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# An inequality for the compositions of convex functions with convolutions and
an alternative proof of the Brunn–Minkowski–Kemperman inequality

Takashi Satomi

###### Abstract

Let $m(G)$ be the infimum of the volumes of all open subgroups of a unimodular
locally compact group $G$. Suppose integrable functions
$\phi_{1},\phi_{2}\colon G\to[0,1]$ satisfy $\|\phi_{1}\|\leq\|\phi_{2}\|$ and
$\|\phi_{1}\|+\|\phi_{2}\|\leq m(G)$, where $\|\cdot\|$ denotes the
$L^{1}$-norm with respect to a Haar measure $dg$ on $G$. We have the following
inequality for any convex function $f\colon[0,\|\phi_{1}\|]\to\mathbb{R}$ with
$f(0)=0$:

$\displaystyle\int_{G}f\circ(\phi_{1}*\phi_{2})(g)dg\leq
2\int_{0}^{\|\phi_{1}\|}f(y)dy+(\|\phi_{2}\|-\|\phi_{1}\|)f(\|\phi_{1}\|).$

As a corollary, we have a slightly stronger version of
Brunn–Minkowski–Kemperman inequality. That is, we have

$\displaystyle\mathrm{vol}_{*}(B_{1}B_{2})\geq\mathrm{vol}(\\{g\in G\mid
1_{B_{1}}*1_{B_{2}}(g)>0\\})\geq\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2})$

for any non-null measurable sets $B_{1},B_{2}\subset G$ with
$\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2})\leq m(G)$, where $\mathrm{vol}_{*}$
denotes the inner measure and $1_{B}$ the characteristic function of $B$.

Keywords: convolution, convexity, locally compact group, combinatorial
inequality, geometric measure theory.

MSC2020: Primary 43A05; Secondary 05A20, 22D05, 28C10, 39B62.

##  1 Introduction

Let $\mathcal{A}$ be the set of all open subgroups of a unimodular locally
compact group $G$ and we set

$\displaystyle m(G):=\inf_{G^{\prime}\in\mathcal{A}}\mathrm{vol}(G^{\prime})$

($m(G)$ may be $\infty$). Let $\|\cdot\|$ be the $L^{1}$-norm with respect to
a Haar measure $dg$ on $G$. For two measurable functions
$\phi_{1},\phi_{2}\colon G\to\mathbb{R}_{\geq 0}$, we define the convolution
$\phi_{1}*\phi_{2}$ as

$\displaystyle\phi_{1}*\phi_{2}(g):=\int_{G}\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)dg^{\prime}.$

The main result of this paper is the following theorem.

######  Theorem 1.1.

Suppose integrable functions $\phi_{1},\phi_{2}\colon G\to[0,1]$ on a
unimodular locally compact group $G$ satisfy $\|\phi_{1}\|\leq\|\phi_{2}\|$
and

$\|\phi_{1}\|+\|\phi_{2}\|\leq m(G).$ (1.1)

Then, for any convex function $f\colon[0,\|\phi_{1}\|]\to\mathbb{R}$ with
$f(0)=0$, one has

$\int_{G}f\circ(\phi_{1}*\phi_{2})(g)dg\leq
2\int_{0}^{\|\phi_{1}\|}f(y)dy+(\|\phi_{2}\|-\|\phi_{1}\|)f(\|\phi_{1}\|).$
(1.2)

For example, by using the $L^{1}$-norm (and $L^{\infty}$-norm) of $\phi_{1}$
and $\phi_{2}$, Theorem 1.1 bounds the $L^{p}$-norm of the convolution
$\phi_{1}*\phi_{2}$ when $f(y)=y^{p}$ holds (Example 2.1 (2) and (3)), and the
entropy of $\phi_{1}*\phi_{2}$ from below when $f(y)=y\log y$ (Example 2.1
(4)). Actually, Theorem 1.1 is best possible for $G=\mathbb{R}$ (Example 2.1
(1)).

Here are some other examples of Theorem 1.1. When $t\geq 0$ is fixed,

$\displaystyle f_{t}(y):=\left\\{\begin{aligned} &y-t&&\text{if}\;y\geq t\\\
&0&&\text{if}\;y\leq t\end{aligned}\right.$ (1.3)

is convex. Since

$\displaystyle
2\int_{0}^{\|\phi_{1}\|}f_{t}(y)dy+(\|\phi_{2}\|-\|\phi_{1}\|)f_{t}(\|\phi_{1}\|)=(\|\phi_{1}\|-t)(\|\phi_{2}\|-t)$

holds for any $0\leq t\leq\|\phi_{1}\|$, we have

$\|f_{t}\circ(\phi_{1}*\phi_{2})\|\leq(\|\phi_{1}\|-t)(\|\phi_{2}\|-t)$ (1.4)

by applying $f=f_{t}$ for Theorem 1.1. Actually, (1.4) holds even when the
assumption (1.1) is weakened by

$\|\phi_{1}\|+\|\phi_{2}\|\leq m(G)+t.$ (1.5)

That is, the following lemma holds.

######  Lemma 1.2.

One has (1.4) for any integrable functions $\phi_{1},\phi_{2}\colon G\to[0,1]$
on a unimodular locally compact group $G$ and $0\leq
t\leq\min(\|\phi_{1}\|,\|\phi_{2}\|)$ with (1.5).

For any characteristic functions $\phi_{1}$ and $\phi_{2}$ on a compact
connected group $G$, Lemma 1.2 was essentially proved by Tao [Tao12, Theorem
3.2.3] and explicitly given by Christ [Chr19, Section 6]. Christ also proved
Lemma 1.2 for any characteristic functions $\phi_{1}$ and $\phi_{2}$ on
$G=\mathbb{R}^{n}$, and discussed a relation between Lemma 1.2 and
Riesz–Sobolev inequality [Rie30] [Sob38].

The similar claim to Lemma 1.2 for any cyclic group of prime order was proved
by Pollard [Pol74], and for any finite abelian group by Green–Ruzsa [GR05,
Proposition 6.1]. By using this claim, Green–Ruzsa explicitly computed the
maximal cardinality of all sum-free subsets for any finite abelian group
[GR05, Theorem 1.5].

Besides, when $\phi_{1}$ and $\phi_{2}$ are characteristic functions, the
following corollary holds by applying Theorem 1.1 for

$\displaystyle f(y):=\left\\{\begin{aligned} &0&&\text{if}\;y=0\\\
&-1&&\text{if}\;y>0\end{aligned}\right..$ (1.6)

######  Corollary 1.3.

Let $B_{1},B_{2}\subset G$ be measurable sets on a unimodular locally compact
group $G$ with $\mathrm{vol}(B_{1}),\mathrm{vol}(B_{2})>0$. If
$\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2})\leq m(G)$ holds, then one has

$\displaystyle\mathrm{vol}(\\{g\in G\mid
1_{B_{1}}*1_{B_{2}}(g)>0\\})\geq\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2}),$
(1.7)

where $1_{B}$ denotes the characteristic function of $B$. In particular,

$\mathrm{vol}_{*}(B_{1}B_{2})\geq\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2})$
(1.8)

holds, where $\mathrm{vol}_{*}$ denotes the inner measure.

The Brunn–Minkowski–Kemperman inequality (1.8) is followed from (1.7) and
$\\{g\in G\mid 1_{B_{1}}*1_{B_{2}}(g)>0\\}\subset B_{1}B_{2}$. Corollary 1.3
was proved by Tao for any compact connected group $G$ by using Lemma 1.2. This
argument is essentially equivalent to applying Theorem 1.1 for (1.6). (1.8)
was proved by Kemperman for any general unimodular locally compact group $G$,
and a number of people for particular cases as follows.

a unimodular locally compact group $G$ |
---|---
$\mathbb{R}^{n}$ ($B_{1}$ and $B_{2}$ are convex) | Brunn [Bru87], Minkowski [Min10]
$\mathbb{R}^{n}$ | Lusternik [Lus35],
Henstock–Macbeath [HM53, Theorem 1]
$S^{1}$ | Raikov [Rai39]
$(S^{1})^{n}$ | Macbeath [Mac53, Theorem 1]
compact connected second countable abelian | Shields [Shi55]
abelian | Kneser [Kne56]
general | Kemperman [Kem64, Theorem 1.2]
connected or $m(G)=\infty$ | Ruzsa [Ruz92, Theorem 2]

For $G=\mathbb{R}^{n}$, (1.8) can be improved to

$\displaystyle\mathrm{vol}_{*}(B_{1}B_{2})^{1/n}\geq\mathrm{vol}(B_{1})^{1/n}+\mathrm{vol}(B_{2})^{1/n}.$
(1.9)

For any convex sets $B_{1}$ and $B_{2}$ on $G=\mathbb{R}^{n}$, Brunn proved
(1.9), and Minkowski proved that the equality is attained in (1.9) if and only
if $B_{1}$ and $B_{2}$ are equal up to dilatation. For arbitrary measurable
sets $B_{1}$ and $B_{2}$ on $G=\mathbb{R}^{n}$, the inequality (1.9) was
stated by Lusternik. However, Lusternik’s proof is defective, and corrected by
Henstock–Macbeath. A generalization of (1.9) for any locally compact group $G$
was studied by McCrudden [McC69, Theorem 1.2] and Jing–Tran–Zhang [JTZ21,
Theorem 1.1].

When $G$ is connected or $m(G)=\infty$ holds, Ruzsa gave an alternative proof
of (1.8). Kemperman’s idea is studying the sets $B_{1}$ and $B_{2}$ such that
the maximum of
$\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2})-\mathrm{vol}_{*}(B_{1}B_{2})$ is
attained. Ruzsa’s idea is showing that
$T(I):=\inf_{\mathrm{vol}(B_{1})=I}\mathrm{vol}_{*}(B_{1}B_{2})$ is a concave
function for any fixed $B_{2}$.

For any compact connected abelian group $G$, Tao studied a property of
measurable sets for which the equality of Corollary 1.3 is almost attained
[Tao18, Theorem 1.3]. This result is generalized for disconnected group by
Griesmer [Gri19, Theorem 1.1], and for non-abelian group by Jing-Tran [JT20,
Theorem 1.2]. A sharper result for any compact abelian group is given by
Christ–Iliopoulou [CI18, Theorem 1.7].

Here is the strategy of proof of Theorem 1.1. As mentioned above, Lemma 1.2
under the assumption (1.1) is an example of Theorem 1.1. However, we will show
Lemma 1.2 without using Theorem 1.1 at first, and obtain Theorem 1.1 by using
Lemma 1.2. We will show Lemma 1.2 by generalizing the argument of Tao [Tao12,
Theorem 3.2.3], and Theorem 1.1 by approximating $f$ by a linear combination
of $f_{t}$.

The organization of this paper is as follows. In Section 2, we will give some
examples and remarks of Theorem 1.1. Sections 3-6 are devoted to the proof of
Lemma 1.2. In Section 3, we will discuss the necessity of (1.1). In Section 4,
we will give a key lemma (Lemma 4.1) to prove Lemma 1.2. By using the
arguments in Sections 3 and 4, we will show Lemma 1.2 in the case of
$m(G)=\infty$ in Section 5, and $m(G)<\infty$ in Section 6. In Section 7, we
will complete the proof of Theorem 1.1 by Lemma 1.2.

##  2 Some examples and remarks of Theorem 1.1

In this section, we give some examples and remarks of Theorem 1.1 (and related
results). At first, we give some examples of Theorem 1.1.

######  Example 2.1.

  1. (1)

One has $m(G)=\infty$ for $G=\mathbb{R}$. The function
$\phi_{(I)}\colon\mathbb{R}\to\\{0,1\\}$ denotes the characteristic function
$\phi_{(I)}:=1_{(-I/2,I/2)}$ of the open interval $(-I/2,I/2)$ for $I>0$. Let
$\psi_{I_{1},I_{2}}:=\phi_{(I_{1})}*\phi_{(I_{2})}$ for $I_{1},I_{2}>0$. If
$I_{1}\leq I_{2}$ holds, then one has

$\displaystyle\psi_{I_{1},I_{2}}(x)=\left\\{\begin{aligned}
&I_{1}&&\text{if}\;|x|\leq c\\\ &c^{\prime}-|x|&&\text{if}\;c\leq|x|\leq
c^{\prime}\\\ &0&&\text{if}\;c^{\prime}\leq|x|\end{aligned}\right.,\quad
c:=\frac{I_{2}-I_{1}}{2},\quad c^{\prime}:=\frac{I_{1}+I_{2}}{2}.$

Thus, if $\phi_{1}=\phi_{(I_{1})}$ and $\phi_{2}=\phi_{(I_{2})}$ hold, then
one has

$\displaystyle\int_{\mathbb{R}}f\circ\psi_{I_{1},I_{2}}(x)dx$
$\displaystyle=\int_{-c^{\prime}}^{-c}f\left(x+c^{\prime}\right)dx+\int_{-c}^{c}f(I_{1})dx+\int_{c}^{c^{\prime}}f\left(c^{\prime}-x\right)dx$
$\displaystyle=2\int_{0}^{I_{1}}f(y)dy+(I_{2}-I_{1})f(I_{1}).$ (2.1)

Since $I_{1}=\|\phi_{1}\|$ and $I_{2}=\|\phi_{2}\|$ hold, the equality is
attained in (1.2). That is, (1.2) can be expressed as

$\displaystyle\int_{G}f\circ(\phi_{1}*\phi_{2})(g)dg\leq\int_{\mathbb{R}}f\circ\psi_{\|\phi_{1}\|,\|\phi_{2}\|}(x)dx$

(We note that $\phi_{1}$ and $\phi_{2}$ are not the same as
$\phi_{1}=\phi_{(I_{1})}$ and $\phi_{2}=\phi_{(I_{2})}$ stated before and are
already general functions on a general unimodular locally compact group $G$).
In particular, (1.4) can be expressed as

$\|f_{t}\circ(\phi_{1}*\phi_{2})\|\leq\|f_{t}\circ\psi_{\|\phi_{1}\|,\|\phi_{2}\|}\|.$
(2.2)

  2. (2)

The function $f(y):=y^{p}$ is convex for $p\geq 1$. Thus, one has

$\displaystyle\int_{G}(\phi_{1}*\phi_{2}(g))^{p}dg\leq
2\int_{0}^{\|\phi_{1}\|}y^{p}dy+(\|\phi_{2}\|-\|\phi_{1}\|){\|\phi_{1}\|}^{p}={\|\phi_{1}\|}^{p}\left(\|\phi_{2}\|-\frac{(p-1)\|\phi_{1}\|}{p+1}\right)$

by applying Theorem 1.1 in this case.

  3. (3)

The function $f(y):=-y^{p}$ is convex for $0<p\leq 1$. Thus, one has

$\displaystyle\int_{G}(\phi_{1}*\phi_{2}(g))^{p}dg\geq
2\int_{0}^{\|\phi_{1}\|}y^{p}dy+(\|\phi_{2}\|-\|\phi_{1}\|){\|\phi_{1}\|}^{p}={\|\phi_{1}\|}^{p}\left(\|\phi_{2}\|+\frac{(1-p)\|\phi_{1}\|}{p+1}\right)$

by applying Theorem 1.1 in this case.

  4. (4)

The function $f(y):=y\log y$ is convex. Thus, one has

$\displaystyle\int_{G}\phi_{1}*\phi_{2}(g)\log\circ(\phi_{1}*\phi_{2})(g)dg$
$\displaystyle\leq 2\int_{0}^{\|\phi_{1}\|}y\log
ydy+(\|\phi_{2}\|-\|\phi_{1}\|)\|\phi_{1}\|\log\|\phi_{1}\|$
$\displaystyle=\|\phi_{1}\|\left(\|\phi_{2}\|\log\|\phi_{1}\|-\frac{\|\phi_{1}\|}{2}\right)$

by applying Theorem 1.1 in this case.

  5. (5)

Since $f_{t}(y)=y$ holds for $t=0$, one has
$\|\phi_{1}*\phi_{2}\|\leq\|\phi_{1}\|\|\phi_{2}\|$ by Lemma 1.2. Besides,
$-\|\phi_{1}*\phi_{2}\|\leq-\|\phi_{1}\|\|\phi_{2}\|$ holds by applying
Theorem 1.1 for the convex function $f(y):=-y$. Thus, one has

$\|\phi_{1}*\phi_{2}\|=\|\phi_{1}\|\|\phi_{2}\|.$ (2.3)

By Fubini’s theorem, (2.3) holds for any $\phi_{1},\phi_{2}\colon
G\to\mathbb{R}_{\geq 0}$ even when the assumption (1.1) is not satisfied. That
is, the following lemma holds.

######  Lemma 2.2.

One has (2.3) for any integrable functions $\phi_{1},\phi_{2}\colon
G\to\mathbb{R}_{\geq 0}$ on a unimodular locally compact group $G$.

The equalities are attained in Lemma 1.2, Example 2.1 (2), (3) and (4) for
$G=\mathbb{R}$, $\phi_{1}=\phi_{(\|\phi_{1}\|)}$ and
$\phi_{2}=\phi_{(\|\phi_{2}\|)}$ by Example 2.1 (1).

Next, we define the convex function used in Theorem 1.1.

######  Definition 2.3.

For a function $f\colon A\to\mathbb{R}$ on an interval $A$ and $y_{1},y_{2}\in
A$ with $y_{1}<y_{2}$, the functions
$\tilde{f}_{y_{1}}^{y_{2}},\hat{f}_{y_{1}}^{y_{2}}\colon A\to\mathbb{R}$ are
defined as

$\displaystyle\tilde{f}_{y_{1}}^{y_{2}}(y):=\frac{(y_{2}-y)f(y_{1})+(y-y_{1})f(y_{2})}{y_{2}-y_{1}},\quad\hat{f}_{y_{1}}^{y_{2}}(y):=\tilde{f}_{y_{1}}^{y_{2}}(y)-f(y).$

A function $f\colon A\to\mathbb{R}$ is called convex if
$\hat{f}_{y_{1}}^{y_{2}}(y)\geq 0$ (that is,
$f(y)\leq\tilde{f}_{y_{1}}^{y_{2}}(y)$) holds for any $y_{1},y_{2},y\in A$
with $y_{1}<y_{2}$ and $y_{1}\leq y\leq y_{2}$.

Here are some remarks on Theorem 1.1 and Corollary 1.3.

######  Remark 2.4.

  1. (1)

The composition $f\circ(\phi_{1}*\phi_{2})$ in Theorem 1.1 is well-defined
because

$\phi_{1}*\phi_{2}(g)=\int_{G}\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)dg^{\prime}\leq\int_{G}\phi_{1}(g^{\prime})dg^{\prime}=\|\phi_{1}\|$
(2.4)

holds for any $g\in G$.

  2. (2)

When $f$ is of $C^{2}$-class, $f$ is convex if and only if
$f^{\prime\prime}\geq 0$ holds. However, Theorem 1.1 does not require $f$ to
be differentiable or even continuous (at $0$ and $\|\phi_{1}\|$).

  3. (3)

The following three conditions (3)(i), (3)(ii) and (3)(iii) are equivalent for
any locally compact group $G$.

    1. (i)

One has $m(G)<\infty$.

    2. (ii)

The locally compact group $G$ has an open compact subgroup.

    3. (iii)

The identity component $G_{0}$ of $G$ is compact.

The equivalence of (3)(i) and (3)(ii) is obtained by the fact that
$\mathrm{vol}(G^{\prime})<\infty$ holds if and only if $G^{\prime}$ is compact
[Wei40]. We show (3)(iii) by (3)(ii). There is an open compact subgroup
$G^{\prime}\subset G$ by (3)(ii). Since $G_{0}\subset G^{\prime}$ holds by the
connectedness of $G_{0}$, one has (3)(iii) by the compactness of $G^{\prime}$.
We show (3)(ii) by (3)(iii). Since $G/G_{0}$ is a totally disconnected locally
compact group [HR79, Theorem 7.3], there is an open compact subgroup $H\subset
G/G_{0}$ [VD31] [HR79, Theorem 7.7]. Then $\pi^{-1}(H)\subset G$ is the open
subgroup, where $\pi\colon G\to G/G_{0}$ be the projection. Since
$\pi^{-1}(H)$ is compact by (3)(iii) [HR79, Note 5.24 (a)], we obtain (3)(ii).

Besides,

$\displaystyle G_{0}=\bigcap_{G^{\prime}\in\mathcal{A}}G^{\prime}$

holds for any $G$ [HR79, Theorem 7.8]. Thus, if the equivalent conditions
(3)(i), (3)(ii) and (3)(iii) are satisfied, then we have
$m(G)=\mathrm{vol}(G_{0})$. Thus, (1.1) holds if and only if at least one of
the following two conditions (3)(a) and (3)(b) are satisfied.

    1. (a)

One has $\|\phi_{1}\|+\|\phi_{2}\|\leq\mathrm{vol}(G_{0})$.

    2. (b)

The identity component $G_{0}$ is not compact.

In particular, if $G_{0}$ is open (e.g., $G$ is a Lie group), then (1.1) and
(3)(a) are equivalent.

  4. (4)

A similar result to Theorem 1.1 holds in the case of
$\|\phi_{2}\|\leq\|\phi_{1}\|$ as follows. For integrable functions
$\phi_{1},\phi_{2}\colon G\to[0,1]$ with $\|\phi_{2}\|\leq\|\phi_{1}\|$ and
(1.1),

$\phi_{1}*\phi_{2}(g)=\int_{G}\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)dg^{\prime}\leq\int_{G}\phi_{2}(g^{\prime-1}g)dg^{\prime}=\|\phi_{2}\|$
(2.5)

holds and hence $f\circ(\phi_{1}*\phi_{2})$ is well-defined for any function
$f\colon[0,\|\phi_{2}\|]\to\mathbb{R}$. Then, one has

$\displaystyle\int_{G}f\circ(\phi_{1}*\phi_{2})(g)dg\leq
2\int_{0}^{\|\phi_{2}\|}f(y)dy+(\|\phi_{1}\|-\|\phi_{2}\|)f(\|\phi_{2}\|)$

for any convex function $f\colon[0,\|\phi_{2}\|]\to\mathbb{R}$ with $f(0)=0$
by the argument similar to Theorem 1.1.

  5. (5)

In Corollary 1.3, we set $U:=\\{g\in G\mid\phi_{1}*\phi_{2}(g)>0\\}$. In fact,
there is a case of $\mathrm{vol}_{*}(B_{1}B_{2})>\mathrm{vol}(U)$. For
example, when $G=\mathbb{R}$, $B_{1}=(0,1)\cup\\{2\\}$ and $B_{2}=(0,1)$ hold,
we have $B_{1}B_{2}=(0,3)\setminus\\{2\\}$ and $U=(0,2)$. Since
$\mathrm{vol}(B_{1})=\mathrm{vol}(B_{2})=1$, $\mathrm{vol}_{*}(B_{1}B_{2})=3$
and $\mathrm{vol}(U)=2$ hold, we have

$\displaystyle\mathrm{vol}_{*}(B_{1}B_{2})>\mathrm{vol}(U)=\mathrm{vol}(B_{1})+\mathrm{vol}(B_{2}).$

  6. (6)

If $\|\phi_{1}\|=0$ holds, then we have $\phi_{1}*\phi_{2}=0$ and hence
Theorem 1.1 and Lemma 1.2 hold. Thus, it suffices to show Theorem 1.1 and
Lemma 1.2 for $\|\phi_{1}\|>0$. In this case, one can reduce the proof to the
case of $\|\phi_{1}\|=1$ by replacing $dg$ and $f(y)$ with $dg/\|\phi_{1}\|$
and $f(y/\|\phi_{1}\|)$ respectively.

##  3 The necessity of the assumption (1.1)

In this section, we discuss the necessity of the assumption (1.1) when $G$ is
connected. That is, if (1.1) does not hold, then (1.4) does not hold for any
$0<t<\|\phi_{1}\|+\|\phi_{2}\|-m(G)$ by the following lemma.

######  Lemma 3.1.

Let $\phi_{1},\phi_{2}\colon G\to[0,1]$ be integrable functions on a
unimodular locally compact group $G$ with $G=G_{0}$ (that is, $G$ is
connected).

  1. (1)

One has $\phi_{1}*\phi_{2}\geq\|\phi_{1}\|+\|\phi_{2}\|-m(G)$.

  2. (2)

If $\|\phi_{1}\|+\|\phi_{2}\|-m(G)\geq 0$ holds, then one has
$\|f_{t}\circ(\phi_{1}*\phi_{2})\|=\|\phi_{1}\|\|\phi_{2}\|-tm(G)$ for any
$0\leq t\leq\|\phi_{1}\|+\|\phi_{2}\|-m(G)$.

###### Proof.

  1. (1)

For any $g\in G$,

$\displaystyle\phi_{1}*\phi_{2}(g)=\int_{G}\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)dg^{\prime}$

holds.

$\displaystyle\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)$
$\displaystyle=\phi_{1}(g^{\prime})+\phi_{2}(g^{\prime-1}g)-1+(1-\phi_{1}(g^{\prime}))(1-\phi_{2}(g^{\prime-1}g))$
$\displaystyle\geq\phi_{1}(g^{\prime})+\phi_{2}(g^{\prime-1}g)-1$

holds by $\phi_{1}(g^{\prime}),\phi_{2}(g^{\prime-1}g)\leq 1$ and hence one
has

$\displaystyle\int_{G}\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)dg^{\prime}\geq\int_{G}(\phi_{1}(g^{\prime})+\phi_{2}(g^{\prime-1}g)-1)dg^{\prime}=\|\phi_{1}\|+\|\phi_{2}\|-\mathrm{vol}(G).$

Since $m(G)=\mathrm{vol}(G)$ holds by $G=G_{0}$ and Remark 2.4 (3), we obtain

$\displaystyle\phi_{1}*\phi_{2}(g)=\int_{G}\phi_{1}(g^{\prime})\phi_{2}(g^{\prime-1}g)dg^{\prime}\geq\|\phi_{1}\|+\|\phi_{2}\|-\mathrm{vol}(G)=\|\phi_{1}\|+\|\phi_{2}\|-m(G).$

  2. (2)

Since $f_{t}\circ(\phi_{1}*\phi_{2})=\phi_{1}*\phi_{2}-t$ holds by (1), we
obtain

$\displaystyle\|f_{t}\circ(\phi_{1}*\phi_{2})\|=\|\phi_{1}*\phi_{2}-t\|=\|\phi_{1}\|\|\phi_{2}\|-tm(G)$

by Lemma 2.2. ∎

By Lemma 3.1, the inequality (1.4) does not hold for any
$0<t<\|\phi_{1}\|+\|\phi_{2}\|-m(G)$ when $G$ is connected and (1.1) does not
hold. In fact,

$\displaystyle\|f_{t}\circ(\phi_{1}*\phi_{2})\|=\|\phi_{1}\|\|\phi_{2}\|-tm(G)=(\|\phi_{1}\|-t)(\|\phi_{2}\|-t)+t(\|\phi_{1}\|+\|\phi_{2}\|-m(G)-t)$

holds by Lemma 3.1 (2) and hence we have
$(\|\phi_{1}\|-t)(\|\phi_{2}\|-t)<\|f_{t}\circ(\phi_{1}*\phi_{2})\|$.

##  4 The first step of the proof of Lemma 1.2

In this section, we give the key lemma (Lemma 4.1) to prove Lemma 1.2. This
idea is based on the submodularity inequality by Pollard [Pol74], Ruzsa
[Ruz92, Lemma 2.2], Green–Ruzsa [GR05, Proposition 6.1] and Tao [Tao12,
Theorem 3.2.3] [Tao13, Lemma 2.1].

It suffices to show Lemma 1.2 for $\|\phi_{1}\|=1$ by Remark 2.4 (6). We fix
$\phi_{1}$ with $\|\phi_{1}\|=1$ and simply write $f$ for $f_{t}$ in (1.3).
The function $S\colon[0,\mathrm{vol}(G)]\to\mathbb{R}_{\geq 0}$ is defined as

$\displaystyle S(I):=\sup_{\phi\in\mathcal{B}(I)}\|\xi_{\phi}\|,$

where $\mathcal{B}(I):=\\{\phi\colon G\to[0,1]\mid\|\phi\|=I\\}$ and
$\xi_{\phi}:=f\circ(\phi_{1}*\phi)$. Lemma 1.2 can be expressed as

$S(I)\leq(1-t)(I-t)$ (4.1)

for any $t\leq I\leq m(G)+t-1$. We show the following lemma to prove Lemma
1.2.

######  Lemma 4.1.

  1. (1)

One has $0\leq S(I^{\prime})-S(I)\leq I^{\prime}-I$ for any $0\leq I\leq
I^{\prime}\leq\mathrm{vol}(G)$. In particular, $S$ is Lipschitz continuous.

  2. (2)

The function $S$ is convex on $[0,m(G)]$ (or $[0,\infty)$ in the case of
$m(G)=\infty$).

  3. (3)

One has $S(t)=0$.

  4. (4)

One has $S(I)\leq(1-t)I$ for any $0\leq I\leq\mathrm{vol}(G)$.

Lemma 4.1 (2) is deduced from the following lemma.

######  Lemma 4.2.

Let $I>0$ and $\phi\in\mathcal{B}(I)$. We define $R_{g}\phi\colon G\to[0,1]$
as $R_{g}\phi(g^{\prime}):=\phi(g^{\prime}g)$ for $g\in G$. Let
$I^{2}/m(G)<I_{1}^{\prime}\leq I$ and $I_{2}^{\prime}:=2I-I_{1}^{\prime}$.

  1. (1)

We define a continuous function $h\colon G\to\mathbb{R}_{\geq 0}$ as
$h(g):=\|\phi R_{g}\phi\|$. If $I_{1}^{\prime}\leq h(e)$ holds, then there is
$g\in G_{0}$ with $h(g)=I_{1}^{\prime}$, where $e\in G$ is the identity.

  2. (2)

There are $g\in G_{0}$ and $\nu_{1}\in\mathcal{B}(I_{1}^{\prime})$ such that
$\phi R_{g}\phi\leq\nu_{1}\leq\min(\phi,R_{g}\phi)$ holds.

  3. (3)

Let $g\in G_{0}$ and $\nu_{1}\in\mathcal{B}(I_{1}^{\prime})$ as in (2). One
has $\max(\phi,R_{g}\phi)\leq\nu_{2}:=\phi+R_{g}\phi-\nu_{1}\leq 1$ and
$\nu_{2}\in\mathcal{B}(I_{2}^{\prime})$.

  4. (4)

Let $g\in G_{0}$ and $\nu_{1}\in\mathcal{B}(I_{1}^{\prime})$ as in (2), and
$\nu_{2}\in\mathcal{B}(I_{2}^{\prime})$ as in (3). One has
$\xi_{\phi}+\xi_{R_{g}\phi}\leq\xi_{\nu_{1}}+\xi_{\nu_{2}}$.

  5. (5)

Let $\nu_{1}\in\mathcal{B}(I_{1}^{\prime})$ as in (2), and
$\nu_{2}\in\mathcal{B}(I_{2}^{\prime})$ as in (3). One has
$2\|\xi_{\phi}\|\leq\|\xi_{\nu_{1}}\|+\|\xi_{\nu_{2}}\|$. In particular,
$2S(I)\leq S(I_{1}^{\prime})+S(I_{2}^{\prime})$ holds.

###### Proof.

  1. (1)

Since $h$ is continuous and $G_{0}$ is connected, it suffices to show that
there is $g^{\prime}\in G_{0}$ with $h(g^{\prime})\leq I_{1}^{\prime}$. We
divide the proof into the following two cases (1)(i) and (1)(ii).

    1. (i)

The case of $m(G)<\infty$:

$\displaystyle\|h\|_{G_{0}}=\int_{G_{0}}\int_{G}\phi(g^{\prime\prime})\phi(g^{\prime\prime}g^{\prime})dg^{\prime\prime}dg^{\prime}\leq
I^{2}$

holds, where $\|h\|_{G_{0}}$ denotes the $L^{1}$-norm of $h$ on $G_{0}$. Since
$\mathrm{vol}(G_{0})=m(G)$ holds by Remark 2.4 (3), there is $g^{\prime}\in
G_{0}$ with

$\displaystyle h(g^{\prime})\leq\frac{I^{2}}{m(G)}\leq I_{1}^{\prime}.$

    2. (ii)

The case of $m(G)=\infty$:

There is a compact set $K\subset G$ with $\|\phi\|_{K}\geq
I-I_{1}^{\prime}/2$. Since $G_{0}$ is not compact by Remark 2.4 (3), there is
$g^{\prime}\in G_{0}\setminus K^{-1}K$.

$\displaystyle h(g^{\prime})=\|\phi R_{g^{\prime}}\phi\|_{K}+\|\phi
R_{g^{\prime}}\phi\|_{G\setminus
K}\leq\|R_{g^{\prime}}\phi\|_{K}+\|\phi\|_{G\setminus K}$

holds by $\phi,R_{g^{\prime}}\phi\leq 1$. Since $Kg^{\prime}\subset G\setminus
K$ holds by $g^{\prime}\in G_{0}\setminus K^{-1}K$, one has

$\displaystyle\|R_{g^{\prime}}\phi\|_{K}=\|\phi\|_{Kg^{\prime}}\leq\|\phi\|_{G\setminus
K}.$

Thus, we obtain

$\displaystyle
h(g^{\prime})\leq\|R_{g^{\prime}}\phi\|_{K}+\|\phi\|_{G\setminus K}\leq
2\|\phi\|_{G\setminus K}\leq I_{1}^{\prime}.$

  2. (2)

Let $e\in G$ be the identity. When $I_{1}^{\prime}\leq h(e)$ holds, we obtain

$\displaystyle\nu_{1}:=\phi R_{g}\phi\leq\min(\phi,R_{g}\phi)$

by taking $g\in G_{0}$ in (1). When $I_{1}^{\prime}\geq h(e)$ holds, there is
$\nu_{1}\in\mathcal{B}(I_{1}^{\prime})$ with $\phi^{2}\leq\nu_{1}\leq\phi$.
Thus,

$\displaystyle\phi R_{g}\phi=\phi^{2}\leq\nu_{1}\leq\min(\phi,R_{g}\phi)=\phi$

holds by taking $g=e$.

  3. (3)

Since $\nu_{1}\leq\min(\phi,R_{g}\phi)$ holds, one has

$\displaystyle\nu_{2}\geq\phi+R_{g}\phi-\min(\phi,R_{g}\phi)=\max(\phi,R_{g}\phi).$

As

$\displaystyle\nu_{2}\leq\phi+R_{g}\phi-\phi
R_{g}\phi=1-(1-\phi)(1-R_{g}\phi)$

holds by $\nu_{1}\geq\phi R_{g}\phi$, one has $\nu_{2}\leq 1$ by
$\phi,R_{g}\phi\leq 1$. Thus, we obtain
$\nu_{2}\in\mathcal{B}(I_{2}^{\prime})$ by

$\displaystyle\|\nu_{2}\|=\|\phi\|+\|R_{g}\phi\|-\|\nu_{1}\|=I_{2}^{\prime}.$

  4. (4)

We fix $g^{\prime}\in G$ and define $u(\nu):=\phi_{1}*\nu(g^{\prime})$ for an
integrable function $\nu\colon G\to[0,1]$. Since $\nu_{1}\leq\phi\leq\nu_{2}$
and $\nu_{1}\leq R_{g}\phi\leq\nu_{2}$ hold by (2) and (3), one has

$u(\nu_{1})\leq u(\phi)\leq u(\nu_{2}),\quad u(\nu_{1})\leq u(R_{g}\phi)\leq
u(\nu_{2}).$ (4.2)

If $u(\nu_{1})=u(\nu_{2})$ holds, then the equalities are attained in (4.2)
and hence one has
$\xi_{\phi}(g^{\prime})+\xi_{R_{g}\phi}(g^{\prime})=\xi_{\nu_{1}}(g^{\prime})+\xi_{\nu_{2}}(g^{\prime})$.
Thus, it suffices to show the result when $u(\nu_{1})<u(\nu_{2})$ holds. One
has

$\displaystyle\xi_{\phi}(g^{\prime})$
$\displaystyle\leq\tilde{f}_{u(\nu_{1})}^{u(\nu_{2})}\circ u(\phi)$
$\displaystyle=\frac{u(\nu_{2}-\phi)\xi_{\nu_{1}}(g^{\prime})+u(\phi-\nu_{1})\xi_{\nu_{2}}(g^{\prime})}{u(\nu_{2}-\nu_{1})},$
$\displaystyle\xi_{R_{g}\phi}(g^{\prime})$
$\displaystyle\leq\tilde{f}_{u(\nu_{1})}^{u(\nu_{2})}\circ u(R_{g}\phi)$
$\displaystyle=\frac{u(\nu_{2}-R_{g}\phi)\xi_{\nu_{1}}(g^{\prime})+u(R_{g}\phi-\nu_{1})\xi_{\nu_{2}}(g^{\prime})}{u(\nu_{2}-\nu_{1})}$

by (4.2) and the convexity of $f$. Since $\nu_{2}=\phi+R_{g}\phi-\nu_{1}$
holds, one has

$\displaystyle\xi_{\phi}(g^{\prime})+\xi_{R_{g}\phi}(g^{\prime})$
$\displaystyle\leq\frac{u(2\nu_{2}-\phi-
R_{g}\phi)\xi_{\nu_{1}}(g^{\prime})+u(\phi+R_{g}\phi-2\nu_{1})(g^{\prime})\xi_{\nu_{2}}(g^{\prime})}{u(\nu_{2}-\nu_{1})}$
$\displaystyle=\xi_{\nu_{1}}(g^{\prime})+\xi_{\nu_{2}}(g^{\prime}).$

  5. (5)

Since $\phi_{1}*R_{g}\phi(g^{\prime})=\phi_{1}*\phi(g^{\prime}g)$ holds for
any $g^{\prime}\in G$, one has
$\xi_{R_{g}\phi}(g^{\prime})=\xi_{\phi}(g^{\prime}g)$. Thus,
$\|\xi_{R_{g}\phi}\|=\|\xi_{\phi}\|$ holds and hence we obtain
$2\|\xi_{\phi}\|\leq\|\xi_{\nu_{1}}\|+\|\xi_{\nu_{2}}\|$ by (4). In
particular, $2S(I)\leq S(I_{1}^{\prime})+S(I_{2}^{\prime})$ holds by taking
the upper bound on $\phi\in\mathcal{B}(I)$. ∎

###### Proof of Lemma 4.1.

  1. (1)

For any $\phi\in\mathcal{B}(I)$, there is
$\bar{\phi}\in\mathcal{B}(I^{\prime})$ with $\phi\leq\bar{\phi}$. Thus, one
has $S(I)\leq S(I^{\prime})$ by $\xi_{\phi}\leq\xi_{\bar{\phi}}$. On the other
hand, for any $\bar{\phi}\in\mathcal{B}(I^{\prime})$, there is
$\phi\in\mathcal{B}(I)$ with $\phi\leq\bar{\phi}$. Since
$\xi_{\bar{\phi}}-\xi_{\phi}\leq\phi_{1}*(\bar{\phi}-\phi)$ holds, one has

$\displaystyle\|\xi_{\bar{\phi}}\|-\|\xi_{\phi}\|\leq\|\phi_{1}*(\bar{\phi}-\phi)\|=I^{\prime}-I$

by Lemma 2.2. Thus, $S(I^{\prime})-S(I)\leq I^{\prime}-I$ holds. In
particular, $S$ is Lipschitz continuous.

  2. (2)

Let $0\leq J_{1}<J_{2}\leq m(G)$. Since $[J_{1},J_{2}]$ is compact, there is

$\displaystyle M:=\min_{J_{1}\leq I\leq J_{2}}\hat{S}_{J_{1}}^{J_{2}}(I)$

by (1). It suffices to show that $M=0$.

$\displaystyle\mathcal{C}:=\left\\{J_{1}\leq I\leq
J_{2}\;\middle|\;\hat{S}_{J_{1}}^{J_{2}}(I)=M\right\\}$

is closed by (1).

We will show that $\mathcal{C}\cap(J_{1},J_{2})$ is open. Suppose
$I\in\mathcal{C}$,

$\displaystyle\max(J_{1},I^{2}/m(G),2I-J_{2})<I_{1}^{\prime}\leq I$ (4.3)

and $I_{2}^{\prime}:=2I-I_{1}^{\prime}$. It suffices to show that
$\hat{S}_{J_{1}}^{J_{2}}(I_{1}^{\prime})=\hat{S}_{J_{1}}^{J_{2}}(I_{2}^{\prime})=M$.
Since $J_{1}<I_{1}^{\prime}\leq I_{2}^{\prime}<J_{2}$ holds by (4.3), one has

$2M\leq\hat{S}_{J_{1}}^{J_{2}}(I_{1}^{\prime})+\hat{S}_{J_{1}}^{J_{2}}(I_{2}^{\prime})=\tilde{S}_{J_{1}}^{J_{2}}(I_{1}^{\prime})-S(I_{1}^{\prime})+\tilde{S}_{J_{1}}^{J_{2}}(I_{2}^{\prime})-S(I_{2}^{\prime}).$
(4.4)

As

$\displaystyle\tilde{S}_{J_{1}}^{J_{2}}(I_{1}^{\prime})+\tilde{S}_{J_{1}}^{J_{2}}(I_{2}^{\prime})=\frac{(2J_{2}-I_{1}^{\prime}-I_{2}^{\prime})S(J_{1})+(I_{1}^{\prime}+I_{2}^{\prime}-2J_{1})S(J_{2})}{J_{2}-J_{1}}=2\tilde{S}_{J_{1}}^{J_{2}}(I)$

holds by $I_{1}^{\prime}+I_{2}^{\prime}=2I$, one has

$\tilde{S}_{J_{1}}^{J_{2}}(I_{1}^{\prime})-S(I_{1}^{\prime})+\tilde{S}_{J_{1}}^{J_{2}}(I_{2}^{\prime})-S(I_{2}^{\prime})\leq
2(\tilde{S}_{J_{1}}^{J_{2}}(I)-S(I))=2\hat{S}_{J_{1}}^{J_{2}}(I)$ (4.5)

by Lemma 4.2 (5). Since $\hat{S}_{J_{1}}^{J_{2}}(I)=M$ holds by
$I\in\mathcal{C}$, the equalities are attained in (4.4) and (4.5). Thus, one
has
$\hat{S}_{J_{1}}^{J_{2}}(I_{1}^{\prime})=\hat{S}_{J_{1}}^{J_{2}}(I_{2}^{\prime})=M$
and hence $\mathcal{C}\cap(J_{1},J_{2})$ is open.

Since $\mathcal{C}\cap(J_{1},J_{2})$ is clopen on the connected set
$(J_{1},J_{2})$, one has $\mathcal{C}\cap(J_{1},J_{2})=\emptyset$ or
$(J_{1},J_{2})\subset\mathcal{C}$. In the case of
$\mathcal{C}\cap(J_{1},J_{2})=\emptyset$, one has
$M=\hat{S}_{J_{1}}^{J_{2}}(J_{1})$ or $M=\hat{S}_{J_{1}}^{J_{2}}(J_{2})$ and
hence we obtain $M=0$ by

$\hat{S}_{J_{1}}^{J_{2}}(J_{1})=\hat{S}_{J_{1}}^{J_{2}}(J_{2})=0.$ (4.6)

In the case of $(J_{1},J_{2})\subset\mathcal{C}$, one has
$\mathcal{C}=[J_{1},J_{2}]$ because $\mathcal{C}$ is closed. Thus, we obtain
$M=0$ by (4.6).

  3. (3)

Since $\xi_{\phi}=0$ holds for any $\phi\in\mathcal{B}(t)$ by (2.5), one has
$S(t)=0$.

  4. (4)

As $\phi_{1}*\phi\leq 1$ holds by (2.4), one has
$\xi_{\phi}\leq\tilde{f}_{0}^{1}\circ(\phi_{1}*\phi)=(1-t)(\phi_{1}*\phi)$ by
the convexity of $f$. Since $\|\xi_{\phi}\|\leq(1-t)\|\phi_{1}*\phi\|=(1-t)I$
holds by Lemma 2.2, we obtain $S(I)\leq(1-t)I$. ∎

##  5 The proof of Lemma 1.2 when $m(G)=\infty$

In this section, we show (4.1) when $m(G)=\infty$ holds. It suffices to show

$S(I_{2})\leq(1-t)(I_{2}-t)$ (5.1)

for any $I_{2}\geq t$ by (4.1). If $I\geq I_{2}$ holds, then

$\displaystyle
S(I_{2})\leq\tilde{S}_{t}^{I}(I_{2})=\frac{(I-I_{2})S(t)+(I_{2}-t)S(I)}{I-t}$

holds by Lemma 4.1 (2). Besides, one has

$\displaystyle\frac{(I-I_{2})S(t)+(I_{2}-t)S(I)}{I-t}\leq\frac{(1-t)(I_{2}-t)I}{I-t}$

by Lemma 4.1 (3) and (4). Thus,

$\displaystyle
S(I_{2})\leq\lim_{I\to\infty}\frac{(1-t)(I_{2}-t)I}{I-t}=(1-t)(I_{2}-t)$

holds because $I_{2}\leq I<\infty(=m(G))$ is arbitrary. Therefore, when
$m(G)=\infty$ holds, we obtain (5.1) and hence Lemma 1.2.

##  6 The proof of Lemma 1.2 when $m(G)<\infty$

In this section, we show (4.1) when $m(G)<\infty$ holds. We simply write $f$
for $f_{t}$ in (1.3). It suffices to show (4.1) when $\|\phi_{1}\|=1$ and
$\|\phi_{2}\|>0$ hold by Remark 2.4 (6). In this case, $G_{0}\subset G$ is an
open subgroup because

$\displaystyle 0\leq 1-t\leq m(G)-\|\phi_{2}\|<m(G)=\mathrm{vol}(G_{0})$

holds by Remark 2.4 (3) and (1.5) (We note that $G^{\prime}$ is open if and
only if $\mathrm{vol}(G^{\prime})>0$ for any closed subgroup
$G^{\prime}\subset G$). Thus, we can take the Haar measure of $G_{0}$ to be
the same as of $G$. That is,

$\displaystyle\|\phi\|=\sum_{gG_{0}\in G/G_{0}}\|\phi\|_{gG_{0}}$ (6.1)

holds for any integrable function $\phi\colon G\to\mathbb{R}_{\geq 0}$.

We give a lemma (Lemma 6.2) to prove Lemma 1.2. For a measurable function
$\phi\colon G\to\mathbb{R}_{\geq 0}$, we define the function
$\phi^{*}\colon\mathbb{R}\times G/G_{0}\to\\{0,1\\}$.

######  Definition 6.1.

For a measurable function $\phi\colon G\to\mathbb{R}_{\geq 0}$, the function
$\phi^{*}\colon\mathbb{R}\times G/G_{0}\to\\{0,1\\}$ is defined as

$\displaystyle\phi^{*}(x,gG_{0}):=\phi_{(P(g))}(x),\quad
P(g):=\|\phi\|_{gG_{0}}=\int_{gG_{0}}\phi(g^{\prime})dg^{\prime},$

where $\phi_{(I)}$ is the same as in Example 2.1 (1).

Lemma 1.2 is deduced by the following lemma.

######  Lemma 6.2.

If integrable functions $\phi_{1},\phi_{2}\colon G\to[0,1]$ satisfy (1.5),
then

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|_{gG_{0}}\leq\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{gG_{0}\\}}$

holds for any $g\in G$. In particular, one has
$\|f\circ(\phi_{1}*\phi_{2})\|\leq\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|$.

The estimate
$\|f\circ(\phi_{1}*\phi_{2})\|\leq\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|$ is
followed from

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|$ $\displaystyle=\sum_{gG_{0}\in
G/G_{0}}\|f\circ(\phi_{1}*\phi_{2})\|_{gG_{0}}$
$\displaystyle\leq\sum_{gG_{0}\in
G/G_{0}}\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{gG_{0}\\}}$
$\displaystyle=\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|$

by (6.1).

######  Example 6.3.

If $G=G_{0}$ holds, then one has $\phi_{1}^{*}=\phi_{(1)}$ and
$\phi_{2}^{*}=\phi_{(\|\phi_{2}\|)}$ by identifying $\mathbb{R}\times G/G_{0}$
with $\mathbb{R}$. Thus,
$\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|=\|f\circ\psi_{1,\|\phi_{2}\|}\|$ holds
and hence Lemma 6.2 is equivalent to Lemma 1.2 and (4.1) by (2.2).

Since $m(\mathbb{R}\times G/G_{0})=\infty$, $\|\phi_{1}\|=\|\phi_{1}^{*}\|=1$
and $\|\phi_{2}\|=\|\phi_{2}^{*}\|$ hold, one has

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|\leq\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|\leq(1-t)(\|\phi_{2}\|-t)$

by the argument in Section 5. Thus, we obtain Lemma 1.2 by Lemma 6.2.

Lemma 6.2 is shown in two steps. We will show Lemma 6.2 in the case of
$G=G_{0}$ (that is, $G$ is connected) in Subsection 6.1 at first, and for
arbitrary $G$ in Subsection 6.2 by using the case of $G=G_{0}$.

###  6.1 The proof of Lemma 6.2 when $G=G_{0}$

In this subsection, we show Lemma 6.2 when $G=G_{0}$ holds. It suffices to
show (4.1) by Example 6.3.

$\displaystyle
S(I)\leq\tilde{S}_{t}^{I^{\prime}}(I)=\frac{(I^{\prime}-I)S(t)+(I-t)S(I^{\prime})}{I^{\prime}-t}$

holds by Lemma 4.1 (2) for $I^{\prime}:=m(G)+t-1$. Since $G=G_{0}$ and
$t=1+I^{\prime}-m(G)$ hold, one has

$\displaystyle S(I^{\prime})=I^{\prime}-tm(G)=(1-t)(I^{\prime}-t)$

by Lemma 3.1 (2). Thus,

$\displaystyle
S(I)\leq\frac{(I^{\prime}-I)S(t)+(I-t)S(I^{\prime})}{I^{\prime}-t}=(1-t)(I-t)$

holds by Lemma 4.1 (3). Therefore, we obtain (4.1) and Lemma 6.2 when
$G=G_{0}$ holds.

###  6.2 The proof of Lemma 6.2 for any $G$

In this subsection, we show Lemma 6.2 for any $G$. It suffices to show it when
$g=e$ by the following lemma and remark.

######  Lemma 6.4.

One has $(R_{g}\phi)^{*}=R_{(0,gG_{0})}(\phi^{*})$ for any integrable function
$\phi\colon G\to\mathbb{R}_{\geq 0}$ and $g\in G$, where $R_{g}\phi$ denotes
the same as in Lemma 4.2.

###### Proof.

Let $P(g)$ be the same as in Definition 6.1. One has

$\displaystyle(R_{g}\phi)^{*}(x,g^{\prime}G_{0})=\phi_{(\|R_{g}\phi\|_{g^{\prime}G_{0}})}(x)=\phi_{(P(g^{\prime}g))}(x)=\phi^{*}(x,g^{\prime}gG_{0})=R_{(0,gG_{0})}(\phi^{*})(x,g^{\prime}G_{0})$

for any $x\in\mathbb{R}$ and $g^{\prime}\in G$. ∎

######  Remark 6.5.

In Lemma 6.2,

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|_{gG_{0}}=\|f\circ
R_{g}(\phi_{1}*\phi_{2})\|_{G_{0}}=\|f\circ(\phi_{1}*R_{g}\phi_{2})\|_{G_{0}}$

and similarly

$\displaystyle\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{gG_{0}\\}}=\|f\circ(\phi_{1}^{*}*R_{(0,gG_{0})}(\phi_{2}^{*}))\|_{\mathbb{R}\times\\{G_{0}\\}}$

hold. Thus, one has

$\displaystyle\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{gG_{0}\\}}=\|f\circ(\phi_{1}^{*}*(R_{g}\phi_{2})^{*})\|_{\mathbb{R}\times\\{G_{0}\\}}$

by Lemma 6.4 and hence it suffices to show Lemma 6.2 in the case of $g=e$ by
replacing $\phi_{2}$ with $R_{g}\phi_{2}$.

We show Lemma 6.2 when $g=e$ holds. We define

$\displaystyle\phi_{1}*_{G_{0}}\phi_{2}(g^{\prime}):=\int_{G_{0}}\phi_{1}(g^{\prime\prime})\phi_{2}(g^{\prime\prime-1}g^{\prime})dg^{\prime\prime}$

for $g^{\prime}\in G_{0}$. The integrable functions
$L_{g^{\prime\prime}}\phi,R_{g^{\prime\prime}}\phi\colon G\to\mathbb{R}_{\geq
0}$ is defined as

$\displaystyle
L_{g^{\prime\prime}}\phi(g^{\prime}):=\phi(g^{\prime\prime-1}g^{\prime}),\quad
R_{g^{\prime\prime}}\phi(g^{\prime}):=\phi(g^{\prime}g^{\prime\prime})$

for $g^{\prime\prime}\in G$ and an integrable function $\phi\colon
G\to\mathbb{R}_{\geq 0}$. We give the following lemma to prove Lemma 6.2.

######  Lemma 6.6.

For any $g^{\prime}\in G_{0}$, one has

$\displaystyle\phi_{1}*\phi_{2}(g^{\prime})=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}(L_{g^{\prime\prime}}\phi_{1})*_{G_{0}}(R_{g^{\prime\prime}}\phi_{2})(g^{\prime\prime}g^{\prime}g^{\prime\prime-1}).$

###### Proof.

Since $g^{\prime\prime}G_{0}g^{\prime\prime-1}=G_{0}$ holds for any
$g^{\prime\prime}\in G$,

$\displaystyle\phi_{1}*\phi_{2}(g^{\prime})$
$\displaystyle=\int_{G}\phi_{1}(g^{\prime\prime\prime})\phi_{2}(g^{\prime\prime\prime-1}g^{\prime})dg^{\prime\prime\prime}$
$\displaystyle=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}\int_{G_{0}}\phi_{1}(g^{\prime\prime-1}g^{\prime\prime\prime\prime})\phi_{2}(g^{\prime\prime\prime\prime-1}g^{\prime\prime}g^{\prime})dg^{\prime\prime\prime\prime}$
$\displaystyle=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}\int_{G_{0}}L_{g^{\prime\prime}}\phi_{1}(g^{\prime\prime\prime\prime})R_{g^{\prime\prime}}\phi_{2}(g^{\prime\prime\prime\prime-1}g^{\prime\prime}g^{\prime}g^{\prime\prime-1})dg^{\prime\prime\prime\prime}$
$\displaystyle=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}(L_{g^{\prime\prime}}\phi_{1})*_{G_{0}}(R_{g^{\prime\prime}}\phi_{2})(g^{\prime\prime}g^{\prime}g^{\prime\prime-1})$

is obtained by (6.1). ∎

######  Example 6.7.

Let $P_{1}(g^{\prime\prime}):=\|\phi_{1}\|_{g^{\prime\prime}G_{0}}$ and
$P_{2}(g^{\prime\prime}):=\|\phi_{2}\|_{g^{\prime\prime}G_{0}}$ for
$g^{\prime\prime}\in G$. The identity component of $\mathbb{R}\times G/G_{0}$
is $\mathbb{R}\times\\{G_{0}\\}$ and hence

$\displaystyle\phi_{1}^{*}*\phi_{2}^{*}(x,G_{0})=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}\phi_{(P_{1}(g^{\prime\prime-1}))}*\phi_{(P_{2}(g^{\prime\prime}))}(x)=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}(x)$

holds for any $x\in\mathbb{R}$ by Lemma 6.6, where $\psi_{I_{1},I_{2}}$
denotes the same as in Example 2.1 (1). Since
$\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}(x)$ is an even
function and monotonically decreasing on $x\geq 0$, the convolution
$\phi_{1}^{*}*\phi_{2}^{*}(x,G_{0})$ is also an even function and
monotonically decreasing on $x\geq 0$ with respect to $x\in\mathbb{R}$.

We show Lemma 6.2 by Example 6.3, Remark 6.5, Lemma 6.6 and Example 6.7. It
suffices to show it when $g=e$ holds by Remark 6.5. We set
$t_{0}:=\phi_{1}^{*}*\phi_{2}^{*}(0,G_{0})$ and divide the proof into the
following two cases (1) and (2).

  1. (1)

The case of $t\leq t_{0}$:

Since $\phi_{1}^{*}*\phi_{2}^{*}(1+\|\phi_{2}\|,G_{0})=0\leq t\leq t_{0}$
holds and $\phi_{1}^{*}*\phi_{2}^{*}$ is continuous, there is $x_{0}\geq 0$
with $\phi_{1}^{*}*\phi_{2}^{*}(x_{0},G_{0})=t$ by the intermediate value
theorem. We set
$t_{g^{\prime\prime}}:=\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}(x_{0})$
for $g^{\prime\prime}\in G$.

$\displaystyle t=\sum_{g^{\prime\prime}G_{0}\in G/G_{0}}t_{g^{\prime\prime}}$

holds by Example 6.7 and hence one has

$\displaystyle\phi_{1}*\phi_{2}(g^{\prime})-t=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}(\omega_{g^{\prime\prime}}(g^{\prime\prime}g^{\prime}g^{\prime\prime-1})-t_{g^{\prime\prime}})$

for any $g^{\prime}\in G_{0}$ by Lemma 6.6, where we write
$\omega_{g^{\prime\prime}}:=(L_{g^{\prime\prime}}\phi_{1})*_{G_{0}}(R_{g^{\prime\prime}}\phi_{2})$.
Thus, one has

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|_{G_{0}}=\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}\int_{\\{g^{\prime}\in G_{0}\mid\phi_{1}*\phi_{2}(g^{\prime})\geq
t\\}}(\omega_{g^{\prime\prime}}(g^{\prime\prime}g^{\prime}g^{\prime\prime-1})-t_{g^{\prime\prime}})dg^{\prime}.$
(6.2)

######  Lemma 6.8.

For any $g^{\prime\prime}\in G$, one has

$\displaystyle\int_{\\{g^{\prime}\in
G_{0}\mid\phi_{1}*\phi_{2}(g^{\prime})\geq
t\\}}(\omega_{g^{\prime\prime}}(g^{\prime\prime}g^{\prime}g^{\prime\prime-1})-t_{g^{\prime\prime}})dg^{\prime}\leq\|\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}-t_{g^{\prime\prime}}\|_{(-x_{0},x_{0})}.$

###### Proof.

Since $\omega_{g^{\prime\prime}}-t_{g^{\prime\prime}}\leq
f_{t_{g^{\prime\prime}}}\circ\omega_{g^{\prime\prime}}$ and
$g^{\prime\prime}G_{0}g^{\prime\prime-1}=G_{0}$ hold, we have

$\int_{\\{g^{\prime}\in G_{0}\mid\phi_{1}*\phi_{2}(g^{\prime})\geq
t\\}}(\omega_{g^{\prime\prime}}(g^{\prime\prime}g^{\prime}g^{\prime\prime-1})-t_{g^{\prime\prime}})dg^{\prime}\leq\|f_{t_{g^{\prime\prime}}}\circ\omega_{g^{\prime\prime}}\|_{G_{0}}.$
(6.3)

$\|L_{g^{\prime\prime}}\phi_{1}\|_{G_{0}}\leq\|\phi_{1}\|=1$ and
$\|R_{g^{\prime\prime}}\phi_{2}\|_{G_{0}}\leq\|\phi_{2}\|$ hold and hence one
has

$\displaystyle\|L_{g^{\prime\prime}}\phi_{1}\|_{G_{0}}+\|R_{g^{\prime\prime}}\phi_{2}\|_{G_{0}}\leq
1+\|\phi_{2}\|\leq m(G)+t$

by (1.5). By Example 6.3 and Subsection 6.1, we have

$\displaystyle\|f_{t_{g^{\prime\prime}}}\circ\omega_{g^{\prime\prime}}\|_{G_{0}}\leq\|f_{t_{g^{\prime\prime}}}\circ\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}\|_{\mathbb{R}}=\|\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}-t_{g^{\prime\prime}}\|_{(-x_{0},x_{0})},$

where the last equality is due to the monotonicity of
$\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}$ and the definition
of $t_{g^{\prime\prime}}$. Thus, we obtain

$\displaystyle\int_{\\{g^{\prime}\in
G_{0}\mid\phi_{1}*\phi_{2}(g^{\prime})\geq
t\\}}(\omega_{g^{\prime\prime}}(g^{\prime\prime}g^{\prime}g^{\prime\prime-1})-t_{g^{\prime\prime}})dg^{\prime}$
$\displaystyle\leq\|f_{t_{g^{\prime\prime}}}\circ\omega_{g^{\prime\prime}}\|_{G_{0}}$
$\displaystyle\leq\|\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}-t_{g^{\prime\prime}}\|_{(-x_{0},x_{0})}$

by (6.3). ∎

Since

$\displaystyle\sum_{g^{\prime\prime}G_{0}\in
G/G_{0}}\|\psi_{P_{1}(g^{\prime\prime-1}),P_{2}(g^{\prime\prime})}-t_{g^{\prime\prime}}\|_{(-x_{0},x_{0})}=\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{G_{0}\\}}$

holds by Example 6.7, we obtain

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|_{G_{0}}\leq\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{G_{0}\\}}$

by (6.2) and Lemma 6.8.

  2. (2)

The case of $t\geq t_{0}$:

One has
$\|f\circ(\phi_{1}*\phi_{2})\|_{G_{0}}\leq\|f_{t_{0}}\circ(\phi_{1}*\phi_{2})\|_{G_{0}}$.
Besides,

$\displaystyle\|f_{t_{0}}\circ(\phi_{1}*\phi_{2})\|_{G_{0}}\leq\|f_{t_{0}}\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{G_{0}\\}}$

holds by (1). One has $\phi_{1}^{*}*\phi_{2}^{*}(x,G_{0})\leq t_{0}$ for any
$x\in\mathbb{R}$ by Example 6.7 and hence

$\displaystyle\|f\circ(\phi_{1}*\phi_{2})\|_{G_{0}}=\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|_{\mathbb{R}\times\\{G_{0}\\}}=0.$

Thus, we obtain
$\|f\circ(\phi_{1}*\phi_{2})\|\leq\|f\circ(\phi_{1}^{*}*\phi_{2}^{*})\|$.

##  7 The proof of Theorem 1.1 for any convex function $f$

In this section, we show Theorem 1.1 by Lemma 1.2. It suffices to show it when
$\|\phi_{1}\|=1$ holds by Remark 2.4 (6). The function
$f_{(n)}\colon[0,1]\to\mathbb{R}$ is defined as

$\displaystyle
f_{(n)}(y):=n\left(f\left(\frac{1}{n}\right)y+2\sum_{k=1}^{n-1}\hat{f}_{(k-1)/n}^{(k+1)/n}\left(\frac{k}{n}\right)f_{k/n}(y)\right)$

for $n\in\mathbb{Z}_{\geq 1}$ and function $f\colon[0,1]\to\mathbb{R}$, where
$\hat{f}_{y_{1}}^{y_{2}}$ denotes the same as in Definition 2.3 and $f_{t}$ in
(1.3). We show the following lemma to prove Theorem 1.1.

######  Lemma 7.1.

Let $f\colon[0,1]\to\mathbb{R}$ be a convex function with $f(0)=0$.

  1. (1)

One has $f_{(n)}(y)=\tilde{f}_{(j-1)/n}^{j/n}(y)$ for any $j=1,\cdots,n$ and
$(j-1)/n\leq y\leq j/n$, where $\tilde{f}_{y_{1}}^{y_{2}}$ denotes the same as
in Definition 2.3. In particular, $f(y)\leq f_{(n)}(y)$ holds for any $0\leq
y\leq 1$ and the equality is attained at $y=0,1/n,2/n,\cdots,1-1/n,1$ (Figure
1).

$\displaystyle\frac{j-1}{n}$$\displaystyle\frac{j}{n}$$f(y)$$f_{(n)}(y)$$y$
Figure 1: the graph of $f_{(n)}$

  2. (2)

For any $0<y<1$, one has

$\displaystyle f_{(n)}(y)\leq f(y)+\frac{\hat{f}_{0}^{1}(y)}{4ny(1-y)}.$

  3. (3)

One has $\lim_{n\to\infty}f_{(n)}(y)=f(y)$ for any $0\leq y\leq 1$. That is,
$f_{(n)}$ converges pointwise to $f$.

  4. (4)

One has $f_{(n)}(y)\leq f(1)y$ for any $n\in\mathbb{Z}_{\geq 1}$ and $0\leq
y\leq 1$.

  5. (5)

For any integrable function $\psi\colon G\to[0,1]$, one has

$\displaystyle\lim_{n\to\infty}\int_{G}f_{(n)}\circ\psi(g)dg=\int_{G}f\circ\psi(g)dg.$

###### Proof.

  1. (1)

Since $(j-1)/n\leq y\leq j/n$ holds, one has

$\displaystyle f_{k/n}(y)=\left\\{\begin{aligned}
&y-\frac{k}{n}&&\text{if}\;k\leq j-1\\\ &0&&\text{if}\;k\geq
j\end{aligned}\right.$

for any $k=1,\cdots,n$.

$\displaystyle\hat{f}_{(k-1)/n}^{(k+1)/n}\left(\frac{k}{n}\right)=\frac{1}{2}\left(f\left(\frac{k-1}{n}\right)+f\left(\frac{k+1}{n}\right)\right)-f\left(\frac{k}{n}\right)$

holds and hence one has

$\displaystyle
f_{(n)}(y)=nf\left(\frac{1}{n}\right)y+\sum_{k=1}^{j-1}\left(f\left(\frac{k-1}{n}\right)+f\left(\frac{k+1}{n}\right)-2f\left(\frac{k}{n}\right)\right)(ny-k).$

Since

$\displaystyle\sum_{k=1}^{j-1}f\left(\frac{k-1}{n}\right)(ny-k)=f\left(\frac{j-1}{n}\right)(j-ny)+\sum_{k=1}^{j-1}f\left(\frac{k}{n}\right)(ny-k-1)$

holds by $f(0)=0$ and

$\displaystyle
nf\left(\frac{1}{n}\right)y+\sum_{k=1}^{j-1}f\left(\frac{k+1}{n}\right)(ny-k)=f\left(\frac{j}{n}\right)(ny-j+1)+\sum_{k=1}^{j-1}f\left(\frac{k}{n}\right)(ny-k+1)$

holds, one has

$\displaystyle
f_{(n)}(y)=\tilde{f}_{(j-1)/n}^{j/n}(y)+\sum_{k=1}^{j-1}f\left(\frac{k}{n}\right)(ny-k-1+ny-k+1-2(ny-k))=\tilde{f}_{(j-1)/n}^{j/n}(y).$

In particular, $f(y)\leq f_{(n)}(y)$ holds for any $0\leq y\leq 1$ by the
convexity of $f$, and the equality is attained at
$y=0,1/n,2/n,\cdots,1-1/n,1$.

  2. (2)

It suffices to show the result when $j=1,\cdots,n$ and $(j-1)/n\leq y\leq j/n$
hold. One has

$f_{(n)}(y)\leq\tilde{f}_{(j-1)/n}^{j/n}(y)=(j-ny)f\left(\frac{j-1}{n}\right)+(ny-j+1)f\left(\frac{j}{n}\right)$
(7.1)

by (1).

$\displaystyle
f\left(\frac{j-1}{n}\right)\leq\tilde{f}_{0}^{y}\left(\frac{j-1}{n}\right)=\frac{(j-1)f(y)}{ny}$

holds by $0\leq(j-1)/n\leq y$, $f(0)=0$ and the convexity of $f$ and similarly

$\displaystyle
f\left(\frac{j}{n}\right)\leq\tilde{f}_{y}^{1}\left(\frac{j}{n}\right)=\frac{(n-j)f(y)+(j-ny)f(1)}{n(1-y)}$

holds by $y\leq j/n\leq 1$. Thus, one has

$\displaystyle\quad f_{(n)}(y)$
$\displaystyle\leq\frac{(j-ny)(j-1)f(y)}{ny}+\frac{(ny-j+1)((n-j)f(y)+(j-ny)f(1))}{n(1-y)}$
$\displaystyle=\frac{f(y)-f(1)y}{ny(1-y)}\cdot
j^{2}+\frac{(2ny+1)(f(1)y-f(y))}{ny(1-y)}\cdot
j+f(y)+\frac{(ny+1)(f(y)-yf(1))}{1-y}$
$\displaystyle=f(y)+\frac{f(y)-f(1)y}{ny(1-y)}\left(\left(j-ny-\frac{1}{2}\right)^{2}-\frac{1}{4}\right)$
$\displaystyle=f(y)+\frac{\hat{f}_{0}^{1}(y)(1-(2ny+1-2j)^{2})}{4ny(1-y)}.$

Since $\hat{f}_{0}^{1}(y)\geq 0$ holds by the convexity of $f$, we obtain

$\displaystyle f_{(n)}(y)\leq
f(y)+\frac{\hat{f}_{0}^{1}(y)(1-(2ny+1-2j)^{2})}{4ny(1-y)}\leq
f(y)+\frac{\hat{f}_{0}^{1}(y)}{4ny(1-y)}.$

  3. (3)

If $y=0,1$ holds, then one has $f_{(n)}(y)=f(y)$ by (1). Thus, it suffices to
show the result when $0<y<1$ holds. One has

$\displaystyle f(y)\leq f_{(n)}(y)\leq f(y)+\frac{a}{n},\quad
a:=\frac{\hat{f}_{0}^{1}(y)}{4y(1-y)}$

by (1) and (2).

$\displaystyle\lim_{n\to\infty}\left(f(y)+\frac{a}{n}\right)=f(y)$

holds and hence we obtain

$\displaystyle\lim_{n\to\infty}f_{(n)}(y)=f(y).$

  4. (4)

It suffices to show the result when $j=1,\cdots,n$ and $(j-1)/n\leq y\leq j/n$
hold. One has

$\displaystyle
f_{(n)}(y)\leq(j-ny)f\left(\frac{j-1}{n}\right)+(ny-j+1)f\left(\frac{j}{n}\right)$

by (7.1). Besides,

$\displaystyle
f\left(\frac{j-1}{n}\right)\leq\tilde{f}_{0}^{1}\left(\frac{j-1}{n}\right)=\frac{(j-1)f(1)}{n},\quad
f\left(\frac{j}{n}\right)\leq\tilde{f}_{0}^{1}\left(\frac{j}{n}\right)=\frac{jf(1)}{n}$

holds by $f(0)=0$ and the convexity of $f$. Thus, one has

$\displaystyle
f_{(n)}(y)\leq\frac{(j-ny)(j-1)f(1)}{n}+\frac{(ny-j+1)jf(1)}{n}=f(1)y.$

  5. (5)

Since $f_{(n)}\geq f$ holds by (1), one has

$\liminf_{n\to\infty}\int_{G}f_{(n)}\circ\psi(g)dg\geq\int_{G}f\circ\psi(g)dg.$
(7.2)

One has $f_{(n)}\circ\psi(g)\leq f(1)\psi(g)$ for any $n\in\mathbb{Z}_{\geq
1}$ by (4). Since $\psi$ is integrable,

$\limsup_{n\to\infty}\int_{G}f_{(n)}\circ\psi(g)dg\leq\int_{G}f\circ\psi(g)dg$
(7.3)

holds by applying Fatou’s lemma to $f(1)\psi-f_{(n)}\circ\psi$. Thus, we
obtain

$\displaystyle\lim_{n\to\infty}\int_{G}f_{(n)}\circ\psi(g)dg=\int_{G}f\circ\psi(g)dg$

by (7.2) and (7.3). ∎

###### Proof of Theorem 1.1.

It suffices to show the result when $\|\phi_{1}\|=1$ holds by Remark 2.4 (6).
Since

$\displaystyle\|\phi_{1}*\phi_{2}\|=\|\psi_{1,\|\phi_{2}\|}\|=\|\phi_{2}\|<\infty$
(7.4)

holds by Lemma 2.2, one has

$\displaystyle\int_{G}f\circ(\phi_{1}*\phi_{2})(g)dg=\lim_{n\to\infty}\int_{G}f_{(n)}\circ(\phi_{1}*\phi_{2})(g)dg$
(7.5)

by Lemma 7.1 (5). By the definition of $f_{(n)}$,

$\displaystyle\int_{G}f_{(n)}\circ(\phi_{1}*\phi_{2})(g)dg$
$\displaystyle=n\left(f\left(\frac{1}{n}\right)\|\phi_{1}*\phi_{2}\|+2\sum_{k=1}^{n-1}\hat{f}_{(k-1)/n}^{(k+1)/n}\left(\frac{k}{n}\right)\|f_{k/n}\circ(\phi_{1}*\phi_{2})\|\right)$
(7.6)

holds. Since (2.2) holds by Lemma 1.2 and
$\hat{f}_{(k-1)/n}^{(k+1)/n}(k/n)\geq 0$ holds by the convexity of $f$, one
has

$\displaystyle\hat{f}_{(k-1)/n}^{(k+1)/n}\left(\frac{k}{n}\right)\|f_{k/n}\circ(\phi_{1}*\phi_{2})\|\leq\hat{f}_{(k-1)/n}^{(k+1)/n}\left(\frac{k}{n}\right)\|f_{k/n}\circ\psi_{1,\|\phi_{2}\|}\|$

for any $k=1,\cdots,n-1$. Thus,

$\displaystyle\int_{G}f_{(n)}\circ(\phi_{1}*\phi_{2})(g)dg$ $\displaystyle\leq
n\left(f\left(\frac{1}{n}\right)\|\psi_{1,\|\phi_{2}\|}\|+2\sum_{k=1}^{n-1}\hat{f}_{(k-1)/n}^{(k+1)/n}\left(\frac{k}{n}\right)\|f_{k/n}\circ\psi_{1,\|\phi_{2}\|}\|\right)$
$\displaystyle=\int_{\mathbb{R}}f_{(n)}\circ\psi_{1,\|\phi_{2}\|}(x)dx$ (7.7)

holds by (7.4) and (7.6). Since

$\displaystyle\lim_{n\to\infty}\int_{\mathbb{R}}f_{(n)}\circ\psi_{1,\|\phi_{2}\|}(x)dx=\int_{\mathbb{R}}f\circ\psi_{1,\|\phi_{2}\|}(x)dx$
(7.8)

holds by Lemma 7.1 (5), we obtain

$\displaystyle\int_{G}f\circ(\phi_{1}*\phi_{2})(g)dg\leq\int_{\mathbb{R}}f\circ\psi_{1,\|\phi_{2}\|}(x)dx=2\int_{0}^{1}f(y)dy+(\|\phi_{2}\|-1)f(1)$

by (2.1), (7.5), (7.7) and (7.8). ∎

## Acknowledgement

This work was supported by JSPS KAKENHI Grant Number JP19J22628 and Leading
Graduate Course for Frontiers of Mathematical Sciences and Physics (FMSP). The
author would like to thank his advisor Toshiyuki Kobayashi for his support.
The author is also grateful to Yuichiro Tanaka and Toshihisa Kubo for helpful
comments. The author also would like to thank the reviewers for the careful
comments.

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Takashi Satomi: Graduate School of Mathematical Sciences, The University of
Tokyo, 3-8-1 Komaba Meguro-ku Tokyo 153-8914, Japan.

E-mail<EMAIL_ADDRESS>


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