$\overline{u}(\infty)=\liminf_{\mathsf{x}^{2}+\mathsf{y}^{2}\to\infty}\frac{u(\mathsf{x},\mathsf{y})}{1+\mathsf{x}^{2}+\mathsf{y}^{2}}~{}.$
(31)

######  Remark 7.3.

The definition of $\overline{u}(\infty)$ in (31) is of course motivated by
Proposition 2.27 (see (11)): if $u=s_{\mathrm{par}}$ is the parabolic support
function of a convex entire achronal surface $\Sigma$, then
$\overline{u}(\infty)$ equals the value of the (elliptic) null support
function of $\Sigma$ at $\pi$. We remark that in the remainder of this section
we will assume (up to applying an isometry of ${\mathbb{R}}^{2,1}$) that
$\theta_{0}=\pi$ and $\phi(\theta_{0})=0$ in the statement of Theorem C; hence
we will be particularly interested in the behaviour of surfaces $\Sigma$ in
the direction $\pi$ in relation with the null support function around $\pi$.

######  Lemma 7.4.

Let $\Sigma$ be a convex entire achronal surface in ${\mathbb{R}}^{2,1}$ and
let

$u=s_{\mathrm{par}}:\overline{\mathcal{H}}\to{\mathbb{R}}\cup\\{+\infty\\}$

be its parabolic support function. Suppose that for every function $f$ of the
form

$f(\mathsf{x},\mathsf{y})=a+b\mathsf{x}+c(\mathsf{x}^{2}+\mathsf{y}^{2})$ (32)

such that $f\leq u$ on $\overline{\mathcal{H}}$, it holds that
$\overline{f}<\overline{u}$ on ${\mathbb{R}}\cup\\{\infty\\}$. Then $\Sigma$
is acausal.

###### Proof.

Suppose $\Sigma$ were not acausal. Then we claim that it is touched by a null
support plane at some point $q$. Indeed, it contains points $p$ and $q$ such
that the segment from $p$ to $q$ is lightlike. Notice that $p$ is contained in
the past of every spacelike plane through $q$, so $\Sigma$ can have no
spacelike support plane containing $q$. Therefore the support plane through
$q$ must be lightlike (and indeed its tangent space is the orthogonal to the
vector $q-p$).

Let $f$ be the parabolic support function of the point $q$ (equivalently, of
its future null cone). From Equation (5) and (7), $f$ is a quadratic
polynomial of the form (32) for some $a,b,c\in{\mathbb{R}}$. Since $\Sigma$ is
achronal, the future null cone of $q$ lies in its (causal) future so the
elliptic support function of $q$ is less than or equal to the elliptic support
function of $\Sigma$. But, since both are supported by the null plane $P$,
their support functions are equal in the null direction corresponding to $P$.
Translated into the parabolic support functions using Proposition 2.27, this
says that $f\leq u$ on $\overline{\mathcal{H}}$ but $f=u$ at the point of
${\mathbb{R}}\cup\\{\infty\\}$ corresponding to $P$. ∎

Following the list in Section 7.1, we then need to be able to show that
$\Sigma_{+}$ is $C^{2}$ and to verify property (2) of the surface
$\Sigma_{+}$, namely that its curvature is bounded below. In order to do this,
we will need to parametrize $\Sigma_{+}$ by the inverse of its Gauss map. More
precisely, under a mild regularity assumption on the support function, we will
point out a continuous map $\Gamma:\mathcal{H}\to\Sigma$ which has the
property that the plane though $\Gamma(\mathsf{x},\mathsf{y})$ orthogonal to
$\zeta(\mathsf{x},\mathsf{y})$ is a support plane for $\Sigma$. (Recall that
the map $\zeta$ has been introduced in (5).)

######  Lemma 7.5.

Let $\Sigma$ be a convex entire acausal surface in ${\mathbb{R}}^{2,1}$ and
let

$u=s_{\mathrm{par}}:\overline{\mathcal{H}}\to\mathbb{R}\cup\\{+\infty\\}$

be its parabolic support function. Assume that $u$ is $C^{1}$ in a
neighbourhood $U$ of a point $(\mathsf{x}_{0},\mathsf{y}_{0})\in\mathcal{H}$.
Then

  1. (1)

There exists a unique map $\Gamma:U\to\mathbb{R}^{2,1}$ which satisfies the
following equations:

$\displaystyle\langle\Gamma,\zeta\rangle$ $\displaystyle=u$ (33)
$\displaystyle\langle\Gamma,\zeta_{\mathsf{x}}\rangle$
$\displaystyle=u_{\mathsf{x}}$ (34)
$\displaystyle\langle\Gamma,\zeta_{\mathsf{y}}\rangle$
$\displaystyle=u_{\mathsf{y}}$ (35)

  2. (2)

$\Gamma$ is continuous and $\Gamma(\mathsf{x}_{0},\mathsf{y}_{0})$ is a point
of $\Sigma$ and the affine plane through
$\Gamma(\mathsf{x}_{0},\mathsf{y}_{0})$ orthogonal to
$\zeta(\mathsf{x}_{0},\mathsf{y}_{0})$ meets $\Sigma$ only at
$\Gamma(\mathsf{x}_{0},\mathsf{y}_{0})$.

  3. (3)

If moreover $u\circ\zeta^{-1}$ is strictly convex on $U$, then $\Gamma$ is a
homeomorphism onto an open subset $V$ of $\Sigma$. In this case $\Sigma$ is
$C^{1}$ and spacelike on $V$, and

$G\circ\Gamma=\bar{\zeta}$ (36)

where $G:V\to\mathbb{H}^{2}$ is the Gauss map and
$\bar{\zeta}:=\frac{\zeta}{|\zeta|}$.

###### Proof.

The fact that $\Gamma$ is well-defined and continuous follows from the
observation that
$\zeta(\mathsf{x},\mathsf{y}),\zeta_{\mathsf{x}}(\mathsf{x},\mathsf{y}),\zeta_{\mathsf{y}}(\mathsf{x},\mathsf{y})$
form a basis of $\mathbb{R}^{2,1}$ for every
$(\mathsf{x},\mathsf{y})\in\mathcal{H}$.

Let $\Gamma_{0}=\Gamma(\mathsf{x}_{0},\mathsf{y}_{0})$ and consider the
function $F:\mathcal{H}\to\mathbb{R}$ defined by

$F(\mathsf{x},\mathsf{y})=u(\mathsf{x},\mathsf{y})-\langle\Gamma_{0},\zeta(\mathsf{x},\mathsf{y})\rangle~{},$

Notice that $F\circ\zeta^{-1}$ is convex and $F$ vanishes at
$(\mathsf{x}_{0},\mathsf{y}_{0})$ by (33). Moreover by (34) and (35), $F$ has
a critical point at $(\mathsf{x}_{0},\mathsf{y}_{0})$. It follows that
$F(\mathsf{x},\mathsf{y})\geq 0$ for all
$(\mathsf{x},\mathsf{y})\in\mathcal{H}$, that is
$\langle\Gamma_{0},\zeta(\mathsf{x},\mathsf{y})\rangle\leq
u(\mathsf{x},\mathsf{y})$, with equality at
$\zeta(\mathsf{x}_{0},\mathsf{y}_{0})$. Now, recall from Proposition 2.17 and
Corollary 2.28 that the entire surface $\Sigma$ is entirely determined by its
parabolic support function $u$: more precisely, $\Sigma$ is the boundary of
the convex domain in ${\mathbb{R}}^{2,1}$ consisting of the points $p$ such
that $\langle p,\zeta(\mathsf{x},\mathsf{y})\rangle\leq
u(\mathsf{x},\mathsf{y})$ for all $(\mathsf{x},\mathsf{y})\in\mathcal{H}$.
Hence $\Gamma_{0}\in\Sigma$.

Let now $p\in\mathbb{R}^{2,1}$ be a point in the affine plane defined by
$u(\mathsf{x}_{0},\mathsf{y}_{0})=\langle
p,\zeta(\mathsf{x}_{0},\mathsf{y}_{0})\rangle$, with $p\neq\Gamma_{0}$. Then
at least one among (34) and (35) does not hold so there is a vector $W\in
T_{(\mathsf{x}_{0},\mathsf{y}_{0})}\mathcal{H}$ such that

$du(W)<\langle p,d\zeta(W)\rangle$

and this implies that there is some point
$(\mathsf{x}_{1},\mathsf{y}_{1})\in\mathcal{H}$ such that

$u(\mathsf{x}_{1},\mathsf{y}_{1})<\langle
p,\zeta(\mathsf{x}_{1},\mathsf{y}_{1})\rangle$

showing that $p\notin\Sigma$.

Assume now that $u\circ\zeta^{-1}$ is strictly convex on $U$. Hence the
function $F\circ\zeta^{-1}$ is strictly convex, since the difference between
$u\circ\zeta^{-1}$ and $F\circ\zeta^{-1}$ is an affine function. We have
already observed that $F(\mathsf{x},\mathsf{y})\geq 0$ and
$F(\mathsf{x}_{0},\mathsf{y}_{0})=0$. Hence strict convexity of
$F\circ\zeta^{-1}$ implies that the zero set of $F$ only contains
$(\mathsf{x}_{0},\mathsf{y}_{0})$. Since the zero set of $F$ equals the set of
$(\mathsf{x},\mathsf{y})\in\mathcal{H}$ such that
$\zeta(\mathsf{x},\mathsf{y})$ is orthogonal to a support plane through
$\Gamma_{0}$, this proves that $\Gamma$ is an injective function, and by the
Invariance of domain Theorem, $V=\Gamma(U)$ is an open subset of $\Sigma$.
Moreover each point in $V$ admits only one support plane, hence the surface
$\Sigma$ is $C^{1}$ on $\Gamma(U)$. Since those support planes are spacelike,
$V$ is a spacelike surface. Finally using that the plane through
$\Gamma(\mathsf{x},\mathsf{y})$ orthogonal to $\zeta(\mathsf{x},\mathsf{y})$
is a support plane for $\Sigma$ we conclude that $G\circ\Gamma=\bar{\zeta}$. ∎

Now for a $C^{2}$ function $u$ on $\mathcal{H}$, we give a characterization of
the convexity of $u\circ\zeta^{-1}$ in terms of the derivatives of $u$.

######  Lemma 7.6.

Let $u$ be a $C^{2}$ function on a domain $U$ in the upper half plane
$\mathcal{H}$, then $u\circ\zeta^{-1}$ is locally convex if and only if the
matrix

$A=\frac{1}{2}\begin{bmatrix}\mathsf{y}u_{\mathsf{x}\mathsf{x}}-u_{\mathsf{y}}&\mathsf{y}u_{\mathsf{x}\mathsf{y}}\\\
\mathsf{y}u_{\mathsf{x}\mathsf{y}}&\mathsf{y}u_{\mathsf{y}\mathsf{y}}-u_{\mathsf{y}}\end{bmatrix}$
(37)

is positive semi-definite at every point of $U$. Moreover if $A$ is positive
definite, then $u\circ\zeta^{-1}$ is strictly convex on $\zeta(U)$.

###### Proof.

Consider for every $(\mathsf{x}_{0},\mathsf{y}_{0})\in U$ a vector
$\Gamma_{0}=\Gamma(\mathsf{x}_{0},\mathsf{y}_{0})\in\mathbb{R}^{2,1}$ which
satisfies the system of equations $(\ref{eqn: def v 1},~{}\ref{eqn: def v
2},~{}\ref{eqn: def v 3})$. This vector exists since
$\zeta,\zeta_{x},\zeta_{y}$ are linearly indipendent. As observed before, both
the function
$F_{(\mathsf{x}_{0},\mathsf{y}_{0})}(\mathsf{x},\mathsf{y})=u(\mathsf{x},\mathsf{y})-\langle\Gamma_{0},\zeta(\mathsf{x},\mathsf{y}))$
and its differential vanish at $(\mathsf{x}_{0},\mathsf{y}_{0})$. Since
$F_{(\mathsf{x}_{0},\mathsf{y}_{0})}\circ\zeta^{-1}$ and $u\circ\zeta^{-1}$
differ by an affine function, $u\circ\zeta^{-1}$ is locally convex if and only
if for every $(\mathsf{x}_{0},\mathsf{y}_{0})$ the function
$F_{(\mathsf{x}_{0},\mathsf{y}_{0})}$ attains a local minimum at
$(\mathsf{x}_{0},\mathsf{y}_{0})$. By differentiating (5) we have that

$\zeta_{\mathsf{xx}}=\zeta_{\mathsf{yy}}=\frac{1}{\mathsf{y}}\zeta_{\mathsf{y}}\,,\qquad\zeta_{\mathsf{xy}}=0$

so a simple computation shows that the Hessian of
$F_{(\mathsf{x}_{0},\mathsf{y}_{0})}$ at $(\mathsf{x}_{0},\mathsf{y}_{0})$
with respect to the basis
$\frac{\partial\,}{\partial\mathsf{x}},\frac{\partial\,}{\partial\mathsf{y}}$
is given by $\frac{2}{\mathsf{y}_{0}}A(\mathsf{x}_{0},\mathsf{y}_{0})$ and the
result follows. ∎

In order to relate a higher regularity of the support function to a higher
regularity of the surface $\Sigma$, the following standard observation will be
useful.

######  Lemma 7.7.

Let $\Sigma$ be a spacelike surface in $\mathbb{R}^{2,1}$, and
$G:\Sigma\to\mathbb{H}^{2}$ be its Gauss map. If $G$ is $C^{1}$ around a point
$p_{0}\in\Sigma$, then $\Sigma$ is a $C^{2}$-submanifold around $p_{0}$.

###### Proof.

Recall that a spacelike surface is $C^{1}$ by definition. Hence around $p_{0}$
the surface $\Sigma$ is the graph of a $C^{1}$-function $f$ defined on some
open subset $\Omega$ of $\mathbb{R}^{2}$. The map $\iota(x,y)=(x,y,f(x,y))$
furnishes a $C^{1}$-parameterization of $\Sigma$ around $p_{0}$. In this
parameterization we have

$G(\iota(x,y))=\frac{1}{\sqrt{1-f_{x}^{2}-f_{y}^{2}}}\begin{bmatrix}f_{x}\\\
f_{y}\\\ 1\end{bmatrix}~{}.$

Since the standard immersion of $\mathbb{H}^{2}$ into $\mathbb{R}^{2,1}$ is
smooth, we have that the functions

$g_{1}(x,y)=\frac{f_{x}}{\sqrt{1-f_{x}^{2}-f_{y}^{2}}},\quad
g_{2}(x,y)=\frac{f_{y}}{\sqrt{1-f_{x}^{2}-f_{y}^{2}}},\quad
g_{3}(x,y)=\frac{1}{\sqrt{1-f_{x}^{2}-f_{y}^{2}}},$

are $C^{1}$. Since $f_{x}=g_{1}/g_{3}$, and $f_{y}=g_{2}/g_{3}$ the result
immediately follows. ∎

######  Lemma 7.8.

Let $\Sigma$ be a convex spacelike surface in $\mathbb{R}^{2,1}$, and let
$u:\mathcal{H}\to\mathbb{R}\cup\\{+\infty\\}$ be its parabolic support
function. Assume that $u$ is $C^{2}$ in a neighborhood $U$ of
$(\mathsf{x}_{0},\mathsf{y}_{0})$, and that $A(\mathsf{x}_{0},\mathsf{y}_{0})$
is positive definite, where $A$ is the matrix defined in (37). Denote by
$\Gamma:U\to\Sigma$ the map defined in Lemma 7.5. Then

  * •

$\Sigma$ is $C^{2}$ around $p_{0}=\Gamma(\mathsf{x}_{0},\mathsf{y}_{0})$;

  * •

The representative matrix of the shape operator $\mathsf{B}(p_{0})$ with
respect to the basis of
$T_{p_{0}}\Sigma=T_{\bar{\zeta}_{0}}\mathbb{H}^{2}=\\{\bar{\zeta}_{\mathsf{x}}(\mathsf{x}_{0},\mathsf{y}_{0}),\bar{\zeta}_{\mathsf{y}}(\mathsf{x}_{0},\mathsf{y}_{0})\\}$
is given by $A^{-1}(\mathsf{x}_{0},\mathsf{y}_{0})$.

In particular its intrinsic curvature on $\Gamma(U)$ is strictly negative and
satisfies

$\begin{split}K(\Gamma(\mathsf{x},\mathsf{y}))^{-1}=-\frac{1}{4}\left((\mathsf{y}u_{\mathsf{x}\mathsf{x}}-u_{\mathsf{y}})(\mathsf{y}u_{\mathsf{y}\mathsf{y}}-u_{\mathsf{y}})-\mathsf{y}^{2}u^{2}_{\mathsf{x}\mathsf{y}}\right)\end{split}$
(38)

###### Proof.

By Lemma 7.5 part (2), the surface $\Sigma$ is strictly convex on $\Gamma(U)$.
Moreover by Lemma 7.6 the function $u\circ\zeta^{-1}$ is strictly convex on
$\zeta(U)$ so, applying Lemma 7.5 part (3), $\Sigma$ is $C^{1}$ over
$\Gamma(U)$. Notice from its definition in Lemma 7.5 part (1) that the map
$\Gamma$ is $C^{1}$, moreover by (36) we have that
$T_{\Gamma(\mathsf{x},\mathsf{y})}\Sigma=T_{\bar{\zeta}(\mathsf{x},\mathsf{y})}\mathbb{H}^{2}$
as subspaces of $\mathbb{R}^{2,1}$.

We claim that $A(\mathsf{x},\mathsf{y})$ is the representative matrix of the
endomorphism
$d_{(\mathsf{x},\mathsf{y})}\Gamma\circ(d_{({\mathsf{x},\mathsf{y}})}\bar{\zeta})^{-1}$
with respect to the basis $\bar{\zeta}_{\mathsf{x}},\bar{\zeta}_{\mathsf{y}}$.
In fact, differentiating the second and third equations defining $\Gamma$,
namely (34) and (35), gives (omitting the evaluation point in the sequel for
the sake of simplicity):

$\begin{split}\langle\Gamma_{\mathsf{x}},\zeta_{\mathsf{x}}\rangle+\langle\Gamma,\zeta_{{\mathsf{x}}{\mathsf{x}}}\rangle&=u_{{\mathsf{x}}{\mathsf{x}}}\\\
\langle\Gamma_{\mathsf{x}},\zeta_{\mathsf{y}}\rangle+\langle\Gamma,\zeta_{{\mathsf{x}}{\mathsf{y}}}\rangle&=u_{{\mathsf{x}}{\mathsf{y}}}\\\
\langle\Gamma_{\mathsf{y}},\zeta_{\mathsf{x}}\rangle+\langle\Gamma,\zeta_{{\mathsf{y}}{\mathsf{x}}}\rangle&=u_{{\mathsf{y}}{\mathsf{x}}}\\\
\langle\Gamma_{\mathsf{y}},\zeta_{\mathsf{y}}\rangle+\langle\Gamma,\zeta_{{\mathsf{y}}{\mathsf{y}}}\rangle&=u_{{\mathsf{y}}{\mathsf{y}}}\\\
\end{split}$ (39)

Verifying from the definition of $\zeta$ that
$\zeta_{{\mathsf{x}}{\mathsf{y}}}=0$ and
$\zeta_{\mathsf{x}\mathsf{x}}=\zeta_{{\mathsf{y}}{\mathsf{y}}}=\zeta_{\mathsf{y}}/{\mathsf{y}}$
and using (35), we see that

$\begin{split}\begin{bmatrix}\langle\Gamma_{\mathsf{x}},\zeta_{\mathsf{x}}\rangle&\langle\Gamma_{\mathsf{x}},\zeta_{\mathsf{y}}\rangle\\\
\langle\Gamma_{\mathsf{y}},\zeta_{\mathsf{x}}\rangle&\langle\Gamma_{\mathsf{y}},\zeta_{\mathsf{y}}\rangle\end{bmatrix}=\begin{bmatrix}u_{{\mathsf{x}}{\mathsf{x}}}-u_{\mathsf{y}}/{\mathsf{y}}&u_{{\mathsf{x}}{\mathsf{y}}}\\\
u_{{\mathsf{y}}{\mathsf{x}}}&u_{{\mathsf{y}}{\mathsf{y}}}-u_{\mathsf{y}}/{\mathsf{y}}\end{bmatrix}=\frac{2}{\mathsf{y}}A=\frac{4}{|\zeta|}A\end{split}$
(40)

Now we use that
$\bar{\zeta}_{\mathsf{x}}\in\frac{\zeta_{\mathsf{x}}}{|\zeta|}+\mathrm{Span}(\zeta)$,
$\bar{\zeta}_{\mathsf{y}}\in\frac{\zeta_{\mathsf{y}}}{|\zeta|}+\mathrm{Span}(\zeta)$,
and that $\langle\Gamma_{x},\zeta\rangle=\langle\Gamma_{y},\zeta\rangle=0$
(which is obtained by combining (33), (34) and (35)), to deduce that:

$\begin{split}\begin{bmatrix}\langle\Gamma_{\mathsf{x}},\bar{\zeta}_{\mathsf{x}}\rangle&\langle\Gamma_{\mathsf{x}},\bar{\zeta}_{\mathsf{y}}\rangle\\\
\langle\Gamma_{\mathsf{y}},\bar{\zeta}_{\mathsf{x}}\rangle&\langle\Gamma_{\mathsf{y}},\bar{\zeta}_{\mathsf{y}}\rangle\end{bmatrix}=\frac{4}{|\zeta|^{2}}A\end{split}$
(41)

Since $\bar{\zeta}_{\mathsf{x}},\bar{\zeta}_{\mathsf{y}}$ is an orthogonal
basis of $T_{\bar{\zeta}}\mathbb{H}^{2}$ and
$||\bar{\zeta}_{\mathsf{x}}||^{2}=||\bar{\zeta}_{\mathsf{y}}||^{2}=4/|\zeta|^{2}$,
the coordinates of a vector of $T_{\bar{\zeta}}\mathbb{H}^{2}$ with respect to
this basis are given by the scalar product with the elements of the basis
multiplied by $|\zeta|^{2}/4$. Since $d\Gamma\circ(d\bar{\zeta})^{-1}$ sends
$\bar{\zeta}_{\mathsf{x}}$ to $\Gamma_{\mathsf{x}}$ and
$\bar{\zeta}_{\mathsf{y}}$ to $\Gamma_{\mathsf{y}}$, from (41) we see that its
representative matrix is given by $A$.

A consequence of this computation is that the differential of $\Gamma$ is
injective under the hypothesis that $A$ is invertible, so $\Gamma$ realizes a
$C^{1}$-diffeomorphism onto its image. Since $G\circ\Gamma=\bar{\zeta}$ we see
that $G=\bar{\zeta}\circ\Gamma^{-1}$, so that the restriction of $G$ to
$\Gamma(U)$ is $C^{1}$. We deduce that $\Gamma(U)$ is $C^{2}$ by Lemma 7.7.

Finally using that
$dG=d\bar{\zeta}\circ(d\Gamma)^{-1}=(d\Gamma\circ(d\bar{\zeta})^{-1})^{-1}$ we
conclude that the representative matrix of $dG$ with respect to the basis
$\bar{\zeta}_{\mathsf{x}},\bar{\zeta}_{\mathsf{y}}$ is $A^{-1}$. By the Gauss
equation $K(\Gamma(\mathsf{x},\mathsf{y}))=\det A^{-1}$ and the result
follows. ∎

Before finally giving the formula for our function $u$, we need two additional
lemmas in order to deal with the conditions (3) and (4) in the list of Section
7.1. The first lemma gives a formula for the extension of $u$ to $\infty$ and
a condition for incompleteness of the corresponding surface under the
additional assumption that $u$ is symmetric about the $\mathsf{y}$ axis.

######  Lemma 7.9.

Suppose that $u:\mathcal{H}\to{\mathbb{R}}$ is the parabolic support function
of a convex entire spacelike surface $\Sigma$, and that
$u(\mathsf{x},\mathsf{y})=u(-\mathsf{x},\mathsf{y})$. Suppose furthermore that
the restriction of $u$ to the $\mathsf{y}$ axis is $C^{2}$ and strictly
convex, and satisfies

$\int_{1}^{\infty}\left(u_{\mathsf{y}\mathsf{y}}-\frac{u_{\mathsf{y}}}{\mathsf{y}}\right)d\mathsf{y}<\infty~{}.$

Then $\Sigma$ is incomplete at $\theta_{0}=\pi$ and

$\overline{u}(\infty)=\sup_{(0,\mathsf{y})\in\mathcal{H}}\frac{u_{\mathsf{y}}(0,\mathsf{y})}{2\mathsf{y}}~{}.$

###### Proof.

Because of the bijection established in Corollary 2.28, $\Sigma$ also has the
corresponding reflection symmetry, across the timelike plane $y=0$ spanned by
the vectors $\vec{0}$ and $\vec{\pi}$. Since $\Sigma$ is convex, it follows
that its null support planes in these directions contains the null support
lines of its intersection with this timelike plane. Since $u$ is $C^{2}$ and
strictly convex on the $\mathsf{y}$ axis, we can parametrize the intersection
of $\Sigma$ with this plane by $\Gamma(0,\mathsf{y})$ defined by equations
$(\ref{eqn: def v 1},~{}\ref{eqn: def v 2},~{}\ref{eqn: def v 3})$. Then since
$\overline{u}(\infty)$ is the value of the elliptic support function of
$\Sigma$ at ${{\theta_{0}}=}{\pi}$ by Proposition 2.27 and since
$\vec{\pi}=\zeta_{\mathsf{y}}/2\mathsf{y}$, we have

$\begin{split}\overline{u}(\infty)&=\sup_{\mathsf{y}}\langle\Gamma(0,\mathsf{y}),\vec{\pi}\rangle\\\
&=\sup_{\mathsf{y}}\langle\Gamma(0,\mathsf{y}),\frac{\zeta_{\mathsf{y}}}{2\mathsf{y}}\rangle\\\
&=\sup_{\mathsf{y}}\frac{u_{\mathsf{y}}(0,\mathsf{y})}{2\mathsf{y}}~{}.\end{split}$
(42)

For the incompleteness part of the lemma, it suffices to show that the length
of $\Gamma(0,\mathsf{y})$ is finite as $\mathsf{y}$ approaches infinity, since
we have already argued that this curve approaches the null support plane of
$\Sigma$ in the direction $\vec{\pi}$. By the reflection symmetry,
$\Gamma_{\mathsf{y}}$ is a multiple of $\overline{\zeta}_{\mathsf{y}}$, so its
length is
$\langle\Gamma_{\mathsf{y}},\overline{\zeta}_{\mathsf{y}}\rangle/|\overline{\zeta}_{\mathsf{y}}|$.
From equation (41), which clearly remains true in the $\mathsf{y}$ direction
even though $u$ is not differentiable in the $\mathsf{x}$ direction, this is
equal to

$\frac{1}{2}\left(u_{\mathsf{yy}}-\frac{u_{\mathsf{y}}}{\mathsf{y}}\right).$

Its integral is the length of the curve $\Gamma(0,\mathsf{y})$, so if the
integral from one to infinity is finite, then the surface $\Sigma$ is
incomplete at $\pi$. ∎

Finally, using the relationship (10) between the elliptic and parabolic null
support functions, we can translate the conditions on $\phi_{+}$ to conditions
on the parabolic null support function $\psi_{+}$ of $\Sigma_{+}$.

######  Lemma 7.10.

Let $s$ be a homogeneous function, and let $\phi$ and $\psi$ be respectively
its elliptic dehomogenization and parabolic dehomogenization, with point at
infinity $\pi$. If

$\begin{split}\psi(\mathsf{x})\leq\epsilon|\mathsf{x}|^{2-\alpha}\end{split}$
(43)

in the complement of a compact set of ${\mathbb{R}}$, then

$\phi(\theta)\leq\epsilon|\theta-\pi|^{\alpha}$

in a neighbourhood of $\pi$.

###### Proof.

From Proposition 2.27, we have $\mathsf{x}=\tan(\theta/2)$ and

$\phi(\theta)=\frac{\psi(\mathsf{x})}{1+\mathsf{x}^{2}}\leq\epsilon\frac{|\mathsf{x}|^{2-\alpha}}{1+\mathsf{x}^{2}}<\epsilon|\mathsf{x}|^{-\alpha}~{}.$

Using $1/\mathsf{x}=\tan(\pi/2-\theta/2)$ and a linear bound
$|t|\geq|\tan(t/2)|$ for $t$ sufficiently small, we get
$\phi(\theta)<\epsilon|\theta-\pi|^{\alpha}$ in a neighbourhood of $\pi$. ∎

###  7.3. The explicit surface $\Sigma_{+}$

We are now ready to present the formula for the parabolic support function $u$
of $\Sigma_{+}$. For constants $0<\beta,\gamma<1$ and $M>0$ to be determined,
let

$u(\mathsf{x},\mathsf{y})=-M\mathsf{y}^{\beta}+\epsilon|\mathsf{x}|^{2-\alpha}(1+\mathsf{y})^{-\gamma}~{}.$
(44)

Clearly $u(\mathsf{x},0)$ satisfies (43). Since $u$ is symmetric about the
$\mathsf{y}$ axis and its restriction to the $\mathsf{y}$ axis is $C^{2}$ and
strictly convex, we can use Lemma 7.9 to check incompleteness and the value of
$\overline{u}(\infty)$, once we know that $u$ is convex. We also remark that
$u$ is, inconveniently, not $C^{2}$ on the $\mathsf{y}$-axis.

Let $F$ be the operator

$\begin{split}F(u)=-4K^{-1}=(\mathsf{y}u_{\mathsf{x}\mathsf{x}}-u_{\mathsf{y}})(\mathsf{y}u_{\mathsf{y}\mathsf{y}}-u_{\mathsf{y}})-\mathsf{y}^{2}u^{2}_{\mathsf{x}\mathsf{y}}\end{split}$
(45)

so that $-4/F(u)$ is the curvature of the surface constructed in Lemma 7.8.

######  Lemma 7.11.

Let $u$ be the function defined in (44), and $F$ the operator defined in (45).

  * •

If $\gamma<1-\alpha$, then $F(u)>0$ for all points $(\mathsf{x},\mathsf{y})$
with $\mathsf{x}\neq 0$.

  * •

If $\gamma\leq\beta(1-\alpha)$, then for any $C>0$ there exists $M>0$ such
that $F(u)\geq 4/C$ for all points $(\mathsf{x},\mathsf{y})$ with
$\mathsf{x}\neq 0$.

###### Proof.

Let us compute

$u_{\mathsf{x}}=\epsilon(2-\alpha)\mathsf{x}|\mathsf{x}|^{-\alpha}(1+\mathsf{y})^{-\gamma}\qquad
u_{\mathsf{y}}=-M\beta\mathsf{y}^{-1+\beta}-\epsilon\gamma|\mathsf{x}|^{2-\alpha}(1+\mathsf{y})^{-1-\gamma}$
(46)

and

$\begin{split}u_{\mathsf{x}\mathsf{x}}&=\epsilon(2-\alpha)(1-\alpha)|\mathsf{x}|^{-\alpha}(1+\mathsf{y})^{-\gamma}\\\
u_{\mathsf{x}\mathsf{y}}&=-\epsilon(2-\alpha)\gamma\mathsf{x}|\mathsf{x}|^{-\alpha}(1+\mathsf{y})^{-1-\gamma}\\\
u_{\mathsf{y}\mathsf{y}}&=M\beta(1-\beta)\mathsf{y}^{-2+\beta}+\epsilon\gamma(1+\gamma)|\mathsf{x}|^{2-\alpha}(1+\mathsf{y})^{-2-\gamma}~{}.\end{split}$
(47)

Hence we have:

$\begin{split}F(u)=&(\mathsf{y}{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}u_{\mathsf{x}\mathsf{x}}}-{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}u_{\mathsf{y}}})(\mathsf{y}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}u_{\mathsf{y}\mathsf{y}}}-{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}u_{\mathsf{y}}})-\mathsf{y}^{2}{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}u_{\mathsf{x}\mathsf{y}}}^{2}\\\
=&\left(\mathsf{y}{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}\epsilon(2-\alpha)(1-\alpha)|\mathsf{x}|^{-\alpha}(1+\mathsf{y})^{-\gamma}}+{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}M\beta\mathsf{y}^{-1+\beta}+\epsilon\gamma|\mathsf{x}|^{2-\alpha}(1+\mathsf{y})^{-1-\gamma}}\right)\cdot\\\
&\left(\mathsf{y}({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}M\beta(1-\beta)\mathsf{y}^{-2+\beta}+\epsilon\gamma(1+\gamma)|\mathsf{x}|^{2-\alpha}(1+\mathsf{y})^{-2-\gamma}})+\right.\\\
&\qquad\left.{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}M\beta\mathsf{y}^{-1+\beta}+\epsilon\gamma|\mathsf{x}|^{2-\alpha}(\mathsf{y}+1)^{-1-\gamma}}\right)\\\
&-\mathsf{y}^{2}\left({\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}-\epsilon(2-\alpha)\gamma\mathsf{x}|\mathsf{x}|^{-\alpha}(1+\mathsf{y})^{-1-\gamma}}\right)^{2}\end{split}$

We drop the final term in each of the two parenthetical expressions and
combine two equivalent terms in the second, and recolor:

$\begin{split}F(u)\geq&\left({\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathsf{y}\epsilon(2-\alpha)(1-\alpha)|\mathsf{x}|^{-\alpha}(\mathsf{y}+1)^{-\gamma}}+{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}M\beta\mathsf{y}^{-1+\beta}}\right)\cdot\\\
&\left({\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}M\beta(2-\beta)\mathsf{y}^{-1+\beta}}+{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathsf{y}\epsilon\gamma(1+\gamma)|\mathsf{x}|^{2-\alpha}(1+\mathsf{y})^{-2-\gamma}}\right)\\\
&-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathsf{y}^{2}\left(\epsilon(2-\alpha)\gamma|\mathsf{x}|^{1-\alpha}(1+\mathsf{y})^{-1-\gamma}\right)^{2}}\end{split}$

and now expand, dropping one of the four terms of the product:

$\begin{split}F(u)\geq&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}((1-\alpha)(1+\gamma)-(2-\alpha)\gamma)\left(\mathsf{y}^{2}\epsilon^{2}(2-\alpha)\gamma|\mathsf{x}|^{2-2\alpha}(1+\mathsf{y})^{-2-2\gamma}\right)}\\\
&+{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}\epsilon(2-\alpha)(1-\alpha)M\beta(2-\beta)|\mathsf{x}|^{-\alpha}(1+\mathsf{y})^{-\gamma}\mathsf{y}^{\beta}}\\\
&+{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}M^{2}\beta^{2}(2-\beta)\mathsf{y}^{-2+2\beta}}\end{split}$

As long as $\gamma<1-\alpha$, the first term is positive, and hence $F(u)$ is
strictly positive. Assuming this, we now try to refine our choice of constants
so that $F(u)\geq 4/C$. First of all, if $\mathsf{y}\leq 1$, the last term is
at least $M^{2}\beta^{2}$, so for any $\beta$ we can choose $M$ to make this
at least $4/C$. Hence, so long as we are free to choose $M$ sufficiently
large, we need only bound $F(u)$ from below under the assumption that
$\mathsf{y}\geq 1$. In this case,
$(1+\mathsf{y})^{-1}\geq\frac{1}{2}\mathsf{y}^{-1}$, and so for some
$\delta(\epsilon,\alpha,\gamma)$ small we have

$\begin{split}F(u)\geq&\delta\left(\frac{\alpha}{2-\alpha}|\mathsf{x}|^{2-2\alpha}\mathsf{y}^{-2\gamma}+\frac{2-2\alpha}{2-\alpha}M\beta|\mathsf{x}|^{-\alpha}\mathsf{y}^{\beta-\gamma}\right)\\\
\geq&\delta\mathsf{y}^{(-2\gamma\alpha+2(1-\alpha)(\beta-\gamma))/(2-\alpha)}(M\beta)^{(2-2\alpha)/(2-\alpha)}\end{split}$

where we have used the weighted arithmetic-geometric inequality $\lambda
a+(1-\lambda)b\geq a^{\lambda}b^{1-\lambda}$ in the last line. If we choose
$\gamma$ such that

$-2\gamma\alpha+2(1-\alpha)(\beta-\gamma)\geq 0$

then for $M$ sufficiently large, $F(u)\geq 4/C$ for all $\mathsf{y}\geq 1$. A
little algebra shows that it suffices to take $\gamma\leq\beta(1-\alpha)$. ∎

######  Lemma 7.12.

If $\gamma<1-\alpha$, the function $u$ defined in (44) is the parabolic
support function of a strictly convex entire spacelike surface $\Sigma_{+}$.

The proof below shows that $\Sigma_{+}$ is a $C^{1}$ surface, as in the
definition of spacelike surfaces. We will prove later (Lemma 7.15) that
$\Sigma_{+}$ is in fact $C^{2}$, but this will require a little additional
work.

###### Proof.

By Corollary 2.28, in order to prove the existence of an achronal surface
$\Sigma_{+}$ with parabolic support function $u$, we need to check that the
function $u\circ\zeta^{-1}$ is a convex function of the intersection of the
affine plane $P_{0}$ of equation $x+z=2$ with the future cone $J^{+}$. Since
the intersection is convex, it is sufficient to prove that $u\circ\zeta^{-1}$
is locally convex around any point $(x,y,z)\in P_{0}\cap I^{+}$; then the
function $u$ defined in (44), which is clearly continuous up to the boundary
of $\mathcal{H}$, is the unique closed convex extension to $P_{0}\cap J^{+}$.

The strict convexity of $u\circ\zeta^{-1}$ at $(x,y,z)\in P_{0}\cap I^{+}$ for
$y\neq 0$ follows from Lemma 7.11, which implies that the matrix (37) is
positive definite, and Lemma 7.6.

To prove the strict convexity at points of the form $(x,0,x+2)$, notice that
$u(\mathsf{x},\mathsf{y})\geq u_{0}(\mathsf{x},\mathsf{y})$ where
$u_{0}(\mathsf{x},\mathsf{y})=-M\mathsf{y}^{\beta}$, with $u=u_{0}$ along
$\mathsf{x}=0$. We observe that $F(u_{0})>0$ so $u_{0}\circ\zeta^{-1}$ is
strictly convex again by Lemma 7.6. We conclude that $u\circ\zeta^{-1}$ is
strictly convex at the points of $P_{0}\cap I^{+}$ with $y=0$ as well.

If $\Gamma:\mathcal{H}\to\Sigma_{+}$ is the map defined in Lemma 7.5, we
conclude that $\Sigma_{+}$ is $C^{1}$ and spacelike on the open subset
$\Gamma(\mathcal{H})$, which consists of points admitting spacelike support
planes. In order to prove that $\Gamma$ is surjective, we show that
$\Sigma_{+}$ is acausal as a consequence of Lemma 7.4. Suppose $f$ is a
polynomial of the given form with $f\leq u$ on $\overline{\mathcal{H}}$. Since
$u$ has a singularity like $-\mathsf{y}^{\beta}$ near $\mathsf{y}=0$ and $f$
is smooth, we must certainly have $f<u$ on ${\mathbb{R}}$. Now at $\infty$,
one checks $\overline{f}(\infty)=c$, and by Lemma 7.9, we have
$\overline{u}(\infty)=-\lim_{\mathsf{y}\to\infty}M\beta\mathsf{y}^{\beta-2}=0$.
So if these were equal, then $f$ is independent of $\mathsf{y}$, so cannot be
less than or equal to $-M\mathsf{y}^{\beta}$ on the $\mathsf{y}$ axis. We
conclude that $f<u$ on ${\mathbb{R}}\cup\\{\infty\\}$, and so by Lemma 7.4
that $\Sigma_{+}$ is acausal. Hence by Corollary 2.9, every point of
$\Sigma_{+}$ has a spacelike support plane, that is, $\Gamma$ is surjective.
This concludes the proof. ∎

By symmetry, $\Gamma(\\{\mathsf{x}=0\\})=\\{y=0\\}$ and so
$\Gamma(\mathcal{H}\setminus\\{\mathsf{x}=0\\})=\Sigma_{+}\setminus\\{y=0\\}$.
Since $u$ is $C^{2}$ on $\mathcal{H}\setminus\\{\mathsf{x}=0\\}$ and $F(u)>0$,
Lemma 7.8 implies that $\Sigma_{+}\setminus\\{y=0\\}$ is $C^{2}$. We will
prove now that $\Sigma_{+}$ is in fact $C^{2}$ everywhere. First we prove the
following lemma that roughly states that the shape operator of $\Sigma_{+}$
extends everywhere. In order to make this statement more precise, notice that
$T\Sigma_{+}\to\Sigma_{+}$ is a continuous vector bundle and a priori the
shape operator $B$ is a section of $(T\Sigma_{+})^{*}\otimes(T\Sigma_{+})$
defined only on $\Sigma_{+}\setminus\\{y=0\\}$.

######  Lemma 7.13.

If $\gamma<1-\alpha$, the shape operator continuously extends to a section of
$(T\Sigma_{+})^{*}\otimes(T\Sigma_{+})$ on the whole $\Sigma_{+}$.

###### Proof.

We consider the continuous parameterization $\Gamma:\mathcal{H}\to\Sigma_{+}$
given by the inverse of the Gauss map. Notice that for any
$(\mathsf{x},\mathsf{y})\in\mathcal{H}$ the vectors
$\bar{\zeta}_{\mathsf{x}}(\mathsf{x},\mathsf{y})$,
$\bar{\zeta}_{\mathsf{y}}(\mathsf{x},\mathsf{y})$ form a basis of
$T_{\Gamma(\mathsf{x},\mathsf{y})}\Sigma_{+}=T_{\bar{\zeta}(\mathsf{x},\mathsf{y})}\mathbb{H}^{2}$.
In this way a continuous frame of $T\Sigma_{+}$ is defined everywhere. By
Lemma 7.8 the representative matrix of the shape operator $B$ with respect to
this frame at a point $\Gamma(\mathsf{x},\mathsf{y})$ is given by
$A(\mathsf{x},\mathsf{y})^{-1}$ where

$A(\mathsf{x},\mathsf{y})=\frac{1}{2}\begin{bmatrix}\mathsf{y}u_{\mathsf{x}\mathsf{x}}-u_{\mathsf{y}}&\mathsf{y}u_{\mathsf{x}\mathsf{y}}\\\
\mathsf{y}u_{\mathsf{x}\mathsf{y}}&\mathsf{y}u_{\mathsf{y}\mathsf{y}}-u_{\mathsf{y}}\end{bmatrix}(\mathsf{x},\mathsf{y})$
(48)

So in order to conclude it is sufficient to check that the entries of the
inverse of (48) extend at $\mathsf{x}=0$. From (46) and (47), we observe that
$u_{\mathsf{x}\mathsf{y}}(\mathsf{x},\mathsf{y})\rightarrow 0$ as
$\mathsf{x}\to 0$, while
$u_{\mathsf{y}}(\mathsf{x},\mathsf{y})\to-M\beta\mathsf{y}^{-1+\beta}$ and
$u_{\mathsf{y}\mathsf{y}}(\mathsf{x},\mathsf{y})\to
M\beta(1-\beta)\mathsf{y}^{-2+\beta}$, which implies that
$\mathsf{y}u_{\mathsf{y}\mathsf{y}}-u_{\mathsf{y}}\to
M\beta(2-\beta)\mathsf{y}^{-1+\beta}$. On the other hand
$u_{\mathsf{x}\mathsf{x}}(\mathsf{x},\mathsf{y})\sim|\mathsf{x}|^{-\alpha}$
diverges as $\mathsf{x}\to 0$, so we simply see that the matrix
$A^{-1}(\mathsf{x},\mathsf{y})$ converges to the matrix

$\begin{bmatrix}0&0\\\
0&\frac{2}{M\beta(2-\beta)\mathsf{y}^{-1+\beta}}\end{bmatrix}\,.$

as $(\mathsf{x},\mathsf{y})\to(0,\mathsf{y})$. ∎

We will need to use an elementary lemma for which we give a short proof for
the convenience of the reader.

######  Lemma 7.14.

Let $E=\\{(x,y)\in\mathbb{R}^{2}\,|\,y=0\\}$ be the $x$-axis, and $f\in
C^{0}(\mathbb{R}^{2})\cap C^{1}(\mathbb{R}^{2}\setminus E)$. Let us assume
that $df$ extends to a continuous $1$-form $\omega=h(x,y)dx+k(x,y)dy$ on
$\mathbb{R}^{2}$. Then $f\in C^{1}(\mathbb{R}^{2})$ and $df=\omega$
everywhere.

###### Proof.

Let us consider the continuous function on $\mathbb{R}^{2}$

$K(x,y)=f(x,0)+\int_{0}^{y}k(x,t)dt\,$

so that $K_{y}(x,y)=k(x,y)$ for all $(x,y)\in\mathbb{R}^{2}$. Since
$(K-f)_{y}$ vanishes on $\mathbb{R}^{2}\setminus E$, the function $K-f$ is
constant on the half-lines of the form $\\{x\\}\times(0,+\infty)$ and on the
half-lines of the form $\\{x\\}\times(-\infty,0)$. Since $K-f$ is continuous
and vanishes on $E$, it vanishes everywhere, which implies that $f$ admits
$y$-derivative everywhere, that coincide with $k(x,y)$.

Similarly let us consider the function

$H(x,y)=f(0,y)+\int_{0}^{x}h(s,y)ds\,$

so that $H_{x}(x,y)=h(x,y)$ everywhere. We have that $H-f$ vanishes on
$\mathbb{R}^{2}\setminus E$. Since $H-f$ is continuous on $\mathbb{R}^{2}$ we
conclude that $f$ coincides with $H$ everywhere. ∎

######  Lemma 7.15.

If $\gamma<1-\alpha$, the function $u$ given by (44) is the parabolic support
function of a $C^{2}$ convex entire spacelike surface $\Sigma_{+}$. If
moreover $\gamma\leq\beta(1-\alpha)$ and $M$ is chosen as in the second point
of Lemma 7.11, then $\Sigma_{+}$ has curvature greater than or equal to $-C$.

###### Proof.

Let $f:\mathbb{R}^{2}\to\mathbb{R}$ be the $C^{1}$-function whose graph is
$\Sigma_{+}$, and consider the corresponding $C^{1}$-parametrization
$\sigma(x,y)=(x,y,f(x,y))$. Notice that
$(G\circ\sigma)_{x}=B(\sigma(x,y))\cdot\sigma_{x}$. Since by Lemma 7.13 the
shape operator $B$ extends to a continuous section of
$(T\Sigma_{+})^{*}\otimes(T\Sigma_{+})$, $\sigma_{x}$ extends to $y=0$ we
conclude that the function $(x,y)\to(G(\sigma(x,y)))_{x}$ continuously extends
at $y=0$. Similarly we have that $(x,y)\to(G(\sigma(x,y))_{y}$ continuously
extends at $y=0$.

Lemma 7.14 implies that $G\circ\sigma:\mathbb{R}^{2}\to\mathbb{H}^{2}$ is
$C^{1}$. Since $\sigma$ is a $C^{1}$-parametrization of $\Sigma_{+}$, the map
$G:\Sigma_{+}\to\mathbb{H}^{2}$ is $C^{1}$ as well, so by Lemma 7.7,
$\Sigma_{+}$ is $C^{2}$.

Finally, under the choices of constants as in the second point of Lemma 7.11
we have $F(u)\geq 4/C$. Therefore the curvature of $\Sigma_{+}$ is greater
than or equal to $-C$ away from its intersection with the $y=0$ plane, and
thus it is greater than or equal to $-C$ everywhere by continuity. ∎

###  7.4. Proof of Theorem C

At this point, we have assembled all the necessary components for a proof of
the main theorem of this section, and so we end by simply putting them in
order.

###### Proof of Theorem C.

As already observed at the beginning of Section 7.1 (Remark 7.3), we can
assume $\theta_{0}=\pi$ and $\phi(\pi)=0$. Let $\Sigma_{+}$ be the surface
whose existence is given by Lemma 7.15, where $\gamma\leq\beta(1-\alpha)$ and
the constant $M$ is chosen as in the second point of Lemma 7.11. From (46) and
(47) we find
$u_{\mathsf{y}\mathsf{y}}(0,\mathsf{y})-u_{\mathsf{y}}(0,\mathsf{y})/\mathsf{y}=M\beta(2-\beta)\mathsf{y}^{-2+\beta}$;
moreover we have
$u_{\mathsf{y}}(0,\mathsf{y})/2\mathsf{y}=-M\beta\mathsf{y}^{-2+\beta}/2$.
Hence Lemma 7.9 shows that $\Sigma_{+}$ is incomplete at $\pi$ and its
elliptic null support function $\phi_{+}$ has value 0 at $\pi$. Moreover,
since its parabolic null support function is

$\psi_{+}(\mathsf{x})=\epsilon|\mathsf{x}|^{2-\alpha}~{},$

by Lemma 7.10 and the condition (Inc), we have that
$\phi_{+}(\theta)\leq\epsilon|\theta-\pi|^{\alpha}<\phi(\theta)$ for $\theta$
in a neighbourhood $U$ of $\pi$. In addition, we have
$\phi_{+}(\pi)=\phi(\pi)=0$.

Analogously to the proof of Corollary 7.1, we remark that we are not yet in
the hypothesis of Corollary G, since we need the inequality $\phi_{+}\leq\phi$
to hold globally on $\mathbb{S}^{1}$. This issue is solved by replacing
$\Sigma_{+}$ with a translate in the direction of $\vec{\pi}$, which has the
effect of changing its parabolic null support function $\psi_{+}$ by adding
the constant $-2c$ and leaving the value of the elliptic null support function
$\phi_{+}$ at $\pi$ unchanged. Since $\psi_{+}$ is continuous, hence bounded
above, and $\psi$ is lower semicontinuous, hence bounded below, by choosing
$c$ large enough, the translate $\Sigma^{\prime}_{+}$ of $\Sigma_{+}$
satisfies $\psi^{\prime}_{+}<\psi$, and therefore $\phi_{+}^{\prime}\leq\phi$
on the whole $\mathbb{S}^{1}$. By Corollary G applied to $\Sigma$ and
$\Sigma_{+}^{\prime}$, we conclude that $\Sigma$ is incomplete at $\pi$. ∎

##  8\. Incompleteness II: one-sided superlogarithmic condition

The goal of this section is to prove Theorem D, which we state here as usual
in a slightly stronger version with respect to the introduction, using
directional incompleteness.

######  Theorem D (One-sided superlogarithmic condition – local version).

Let $\phi:\mathbb{S}^{1}\to{\mathbb{R}}\cup\\{+\infty\\}$ be lower
semicontinuous and finite on at least three points. Suppose
$\theta_{0}\in\mathbb{S}^{1}$ is such that $\phi(\theta_{0})<+\infty$ and
there exist a neighborhood $U$ of $\theta_{0}$ and a constant $\epsilon>0$
such that

$\begin{cases}\phi(\theta)=+\infty&\text{ if }\theta\text{ is on one side of
}\theta_{0}\\\
\phi(\theta)\geq\phi(\theta_{0})+\epsilon|(\theta-\theta_{0})\log|\theta-\theta_{0}||&\text{
if }\theta\text{ is on the other one side of }\theta_{0}.\end{cases}$ (Inc’)

for every $\theta\in U$.

If $\Sigma$ is a convex entire spacelike surface in ${\mathbb{R}}^{2,1}$ with
null support function $\phi$ and curvature bounded above by a negative
constant, then $\Sigma$ is incomplete at $\theta_{0}$.

To prove the incompleteness criterion given in Theorem D, we will construct an
incomplete entire hyperbolic surface $\Sigma_{\lambda}$ (depending on a real
parameter $\lambda)$ invariant under a one-parameter glide hyperbolic group,
study its null support function, and apply the comparison principle (Corollary
G).

###  8.1. The invariant surface $\Sigma_{\lambda}$

Let us fix $\lambda\in{\mathbb{R}}$, and consider the one-parameter group
$A^{\lambda}=\\{A^{\lambda}_{s}\\}_{s\in\mathbb{R}}$ of affine isometries of
Minkowski space defined by

$A^{\lambda}_{s}\begin{bmatrix}x_{1}\\\ x_{2}\\\
x_{3}\end{bmatrix}=\begin{bmatrix}x_{1}+\lambda s\\\
\cosh(s)x_{2}+\sinh(s)x_{3}\\\ \sinh(s)x_{2}+\cosh(s)x_{3}\end{bmatrix}~{}.$

Let us define the map
$X^{\lambda}:(0,+\infty)\times{\mathbb{R}}\to{\mathbb{R}}^{2,1}$ given by

$X^{\lambda}(t,s)=\begin{bmatrix}\sqrt{1+\lambda^{2}}(t-\coth(t))+\lambda s\\\
\sqrt{1+\lambda^{2}}{\sinh(s)}/{\sinh(t)}\\\
\sqrt{1+\lambda^{2}}{\cosh(s)}/{\sinh(t)}\end{bmatrix}$

We have that

$A^{\lambda}_{s_{0}}X^{\lambda}(t,s)=X^{\lambda}(t,s+s_{0})~{}.$

######  Remark 8.1.

When $\lambda=0$ in the above expression, we recover the embedding $X=X^{0}$
of the semitrough that we described in Example 2.29. We have already observed
there that the semitrough is complete. We will see that $X^{\lambda}$ is a
proper spacelike immersion, whose image $\Sigma_{\lambda}$ is an entire
spacelike surface of constant curvature $-1$ which is complete if and only if
$\lambda=0$. The incompleteness of $\Sigma_{\lambda}$ for $\lambda\neq 0$ is
the fundamental property that we will apply to prove Theorem D.

######  Remark 8.2.

Observe that $\Sigma_{-\lambda}=R(\Sigma_{\lambda})$, where
$R=\mathrm{diag}(1,-1,1)$ is the reflection in the plane $y=0$. Indeed, we
immediately see from the definition that $X^{-\lambda}(t,s)=R\circ
X^{\lambda}(t,-s)$.

######  Lemma 8.3.

For any $\lambda\in{\mathbb{R}}$, the image of $X^{\lambda}$ is an entire
spacelike surface.

###### Proof.

An explicit computation shows that

$\partial_{t}X^{\lambda}(t,s)=\sqrt{1+\lambda^{2}}\frac{\cosh(t)}{\sinh^{2}(t)}\begin{bmatrix}\cosh(t)\\\
-\sinh(s)\\\
-\cosh(s)\end{bmatrix}\qquad\partial_{s}X^{\lambda}(t,s)=\frac{1}{\sinh(t)}\begin{bmatrix}\lambda\sinh(t)\\\
\sqrt{1+\lambda^{2}}\cosh(s)\\\ \sqrt{1+\lambda^{2}}\sinh(s)\end{bmatrix}$

so that the coefficients of the pull-back of the Minkowski product are

$\displaystyle E$
$\displaystyle=\langle\partial_{t}X^{\lambda},\partial_{t}X^{\lambda}\rangle=(1+\lambda^{2})\coth^{2}(t)$
$\displaystyle F$
$\displaystyle=\langle\partial_{t}X^{\lambda},\partial_{s}X^{\lambda}\rangle=\lambda\sqrt{1+\lambda^{2}}\coth^{2}(t)$
$\displaystyle G$
$\displaystyle=\langle\partial_{s}X^{\lambda},\partial_{s}X^{\lambda}\rangle=\lambda^{2}\coth^{2}(t)+\frac{1}{\sinh^{2}(t)}\,.$

Hence we get

$EG-F^{2}=(1+\lambda^{2})\frac{\coth^{2}(t)}{\sinh^{2}(t)}>0$ (49)

which shows that the pull-back metric is Riemannian, and the map $X^{\lambda}$
is a spacelike immersion.

Let us prove that $X^{\lambda}$ is proper. By contradiction let us assume
there exists a diverging sequence
$(t_{n},s_{n})\in\mathbb{R}_{+}\times\mathbb{R}$ such that
$X^{\lambda}(t_{n},s_{n})$ is converging in $\mathbb{R}^{2,1}$. First let us
assume that $s_{n}\to+\infty$. Since $X^{\lambda}(t_{n},s_{n})$ stays in a
compact region of ${\mathbb{R}}^{2,1}$, from boundedness of the second or
third coordinate we deduce that $e^{s_{n}-t_{n}}$ is bounded. That is, that
there exists $C>0$ such that $t_{n}>s_{n}-C$. Therefore $\coth(t_{n})$ is
bounded above by some constant $C^{\prime}$, and the first component of
$X_{\lambda}(t_{n},s_{n})$ is larger than
$(\sqrt{1+\lambda^{2}}+\lambda)s_{n}-(C+C^{\prime})\sqrt{1+\lambda^{2}}$ for
$n$ sufficiently large. Since this quantity tends to $+\infty$, we get a
contradiction.

Similarly, if $s_{n}\to-\infty$, from the third component we obtain that
$e^{-s_{n}-t_{n}}$ is bounded, hence $t_{n}>-s_{n}-C$, and the third component
would be larger than
$(\sqrt{1+\lambda^{2}}-\lambda)(-s_{n})-(C+C^{\prime})\sqrt{1+\lambda^{2}}$,
which is impossible.

Finally let us consider the case where $s_{n}$ is bounded and $t_{n}$ is
diverging in ${\mathbb{R}}_{+}$. If $t_{n}\to 0$ then the third component of
$X(t_{n},s_{n})$ diverges. On the other hand, if $t_{n}\to+\infty$ then the
first component diverges.

We have thus shown that $X^{\lambda}$ is a proper spacelike immersion. By
[BSS19, Proposition 1.10], $X^{\lambda}$ is a proper embedding, and its image
is an entire spacelike surface. This concludes the proof. ∎

Let us call $\Sigma_{\lambda}$ the image of the embedding $X^{\lambda}$, which
is an entire spacelike surface. We now show that it is hyperbolic:

######  Lemma 8.4.

For any $\lambda\in{\mathbb{R}}$, the surface $\Sigma_{\lambda}$ has constant
curvature $-1$.

###### Proof.

A direct computation shows that the future pointing unit normal vector of
$\Sigma_{\lambda}$ at $X^{\lambda}(t,s)$ is

$N(t,s)=\begin{bmatrix}{-\sqrt{1+\lambda^{2}}}/{\sinh(t)}\\\
\sqrt{1+\lambda^{2}}\coth(t)\sinh(s)+\lambda\cosh(s)\\\
\sqrt{1+\lambda^{2}}\coth(t)\cosh(s)+\lambda\sinh(s)\end{bmatrix}$ (50)

On the other hand the Hessian of $X^{\lambda}$ is the following
${\mathbb{R}}^{2,1}$-valued symmetric form:

$\begin{split}D^{2}X_{\lambda}&=(\partial_{tt}X_{\lambda})dt^{2}+2(\partial_{ts}X_{\lambda})dtds+(\partial_{ss}X_{\lambda})ds^{2}=\\\
&=\sqrt{1+\lambda^{2}}\left(\begin{bmatrix}-2\frac{\coth(t)}{\sinh^{2}(t)}\\\
\frac{1+\cosh^{2}(t)}{\sinh^{3}(t)}\sinh(s)\\\
\frac{1+\cosh^{2}(t)}{\sinh^{3}(t)}\cosh(s)\end{bmatrix}dt^{2}-2\begin{bmatrix}0\\\
\frac{\coth(t)}{\sinh(t)}\cosh(s)\\\
\frac{\coth(t)}{\sinh(t)}\sinh(s)\end{bmatrix}dtds+\begin{bmatrix}0\\\
\frac{1}{\sinh(t)}\sinh(s)\\\
\frac{1}{\sinh(t)}\cosh(s)\end{bmatrix}ds^{2}\right)\end{split}$

Recall that the second fundamental form is defined as
$\mathrm{I}\mathrm{I}=-\langle D^{2}X_{\lambda},N\rangle$. It follows that the
coefficients of the second fundamental form are given by

$\displaystyle e$ $\displaystyle=g=(1+\lambda^{2})\frac{\coth(t)}{\sinh(t)}$
$\displaystyle f$
$\displaystyle=\lambda\sqrt{1+\lambda^{2}}\frac{\coth(t)}{\sinh(t)}~{}.$

We have

$eg-f^{2}=\frac{\coth^{2}(t)}{\sinh^{2}(t)}[(1+\lambda^{2})^{2}-\lambda^{2}(1+\lambda^{2})]=(1+\lambda^{2})\frac{\coth^{2}(t)}{\sinh^{2}(t)}$

By the Gauss equation and Equation (49) we have that the curvature of $\Sigma$
is given by

$K=-\frac{eg-f^{2}}{EG-F^{2}}=-1$

as claimed. ∎

By Proposition 2.10, $\Sigma_{\lambda}$ is either convex or concave; the proof
of Lemma 8.4 actually showed that the second fundamental form is positive
definite, hence $\Sigma_{\lambda}$ is convex (Remark 2.14).

######  Lemma 8.5.

The surface $\Sigma_{\lambda}$ is incomplete at $\theta_{0}=\pi/2$ if
$\lambda<0$, and at $\theta_{0}=-\pi/2$ if $\lambda>0$.

###### Proof.

By Remark 8.2, it suffices to prove the statement for $\lambda>0$. Let us
consider the proper path $\gamma:[1,+\infty)\to\Sigma$ given by

$\gamma(\tau)=X^{\lambda}(\lambda\tau,-\sqrt{1+\lambda^{2}}\tau)$

We claim that $\gamma$ has finite length. Indeed, a direct computation shows
that

$\dot{\gamma}(\tau)=\lambda\partial_{t}X^{\lambda}(\lambda\tau,-\sqrt{1+\lambda^{2}}\tau)-\sqrt{1+\lambda^{2}}\partial_{s}X_{\lambda}(\lambda\tau,-\sqrt{1+\lambda^{2}}\tau)~{}.$

Using the coefficients of the first fundamental form that we derived in the
proof of Lemma 8.3, we have

$\displaystyle||\dot{\gamma}(\tau)||^{2}$
$\displaystyle=\lambda^{2}E(\lambda\tau,-\sqrt{1+\lambda^{2}}\tau)+(1+\lambda^{2})G(\lambda\tau,-\sqrt{1+\lambda^{2}}\tau)-2\lambda\sqrt{1+\lambda^{2}}F(\lambda\tau,-\sqrt{1+\lambda^{2}}\tau)$
$\displaystyle=2\lambda^{2}(1+\lambda^{2})\coth^{2}(\lambda\tau)+\frac{1+\lambda^{2}}{\sinh^{2}(\lambda\tau)}-2\lambda^{2}(1+\lambda^{2})\coth^{2}(\lambda\tau)=\frac{1+\lambda^{2}}{\sinh^{2}(\lambda\tau)}~{}.$

This shows that

$\int_{1}^{+\infty}||\dot{\gamma}(\tau)||d\tau<+\infty~{}.$

It remains to show that the direction of incompleteness is
$\theta_{0}=-\pi/2$. One sees directly from the definition of $X^{\lambda}$
that the null planes $z=\pm y$ are support planes for $\Sigma^{\lambda}$ (for
all $\lambda$). Moreover, we can check that

$\langle\gamma(\tau),(0,-1,1)\rangle=-\sqrt{1+\lambda^{2}}\frac{e^{-\sqrt{1+\lambda^{2}}\tau}}{\sinh(\lambda\tau)}\to
0$

as $\tau\to+\infty$. Hence we can apply Lemma 3.12, and conclude that for
$\lambda>0$, $\Sigma_{\lambda}$ is incomplete at $-\pi/2$. ∎

###  8.2. Null support function

We will now compute the null support function of the surface
$\Sigma_{\lambda}$. We will use again the parabolic null support function.
However, since we have shown in Lemma 8.2 that $\Sigma_{\lambda}$ is
incomplete at $-\pi/2$, we will use a variation of the map $\zeta$ defined in
Equation (5), so as to make the point $\theta_{0}=-\pi/2$ correspond to
$\mathsf{x}_{0}=0$, and the point $\pi/2$ correspond to the direction at
infinity. Namely, we just compose the map
$\zeta:\overline{\mathcal{H}}\to{\mathbb{R}}^{2,1}$ with a rotation of angle
$-\pi/2$. That is, let us define:

${\xi}(\mathsf{x},\mathsf{y}):=\begin{bmatrix}2\mathsf{x}\\\
-1+\mathsf{x}^{2}+\mathsf{y}^{2}\\\
1+\mathsf{x}^{2}+\mathsf{y}^{2}\end{bmatrix}$ (51)

We then define the parabolic null support function _with point at infinity
$\pi/2$_ analogously to (8) with respect to $\xi$ instead of $\zeta$, and we
will denote it by $\varphi$ hereafter:

$\varphi(\mathsf{x}):=\sup_{{p}\in\Sigma}\langle{p},{\xi}(\mathsf{x},0)\rangle~{}.$
(52)

Notice that $\xi(\mathsf{x},0)$ parameterizes all the lightlike directions
$\vec{\theta}$ except $(0,1,1)$, which is however the projective limit of
${\xi}(\mathsf{x},0)$ as $\mathsf{x}\to\infty$. Observe that with this
convention, $\mathsf{x}=\tan(\theta+\pi/2)$.

We are interested in the parabolic null support function of $\Sigma_{\lambda}$
near the point $\mathsf{x}_{0}=0$, which corresponds to $\theta_{0}=-\pi/2$
and to $\xi(0,0)=(0,-1,1)$. The fundamental result we have to achieve is the
following, which holds for any $\lambda\in{\mathbb{R}}$, although we are
mostly concerned with the case where $\Sigma_{\lambda}$ is incomplete at
$-\pi/2$, namely (by Lemma 8.2) when $\lambda>0$.

######  Proposition 8.6.

For any $\lambda\in{\mathbb{R}}$, let $\varphi_{\lambda}$ be the parabolic
null support function of the surface $\Sigma_{\lambda}$, with point at
infinity $\pi/2$. Then

$\varphi_{\lambda}(\mathsf{x})=\left\\{\begin{array}[]{ll}-2\lambda|\mathsf{x}|\log|\mathsf{x}|&\textrm{if
}\mathsf{x}<0\\\ 0&\textrm{if }\mathsf{x}=0\\\ +\infty&\textrm{if
}\mathsf{x}>0.\end{array}\right.$

Before proving Proposition 8.6 we notice that the invariance of
$\Sigma^{\lambda}$ under the action of the one-parameter group
$A^{\lambda}=\\{A^{\lambda}_{s}\\}_{s\in\mathbb{R}}$ gives a strong constraint
on its null support function.

######  Lemma 8.7.

Let $\varphi_{\lambda}$ be the parabolic null support function of any convex
entire achronal surface in ${\mathbb{R}}^{2,1}$ invariant under the one-
parameter group $A^{\lambda}$, with point at infinity $\pi/2$. Then for every
$s\in\mathbb{R}$ and $\mathsf{x}\in\mathbb{R}$ the following identity holds:

$\varphi_{\lambda}(e^{s}\mathsf{x})=e^{s}(\varphi_{\lambda}(\mathsf{x})+2\lambda
s\mathsf{x})$

###### Proof.

Let us call $\Sigma$ the invariant surface. Denote by $L_{s}$ and $\tau_{s}$
the linear and the translational part of $A_{s}$ respectively, namely

$L_{s}=\begin{bmatrix}1&0&0\\\ 0&\cosh(s)&\sinh(s)\\\
0&\sinh(s)&\cosh(s)\end{bmatrix}~{},\qquad\tau_{s}=\begin{bmatrix}\lambda s\\\
0\\\ 0\end{bmatrix}~{}.$

By the invariance under the action of the group $A^{\lambda}$, we have

$\displaystyle\varphi_{\lambda}(\mathsf{x})$
$\displaystyle=\sup_{p\in\Sigma}\langle\xi(\mathsf{x},0),p\rangle$
$\displaystyle=\sup_{p\in\Sigma}\langle\xi(\mathsf{x},0),A_{s}(p)\rangle$
$\displaystyle=\sup_{p\in\Sigma}\langle\xi(\mathsf{x},0),L_{s}(p)+\tau_{s}\rangle$
$\displaystyle=\sup_{p\in\Sigma}\langle\xi(\mathsf{x},0),L_{s}(p)\rangle+\langle\xi(\mathsf{x},0),\tau_{s}\rangle$
$\displaystyle=\sup_{p\in\Sigma}\langle
L_{-s}(\xi(\mathsf{x},0)),p\rangle+2\lambda\mathsf{x}s$

By a direct computation,

$L_{-s}(\xi(\mathsf{x},0))=e^{s}\xi(e^{-s}\mathsf{x},0)~{}.$

Hence we obtain

$\varphi_{\lambda}(\mathsf{x})=e^{s}\varphi_{\lambda}(e^{-s}\mathsf{x})+2\lambda\mathsf{x}s\,$

and the result follows. ∎

###### Proof of Proposition 8.6.

We have already observed in the proof of Lemma that the null planes $z=\pm y$
are support planes of $\Sigma_{\lambda}$. This implies that the elliptic null
support function of $\Sigma^{\lambda}$ at $\theta_{0}=\pm\pi/2$ equals $0$.
Comparing the elliptic and parabolic null support functions as in Proposition
2.27, one has $\varphi_{\lambda}(0)=0$.

Moreover, from Equation (50), the image of the Gauss map is contained in the
half plane defined by $x\geq 0$ in the hyperboloid model of $\mathbb{H}^{2}$,
and therefore the elliptic support function takes the value $+\infty$ on the
half-disc $\\{(x,y,1)\,|\,x\geq 0\\}$. By convexity, the elliptic null support
function equals $+\infty$ at every $\theta\in(-\pi/2,\pi/2)$, that is
$\varphi_{\lambda}(\mathsf{x})=+\infty$ for every $\mathsf{x}>0$.

Finally, we claim that $\varphi_{\lambda}(-1)=0$. From Lemma 8.7, this will
imply that

$\varphi_{\lambda}(-e^{s})=-2\lambda se^{s}$

and the statement will be proved.

By definition of $\varphi_{\lambda}$, since $\xi(-1,0)=(-2,0,2)$, we have

$\varphi_{\lambda}(-1)=\sup_{p\in\Sigma_{\lambda}}\langle
p,(-2,0,2)\rangle~{}.$ (53)

As an additional preliminary remark, we observe that it suffices to show that
$\varphi_{\lambda}(-1)=0$ for $\lambda\geq 0$. Indeed, by Remark 8.2,
$\Sigma_{-\lambda}=R(\Sigma_{\lambda})$ where $R$ is the reflection in the
plane $y=0$. Hence

$\varphi_{-\lambda}(-1)=\sup_{p\in\Sigma_{\lambda}}\langle
R(p),(-2,0,2)\rangle=\sup_{p\in\Sigma_{\lambda}}\langle
p,(-2,0,2)\rangle=\varphi_{\lambda}(-1)$

where we have used that $R$ fixes $(-2,0,2)$.

We are now ready to provide the final computation. From (53) and the
parameterization of $\Sigma_{\lambda}$, we have

$\displaystyle\varphi_{\lambda}(-1)$
$\displaystyle=\sup_{(t,s)\in\mathbb{R}_{+}\times\mathbb{R}}\langle
X^{\lambda}(t,s),(-2,0,2)\rangle$
$\displaystyle=2\sup_{(t,s)\in\mathbb{R}_{+}\times\mathbb{R}}\left(\sqrt{1+\lambda^{2}}\left(\coth(t)-\frac{\cosh(s)}{\sinh(t)}-t\right)-\lambda
s\right)\,.$

For every $t\in\mathbb{R}_{+}$ let us consider the function
$F_{t}:\mathbb{R}\to\mathbb{R}$ defined by

$F_{t}(s)=\sqrt{1+\lambda^{2}}\left(\coth(t)-\frac{\cosh(s)}{\sinh(t)}-t\right)-\lambda
s~{}.$

This function has a global maximum at the point $s_{\max}=s_{\max}(t)$ such
that

$\sinh(s_{\max})=-({\lambda}/{\sqrt{1+\lambda^{2}}})\sinh(t)~{}.$ (54)

The corresponding maximum value is

$\displaystyle G(t):$ $\displaystyle=\sup_{s}F_{t}(s)=F(s_{\max}(t))$
$\displaystyle=\sqrt{1+\lambda^{2}}\left(\coth(t)-\frac{\cosh(s_{\max}(t))}{\sinh(t)}-t\right)-\lambda
s_{\max}(t)$
$\displaystyle=\sqrt{1+\lambda^{2}}(\coth(t)-t)+\lambda(\coth(s_{\max}(t))-s_{\max}(t))$
$\displaystyle=\sqrt{1+\lambda^{2}}g(t)+\lambda g(s_{\max}(t))~{},$

where we have set $g(t):=\coth(t)-t$. We will show that $G^{\prime}(t)<0$. For
this, let us first observe that $g^{\prime}(t)=-\coth^{2}(t)$ and,
differentiating (54), we have the relation

$\cosh(s_{\max}(t))s_{\max}^{\prime}(t)=-\frac{\lambda}{\sqrt{1+\lambda^{2}}}\cosh(t)~{}.$

Using these two relations, we obtain:

$\displaystyle G^{\prime}(t)$
$\displaystyle=-\sqrt{1+\lambda^{2}}\frac{\cosh^{2}(t)}{\sinh^{2}(t)}-\lambda\frac{\cosh^{2}(s_{\max}(t))}{\sinh^{2}(s_{\max}(t))}(s_{\max}^{\prime}(t))$
$\displaystyle={\cosh(t)}\left(-{\sqrt{1+\lambda^{2}}}\frac{\cosh(t)}{\sinh^{2}(t)}+\frac{\lambda^{2}}{{\sqrt{1+\lambda^{2}}}}\frac{\cosh(s_{\max}(t))}{\sinh^{2}(s_{\max}(t))}\right)$
$\displaystyle=\sqrt{1+\lambda^{2}}\frac{\cosh(t)}{\sinh^{2}(t)}(-\cosh(t)+\cosh(s_{\max}(t)))<0$

where in the last inequality we have used that, from (54), $|s_{\max}(t)|<t$.

This shows that $\psi_{\lambda}(-1)=2\lim_{t\to 0^{+}}G(t)$. To compute this
limit, observe that, by (54),

$s_{\max}(t)=-\frac{\lambda}{\sqrt{1+\lambda^{2}}}t+O(t^{2})~{}.$

From the expression of $G(t)$, we then have:

$\displaystyle G(t)$
$\displaystyle=\sqrt{1+\lambda^{2}}\left(\frac{\cosh(t)-\cosh(s_{\max}(t))}{\sinh(t)}\right)-\sqrt{1+\lambda^{2}}t-\lambda
s_{\max}(t)$
$\displaystyle=\sqrt{1+\lambda^{2}}\left(\frac{ct^{2}+O(t^{3})}{t+O(t^{3})}\right)+O(t)~{}.$

This shows that $\lim_{t\to 0^{+}}G(t)=0$ and thus concludes the proof. ∎

###  8.3. The surface $\Sigma_{+}$

By Proposition 8.6, the invariant surface has the correct local behaviour in
order to prove Theorem D, using the comparison result of Corollary G. The
first step is, analogously to Lemmas 6.3 and 7.10, comparing the elliptic and
parabolic null support functions.

######  Lemma 8.8.

Let $s$ be a homogeneous function, and let $\phi$ and $\varphi$ be
respectively its elliptic dehomogenization and parabolic dehomogenization, the
latter with point at infinity $\pi/2$. If

$\varphi(\mathsf{x})\leq\epsilon|\mathsf{x}\log|\mathsf{x}||$

for all $\mathsf{x}$ in some (one-sided) neighbourhood of $\mathsf{x}_{0}=0$,
then

$\phi(\theta)\leq\epsilon|(\theta-\theta_{0})\log|\theta-\theta_{0}||$

for all $\theta$ in some (one-sided) neighbourhood $\theta_{0}=-\pi/2$.

###### Proof.

The proof is very similar to Lemma 6.3, except that we consider the point at
infinity $\pi/2$. Analogously to Proposition 2.27, we have
$\mathsf{x}=\tan((1/2)(\theta-\theta_{0}))$, hence for $\mathsf{x}$ close to
$\mathsf{x}_{0}=0$, the linear bound $|\mathsf{x}|\leq|\theta-\theta_{0}|$
holds. Thus

$\phi(\theta)=\frac{\varphi(\mathsf{x})}{1+\mathsf{x}^{2}}\leq\frac{\epsilon}{1+\mathsf{x}^{2}}|\mathsf{x}\log|\mathsf{x}||\leq\epsilon|(\theta-\theta_{0})\log|\theta-\theta_{0}||~{}.$

This concludes the proof. ∎

So, let $\phi$ be a function as in the hypothesis of Theorem D. Of course, up
to applying an isometry of ${\mathbb{R}}^{2,1}$, we can assume
$\theta_{0}=-\pi/2$ and $\phi(\theta_{0})=0$. Let $\phi_{\lambda}$ the
elliptic null support function of the invariant surface $\Sigma_{\lambda}$,
for $\lambda=\epsilon/2$. We thus have
$\phi_{\lambda}(\theta_{0})=\phi(\theta_{0})$ and $\phi_{\lambda}\leq\phi$ in
a neighbourhood of $\theta_{0}=-\pi/2$. However, in order to apply Corollary
G, with $\phi_{\lambda}$ playing the role of $\phi_{+}$, we would need the
inequality $\phi_{\lambda}\leq\phi$ to hold globally.

This is an issue that we have already taken care of in the proof of Theorem C,
by a translation of the surface. In this setting, however, there is one more
step to do. Indeed, we have that $\phi_{\lambda}$ equals $+\infty$ in the
whole interval $(-\pi/2,\pi/2)$, while the function $\phi$ is by hypothesis
equal to $+\infty$ only in a one-sided neighbourhood (we can assume it is a
right neighbourhood) of $\theta_{0}=-\pi/2$. Translating the surface has an
effect of adding a bounded function to its null support function, hence will
not change this situation. We will instead modify the surface
$\Sigma_{\lambda}$ by applying a parabolic isometry fixing the direction
$\vec{\theta_{0}}$, which has the effect of “shrinking” the interval on which
$\phi_{\lambda}$ is infinite. We provide the details of this argument in the
following proof.

######  Proposition 8.9.

For any $\epsilon,\kappa>0$ and $\theta_{0},\theta_{1}\in\mathbb{S}^{1}$,
there exists a convex entire spacelike surface $\Sigma_{+}$ of constant
curvature $-\kappa$ which is incomplete at $\theta_{0}$ and whose null support
function $\phi_{+}$ satisfies

  * •

$\phi_{+}(\theta_{0})=0$;

  * •

the set $\\{\phi_{+}=+\infty\\}$ equals the right open interval between
$\theta_{0}$ and $\theta_{1}$;

  * •

$\phi_{+}$ is continuous on the subset $\\{\phi_{+}<+\infty\\}$ where it is
finite;

  * •

$\phi_{+}(\theta)\leq\epsilon|(\theta-\theta_{0})\log|\theta-\theta_{0}||$ in
a left one-sided neighbourhood of $\theta_{0}$.

###### Proof.

First, observe that it suffices to prove the statement for $\kappa=1$, since
applying a dilatation has the effect of multiplying the support function by a
constant (Lemma 6.4).

Now, assuming $\theta_{0}=-\pi/2$, we will prove the proposition by applying
to the surface $\Sigma_{\lambda}$ studied above (for $\lambda>0$ to be chosen
later) a parabolic linear isometry $L$ that fixes the direction
$\vec{\theta_{0}}=(0,-1,1)$. Let
$\Sigma_{\lambda}^{\prime}=L(\Sigma_{\lambda})$ and $\phi_{\lambda}^{\prime}$
be the elliptic null support function of $\Sigma_{\lambda}^{\prime}$. Clearly
$L$ fixes the plane $z=-y$, which is a null support plane of
$\Sigma_{\lambda}$, and it is therefore also a null support plane of
$\Sigma_{\lambda}^{\prime}$. Hence
$\phi_{\lambda}^{\prime}(\theta_{0})=\phi_{\lambda}(\theta_{0})=0$. Moreover,
the action induced by $L$ on the unit circle $\mathbb{S}^{1}$ clearly maps
$\\{\theta\,|\,\phi_{\lambda}(\theta)=+\infty\\}=(-\pi/2,\pi/2)$ to
$\\{\theta\,|\,\phi_{\lambda}^{\prime}(\theta)=+\infty\\}$. Choosing $L$
suitably, we can therefore achieve the latter set to be equal to the right
interval between $\theta_{0}=-\pi/2$ and $\theta_{1}$, for any
$\theta_{1}\in\mathbb{S}^{1}$.

Observe that by Lemma 8.8, the null support function $\phi_{\lambda}$ is
smaller than $\lambda|(\theta+\pi/2)\log|\theta+\pi/2||$ on the left of
$-\pi/2$. It remains to show that $\phi_{\lambda}^{\prime}$ still has the same
behaviour in a left one-sided neighbourhood of $-\pi/2$. This is done,
avoiding unnecessary computations, by considering the parabolic null support
function _with point at infinity $-\pi/2$._

For this purpose, let us introduce yet another map
$\widehat{\xi}:\overline{\mathcal{H}}\to{\mathbb{R}}^{2,1}$ by:

${\widehat{\xi}}(\mathsf{x},\mathsf{y})=\begin{bmatrix}2\mathsf{x}\\\
1-\mathsf{x}^{2}-\mathsf{y}^{2}\\\
1+\mathsf{x}^{2}+\mathsf{y}^{2}\end{bmatrix}~{}.$ (55)

This map parameterizes all timelike and lightlike directions in Minkowski
space, except $(0,-1,1)$, which corresponds to $\vec{\theta_{0}}$. Observe
that $\widehat{\xi}=R\circ\xi$, where $\xi$ is the map we have used in this
section (Equation (51)) and $R$ is the reflection in the plane $y=0$. Hence,
defining the parabolic null support function $\widehat{\psi}$ _with respect to
$-\pi/2$_ as

$\widehat{\varphi}_{\lambda}(\mathsf{x})=\sup_{{p}\in\Sigma_{\lambda}}\langle{p},\widehat{\xi}(\mathsf{x},0)\rangle~{},$

we immediately obtain from the identity
$\Sigma_{-\lambda}=R(\Sigma_{\lambda})$ (Remark 8.2) that

$\widehat{\varphi}_{\lambda}(\mathsf{x})=\sup_{{p}\in\Sigma_{\lambda}}\langle{p},\widehat{\xi}(\mathsf{x},0)\rangle=\sup_{{p}\in\Sigma_{\lambda}}\langle{p},R\circ{\xi}(\mathsf{x},0)\rangle=\sup_{{p}\in\Sigma_{\lambda}}\langle
R(p),{\xi}(\mathsf{x},0)\rangle=\varphi_{-\lambda}(\mathsf{x})~{}.$

Therefore from Proposition 8.6, we have:

$\widehat{\varphi}_{\lambda}(\mathsf{x})=2\lambda|x|\log|x|$

for $\mathsf{x}<0$. Similarly to Remark 2.23 (or by a direct check using
(55)), we have
$L\circ\widehat{\xi}(\mathsf{x},\mathsf{y})=\widehat{\xi}(\mathsf{x}+c,\mathsf{y})$.
Hence applying a parabolic isometry fixing the direction $\vec{\theta}_{0}$ to
a surface has the effect of precomposing the parabolic null support function
with point at infinity $\theta_{0}$ by a translation:

$\widehat{\varphi}^{\prime}_{\lambda}(\mathsf{x}):=\sup_{{p}\in\Sigma^{\prime}_{\lambda}=L(\Sigma_{\lambda})}\langle{p},\widehat{\xi}(\mathsf{x},0)\rangle=\sup_{{p}\in\Sigma_{\lambda}}\langle{p},\widehat{\xi}(\mathsf{x}+c,0)\rangle=\widehat{\varphi}_{\lambda}(\mathsf{x}+c)~{}.$

In particular,

$\widehat{\varphi}^{\prime}_{\lambda}(\mathsf{x})=2\lambda|\mathsf{x}+c|\log|\mathsf{x}+c|~{}.$

Since

$\lim_{t\to+\infty}\frac{(t+c)\log(t+c)}{t\log t}=1~{},$

we have that, for any $\epsilon>2\lambda$ and for $|\mathsf{x}|$ large,

$\widehat{\varphi}^{\prime}_{\lambda}(\mathsf{x})\leq\epsilon|\mathsf{x}|\log|\mathsf{x}|$

This shows the null support function of $L(\Sigma_{\lambda})$ is smaller than
that of $\Sigma_{\epsilon/2}$ in a one-sided neighbourhood of
$\theta_{0}=-\pi/2$. Going back to the parabolic null support function with
point at infinity $-\pi/2$, this means (from Proposition 8.6) that
$\varphi^{\prime}_{\lambda}(\mathsf{x})\leq-\epsilon|\mathsf{x}|\log|\mathsf{x}|$
for $\mathsf{x}$ in a left neighbourhood of $0$. Using Lemma 8.8, the proof is
complete. ∎

###  8.4. Proof of Theorem D

We are finally ready to conclude the proof of Theorem D.

###### Proof of Theorem D.

Let $\phi$ be the null support function as in the hypothesis of the theorem.
Assume as usual that $\phi(\theta_{0})=0$, and let’s say that the curvature of
$\Sigma$ is bounded above by $-\kappa$. By Proposition 8.9 (possibly applying
a reflection) let be $\phi_{+}$ the null support function of a convex entire
spacelike surface $\Sigma_{+}$ of curvature $-\kappa$, with the following
properties:

  * •

$\Sigma_{+}$ is incomplete at $\theta_{0}$,

  * •

$\phi_{+}(\theta_{0})=\phi(\theta_{0})=0$,

  * •

the set where $\phi_{+}=+\infty$ is a proper subset of the set where
$\phi=+\infty$,

  * •

$\phi_{+}$ is continuous on the subset of $\mathbb{S}^{1}$ where it is finite,

  * •

$\phi_{+}\leq\phi$ in a one-sided neighbourhood of $\theta_{0}$, on the side
of $\theta_{0}$ where both are finite.

Exactly as in the proof of Theorem C, we must now modify $\Sigma_{+}$ further
so that $\phi_{+}\leq\phi$ holds globally, so that by Corollary G we will
deduce that $\Sigma$ is incomplete at $\theta_{0}$. To achieve this, it now
suffices to translate $\Sigma_{+}$ in the direction of $\vec{\theta}_{0}$.
Indeed, $\phi_{+}$ is continuous on the complement of the open interval where
it is equal to $+\infty$, whereas $\phi$ is lower-semicontinuous, hence
bounded below. The same of course holds for the parabolic null support
functions $\widehat{\psi}_{+}$ and $\widehat{\psi}$ with point at infinity
$\theta_{0}$. This implies that $\widehat{\psi}-\widehat{\psi}_{+}$ is bounded
below on $\\{\widehat{\psi}_{+}<+\infty\\}$. Recall from the proof of Theorem
C that translating $\Sigma_{+}$ in the direction $c\vec{\theta}_{0}$ has the
effect of adding the constant $-2c$ to $\widehat{\psi}_{+}$, and does not
change the elliptic null support function at $\theta_{0}$. Hence choosing $c$
large enough, one can make sure that $\widehat{\psi}_{+}\leq\widehat{\psi}$,
and therefore also $\phi_{+}\leq\phi$ on the whole $\mathbb{S}^{1}$. The proof
is then concluded by Corollary G. ∎

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