverse-associator-map-tensor-product one concludes
\overline{\lambda_{I,A}} \circ \alpha_{I,A} \circ \tau_I =
\overline{\lambda_{I,A}} \circ \tau_A \circ (\tau_a)_{a\in A} \circ \kappa_{I,A} =
\lambda_{I,A} \circ (\tau_a)_{a\in A} \circ \kappa_{I,A} = \tau_I .
Since the image of \tau_I generates \bigotimes_{i\in I} M_i as an R-module,
\overline{\lambda_{I,A}} has to be left inverse to \alpha_{I,A}.
Using Eqs.~\eqrefeq:defining-equation-associator-map-tensor-product and
\eqrefeq:defining-equation-inverse-associator-map-tensor-product again compute
\alpha_{I,A} \circ\overline{\lambda_{I,A}} \circ \tau_A \circ (\tau_a)_{a\in A} =
\alpha_{I,A} \circ \lambda_{I,A} \circ (\tau_a)_{a\in A} = \alpha_{I,A}\circ\tau_A\circ \kappa_{I,A}^{-1}
= \tau_A \circ (\tau_a)_{a\in A} .
Since by \Crefthm:image-generating-system-canoncial-map-finite-tensor-product-generating-system
the image of
\tau_A \circ (\tau_a)_{a\in A} generates \bigotimes_{a \in A} \bigotimes_{i \in I_a} M_i, the equality
\alpha_{I,A} \circ\overline{\lambda_{I,A}}=\mathrm{id}_{\bigotimes_{a \in A} \bigotimes_{i \in I_a} M_i}
follows and the proposition is proved.
Proposition and Definition 7.1.13
Let (A_i)_{i\in I} be a family of R-algebras. Then the tensor product
A = \bigotimes_{i\in I} A_i carries in a natural way the structure of an
R-algebra where the product map is defined by
\cdot : A \times A \to A , \enspace
(\otimes_{i\in I} a_i , \otimes_{i\in I} b_i)\mapsto
\otimes_{i\in I} (a_i\cdot b_i) .
In case each of the algebras A_i is commutative, then A is commutative
as well.
Likewise, if each A_i is unital and 1_i denotes the unit element of
A_i, then A is unital with unit given by 1 = \otimes_{i\in I} 1_i.
One calls A the tensor product algebra of the family of algebras
(A_i)_{i\in I}.
Proof.
The map
\prod_{(i,k)\in I \times \{ 1,2\}} A_i \to A,\enspace
(a_{i,k})_{(i,k) \in I\times \{1,2\}} \mapsto\otimes_{i\in I}(a_{i,1}\cdot a_{i,2})
is multilinear by bilinearity of the product maps on the A_i and multilinearity of \tau_I,
so factors through a linear map
\mu: A \otimes A \cong \bigotimes_{(i,k)\in I\times \{1,2\}} A_i \to A. Composition of
\mu with the canonical bilinear map A \times A \to A\otimes A gives the product map
\cdot : A \times A\to A
and shows that the product on A is well-defined. By construction, the product map \cdot
is bilinear. Given \otimes_{i\in I} a_i, \otimes_{i\in I} b_i, \otimes_{i\in I} c_i \in A one
computes
\big( \otimes_{i\in I} a_i \cdot \otimes_{i\in I} b_i \big) \cdot \otimes_{i\in I} c_i
= \otimes_{i\in I} ((a_i\cdot b_i)\cdot c_i) =
\otimes_{i\in I} (a_i\cdot( b_i\cdot c_i)) =
\otimes_{i\in I} a_i \cdot \big( \otimes_{i\in I} b_i \cdot \otimes_{i\in I} c_i\big) .
This entails that the product on A is associative. In the same way one shows
that A is commutive respectively unital if each of the A_i is.
7.1.14
As we have seen, the infinite tensor product construction works well for objects of
algebraic categories like R-modules, vector spaces or R-algebras. As soon as
a topologies compatible with the algebraic structure come in it becomes difficult and
sometimes even impossible to construct or even define
\bibliographylmlib
\addcontentslinetocpartBibliography
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\iheadBibliography
\ohead
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