# Functions on a convex set which are both $\omega$-semiconvex and
$\omega$-semiconcave II

Václav Kryštof  Charles University, Faculty of Mathematics and Physics,
Sokolovská 83, 186 75 Praha 8, Karlín, Czech Republic
<EMAIL_ADDRESS>
###### Abstract.

In a recent article (2022) we proved with L. Zajíček that if
$G\subset\mathbb{R}^{n}$ is an unbounded open convex set that does not contain
a translation of a convex cone with non-empty interior, then there exist
$f:G\to\mathbb{R}$ and a concave modulus $\omega$ such that
$\lim_{t\to\infty}\omega(t)=\infty$, $f$ is both semiconvex and semiconcave
with modulus $\omega$ and $f\notin C^{1,\omega}(G)$. Here we improve the
previous result as follows: If $G$ is as above and $\omega(t)=t^{\alpha}$ for
some $\alpha\in(0,1)$, then there exists $f:G\to\mathbb{R}$ that is both
semiconvex and semiconcave with modulus $\omega$ and $f\notin
C^{1,\alpha}(G)$. This result has immediate consequences concerning a first-
order quantitative converse Taylor theorem and the problem whether $f\in
C^{1,\alpha}(G)$ whenever $f$ is smooth in a corresponding sense on all lines.

###### Key words and phrases:

Semiconvex function with general modulus, semiconcave function with general
modulus, $C^{1,\alpha}$ function, $C^{1,\omega}$ function, unbounded open
convex set.

###### 1991 Mathematics Subject Classification:

26B25.

##  1\. Introduction

If $G\subset\mathbb{R}^{n}$ is an open convex set, then $f:G\to\mathbb{R}$ is
said to be (classically) semiconvex if there exists $C\geq 0$ such that the
function

$\displaystyle x\mapsto f(x)+C|x|^{2},\quad x\in G,$

is convex. We say that $f:G\to\mathbb{R}$ is (classically) semiconcave if $-f$
is semiconvex. Then it is well-known (see [KZ, Section 3] for references) that

(1) $\displaystyle f:G\to\mathbb{R}\;\textnormal{is both semiconvex and
semiconcave}\;\Longleftrightarrow f\in C^{1,1}(G).$

The notion of semiconvex, resp. semiconcave, functions is generalized to the
notion of funtions that are semiconvex, resp. semiconcave, with modulus
$\omega$ (see [CS, Definition 2.1.1]). We will call these functions
$\omega$-semiconvex, resp. $\omega$-semiconcave, for short (see Definition 2.1
and Definition 2.3 below). Note that a function is semiconvex, resp.
semiconcave, if and only if it is $\omega$-semiconvex, resp.
$\omega$-semiconcave, with $\omega(t)=Ct$ for some $C\geq 0$ (see [CS,
Proposition 1.1.3]). For the theory and application of $\omega$-semiconvex,
resp. $\omega$-semiconcave, functions, see the monograph [CS].

This article deals with functions that are both $\omega$-semiconvex and
$\omega$-semiconcave. Let $G\subset\mathbb{R}^{n}$ be an open convex set and
$\omega$ a modulus. It is easy to show that if $f\in C^{1,\omega}(G)$, then
there exists $C>0$ such that $f$ is both $(C\omega)$-semiconvex and
$(C\omega)$-semiconcave (see [CS, Proposition 2.1.2]). So we will consider the
natural problem whether also the opposite implication holds. This implication
can be equivalently formulated as follows:

(2) $\displaystyle f:G\to\mathbb{R}\;\textnormal{is
both}\;\omega\textnormal{-semiconvex
and}\;\omega\textnormal{-semiconcave}\Longrightarrow f\in C^{1,\omega}(G).$

Recall (see (1)) that implication (2) holds if $\omega(t)=Ct$ for some $C\geq
0$. Further, by Remark 2.4 (i) below and [KZ, p. 839], implication (2) holds
if either $G$ is bounded or contains a translation of a convex cone with non-
empty interior. However, (2) doesn’t hold in general. Namely, we proved (see
[KZ, Theorem 1.9]) that if $G\subset\mathbb{R}^{n}$ is an unbounded open
convex set which does not contain a translation of a convex cone with non-
empty interior, then there exists a concave modulus $\omega$ such that
$\lim_{t\to\infty}\omega(t)=\infty$ and (2) doesn’t hold. The main
disadvantage of [KZ, Theorem 1.9] is that $\omega$ depends on $G$ and may be
irregular. In this article we improve [KZ, Theorem 1.9] in the following way:

######  Theorem 1.1.

Let $G\subset\mathbb{R}^{n}$ ($n\geq 2$) be an unbounded open convex set that
doesn’t contain a translation of a cone with non-empty interior. Let
$\alpha\in(0,1)$ and set $\omega(t):=t^{\alpha}$ for all $t\geq 0$. Then there
exists $f:G\to\mathbb{R}$ that is both $\omega$-semiconvex and
$\omega$-semiconcave and $f\notin C^{1,\alpha}(G)$.

So Theorem 1.1 gives a positive answer to [KZ, Question 6.10 (3), (4)]. By
(31), Theorem 1.1 is a special case of Theorem 5.5 below which works with more
general moduli (namely those with property (* ‣ 5) below).

The strategy of the proof of Theorem 5.5 is similar to that used in [KZ] where
we reduced (using the notion of recessive cones) the case of an open convex
subset of $\mathbb{R}^{n}$ to the case of an open convex subset of
$\mathbb{R}^{2}$ of the special form
$G_{\eta}:=\\{(x,y)\in\mathbb{R}^{2}:x>0,\;|y|<\eta(x)\\}$ where
$\eta:(0,\infty)\to(0,\infty)$ is concave and satisfies
$\lim_{t\to\infty}\eta(t)/t=0$. However, the construction of
$f:G_{\eta}\to\mathbb{R}$ in the proof of Proposition 4.4 is much more
sophisticated than the corresponding construction used in [KZ]. The present
construction is based on the notion of modulus $\omega_{\eta}$ (see Definition
3.1 below and for the motivation Remark 3.2 below).

It’s worth mentioning that if $\omega$ is a modulus, then functions that are
both $(C\omega)$-semiconvex and $(C\omega)$-semiconcave for some $C\geq 0$ can
be characterized via “Taylor approximation of the first order” or via
“$C^{1,\omega}$-smoothness on all lines”. See Lemma 2.5 below for a precise
formulation. Thus Theorem 1.1 can be reformulated as follows:

######  Corollary 1.2.

Let $G\subset\mathbb{R}^{n}$ ($n\geq 2$) be an unbounded open convex set that
doesn’t contain a translation of a cone with non-empty interior. Let
$\alpha\in(0,1)$ and set $\omega(t):=t^{\alpha}$ for all $t\geq 0$. Then there
exists $f:G\to\mathbb{R}$ that is Fréchet differentiable,

  * (i)

for every $a\in G$ and $v\in\mathbb{R}^{n}$, $|v|=1$, the derivative of the
function $t\mapsto f(a+tv)$ is uniformly continuous on
$\\{t\in\mathbb{R}:a+tv\in G\\}$ with modulus $\omega$,

  * (ii)

$|f(x+h)-f(x)-f^{\prime}(x)[h]|\leq|h|^{1+\alpha}$ for all $x,x+h\in G$ and

  * (iii)

$f\notin C^{1,\alpha}(G)$.

Let us remark that Corollary 1.2 improves [KZ, Theorem 6.8]. Note also that
Theorem 5.5 can be reformulated (using Lemma 2.5) in the same way as Theorem
1.1 and so we can get a more general statement than Corollary 1.2.

##  2\. Preliminaries

Throughout this article, the norm on $\mathbb{R}^{n}$ is always Euclidean. For
$A\subset\mathbb{R}^{n}$ the symbol $\operatorname{span}(A)$ denotes the
linear span of $A$. By a cone in $\mathbb{R}^{n}$ we mean a set
$C\subset\mathbb{R}^{n}$ such that $\lambda x\in C$ for each $x\in C$ and
$\lambda>0$. If $G\subset\mathbb{R}^{n}$ is an open set, $f:G\to\mathbb{R}$,
$x\in G$ and $h\in\mathbb{R}^{n}$, then $f^{\prime}(x)$ denotes the Fréchet
derivative of $f$ at $x$ and $f^{\prime}(x)[h]$ denotes its evaluation at $h$.

######  Definition 2.1.

We denote by $\mathcal{M}$ the set of all $\omega:[0,\infty)\to[0,\infty)$
which are non-decreasing and satisfy $\lim_{h\to 0+}\omega(h)=0$. The members
of $\mathcal{M}$ are called moduli.

######  Definition 2.2.

Let $G\subset\mathbb{R}^{n}$ be an open set, $\omega\in\mathcal{M}$ and
$\alpha\in(0,1]$. Then we denote by $C^{1,\omega}(G)$ the set of all Fréchet
differentiable $f:G\to\mathbb{R}$ such that $f^{\prime}$ is uniformly
continuous with modulus $C\omega$ for some $C>0$. If $\omega(t)=t^{\alpha}$
for every $t\geq 0$, then we also write $C^{1,\alpha}(G)$ instead of
$C^{1,\omega}(G)$.

######  Definition 2.3.

Let $G\subset\mathbb{R}^{n}$ be an open convex set and $\omega\in\mathcal{M}$.

  * •

We say that $f:G\to\mathbb{R}$ is semiconvex with modulus $\omega$ (or
$\omega$-semiconvex for short) if

$f(\lambda x+(1-\lambda)y)\leq\lambda
f(x)+(1-\lambda)f(y)+\lambda(1-\lambda)|x-y|\omega(|x-y|)$

for every $x,y\in G$ and $\lambda\in[0,1]$.

  * •

We say that $f:G\to\mathbb{R}$ is semiconcave with modulus $\omega$ (or
$\omega$-semiconcave) if $-f$ is semiconvex with modulus $\omega$.

######  Remark 2.4.

Let $G\subset\mathbb{R}^{n}$ be an open convex set, $f:G\to\mathbb{R}$ and
$\omega\in\mathcal{M}$. Then the following hold (see [CS, Theorem 2.1.7] and
[CS, Theorem 3.3.7]):

  * (i)

If $f$ is $\omega$-semiconvex or $\omega$-semiconcave, then $f$ is continuous.

  * (ii)

If $f$ is both $\omega$-semiconvex and $\omega$-semiconcave, then $f\in
C^{1}(G)$.

As already mentioned in the introduction, we have the following:

######  Lemma 2.5.

Let $G\subset\mathbb{R}^{n}$ be an open convex set, $f:G\to\mathbb{R}$ and
$\omega\in\mathcal{M}$. Then the following are equivalent:

  * (i)

There exists $C_{1}>0$ such that $f$ is both $(C_{1}\omega)$-semiconvex and
$(C_{1}\omega)$-semiconcave.

  * (ii)

There exists $C_{2}>0$ such that for every $a\in G$ and $v\in\mathbb{R}^{n}$,
$|v|=1$, the function $t\mapsto f(a+tv)$ is differentiable on
$\\{t\in\mathbb{R}:a+tv\in G\\}$ and its derivative is uniformly continuous
with modulus $C_{2}\omega$.

  * (iii)

$f$ is Fréchet differentiable and there exists $C_{3}>0$ such that

$\displaystyle|f(x+h)-f(x)-f^{\prime}(x)[h]|\leq C_{3}|h|\omega(|h|),\quad
x,x+h\in G.$

###### Proof.

It follows from Remark 2.4 (i) and [KZ, Proposition 6.4 (i), (iv)] that (i)
implies (ii) and (iii). Further, it follows easily from [K, Lemma 2.3 (i)] and
[K, Proposition 2.4 (ii)] that (ii) implies (i). Finally, (iii) implies (i) by
[KZ, Proposition 6.4 (ii)]. ∎

Recall that $\omega\in\mathcal{M}$ is called sub-additive if
$\omega(x+h)\leq\omega(x)+\omega(h)$ for all $x,h\geq 0$.

######  Remark 2.6.

Let $\omega\in\mathcal{M}$. Then the following hold:

  * (i)

If $\omega$ is concave, then it is sub-additive.

  * (ii)

If $\omega$ is sub-additive, then there exists a concave
$\varphi\in\mathcal{M}$ such that $\omega\leq\varphi\leq 2\omega$.

Both of these statements are well-known. The proof of (i) is easy. Statement
(ii) is due to Stechkin (see [DZ, Lemma 2.3] for references).

######  Remark 2.7.

Let $f:(0,\infty)\to[0,\infty)$ be concave. Then the following hold:

  * (i)

$f$ is non-decreasing.

  * (ii)

The function $t\mapsto f(t)/t$, $t>0$, is non-increasing.

  * (iii)

For every $t>0$ and $c\geq 1$ we have $f(ct)\leq cf(t)$.

All these facts are easy and well-known. Note that (iii) is just a
reformulation of (ii). However, a useful one.

As in [KZ] we will use the following auxiliary notation:

######  Definition 2.8.

Let $n\in\mathbb{N}$ and $\omega\in\mathcal{M}$. Then we denote by
$\mathcal{G}_{n}(\omega)$ the set of all open convex sets
$G\subset\mathbb{R}^{n}$ for which there exist $a\in G$,
$x\in\mathbb{R}^{n}\setminus\\{0\\}$ and $f\in C^{1}(G)$ such that $f$ is both
$\omega$-semiconvex and $\omega$-semiconcave, $M:=\\{a+\lambda x:\lambda\geq
0\\}\subset G$ and $f^{\prime}$ is uniformly continuous on $M$ with modulus
$C\omega$ for no $C>0$.

######  Remark 2.9.

Let $n\in\mathbb{N}$ and $\omega\in\mathcal{M}$. Then:

  * (i)

By Remark 2.4 (ii) we can equivalently write $f:G\to\mathbb{R}$ instead of
$f\in C^{1}(G)$ in the definition of $\mathcal{G}_{n}(\omega)$.

  * (ii)

If $G\in\mathcal{G}_{n}(\omega)$, then (2) clearly doesn’t hold.

  * (iii)

If $b\in\mathbb{R}^{n}$ and $G+b\in\mathcal{G}_{n}(\omega)$, then it is easy
to check that $G\in\mathcal{G}_{n}(\omega)$.

For the proof of the following lemma see [KZ, Example 5.1].

######  Lemma 2.10.

Let $\omega\in\mathcal{M}$ satisfy

$\displaystyle\liminf_{h\to
0+}\frac{\omega(h)}{h}>0,\;\lim_{h\to\infty}\frac{\omega(h)}{h}=0.$

Then $\mathbb{R}\times(0,1)\in\mathcal{G}_{2}(\omega)$.

As in [KZ], we will work with recession cones of unbounded convex sets
$A\subset\mathbb{R}^{n}$ and use the notation $\operatorname{rec}(A)$ (instead
of $0^{+}A$ used in [R]).

######  Definition 2.11.

Let $\emptyset\neq A\subset\mathbb{R}^{n}$ be a convex set. Then we set

$\displaystyle\operatorname{rec}(A):=\\{x\in\mathbb{R}^{n}:(\forall a\in
A)(\forall\lambda\geq 0)\;a+\lambda x\in A\\}.$

Now we recall some basic properties of recession cones. For the proof see [KZ,
Lemma 2.10].

######  Lemma 2.12.

Let $n\in\mathbb{N}$ and let $\emptyset\neq G\subset\mathbb{R}^{n}$ be an open
convex set. Then the following hold:

  * (i)

$\operatorname{rec}(G)=\operatorname{rec}(\overline{G})$.

  * (ii)

$\operatorname{rec}(G)$ is a closed convex cone and
$0\in\operatorname{rec}(G)$.

  * (iii)

$\operatorname{rec}(G)=\\{x\in\mathbb{R}^{n}:(\exists a\in
A)(\forall\lambda\geq 0)\;a+\lambda x\in A\\}$.

  * (iv)

If $G\subset\mathbb{R}^{n}$ is unbounded, then
$1\leq\dim(\operatorname{span}(\operatorname{rec}(G)))$.

  * (v)

$\dim(\operatorname{span}(\operatorname{rec}(G)))=n$ if and only if $G$
contains a translation of a cone with non-empty interior.

We will also need the following special case of [KZ, Lemma 5.11]:

######  Lemma 2.13.

Let $m\in\mathbb{N}$, $m\geq 2$, and $\omega\in\mathcal{M}$. Let
$G\subset\mathbb{R}^{m}$ be an open convex set and
$L:\mathbb{R}^{m}\to\mathbb{R}^{2}$ a linear surjection. Suppose that $\omega$
is concave, $L(G)\in\mathcal{G}_{2}(\omega)$ and
$\operatorname{rec}(L(G))\subset L(\operatorname{rec}(G))$. Then
$G\in\mathcal{G}_{m}(\omega)$.

The following lemma is an easy consequence of [R, Theorem 9.1].

######  Lemma 2.14.

Let $m,n\in\mathbb{N}$, let $G\subset\mathbb{R}^{n}$ be an unbounded open
convex set and let $L:\mathbb{R}^{n}\to\mathbb{R}^{m}$ be a linear surjection
such that

$\displaystyle\operatorname{rec}(G)\cap
L^{-1}(\\{0\\})\subset\operatorname{rec}(G)\cap(-\operatorname{rec}(G)).$

Then $\operatorname{rec}(L(G))=L(\operatorname{rec}(G))$.

###### Proof.

By Lemma 2.12 (i) we have
$\operatorname{rec}(G)=\operatorname{rec}(\overline{G})$. So
$\operatorname{rec}(\overline{G})\cap
L^{-1}(\\{0\\})\subset\operatorname{rec}(\overline{G})\cap(-\operatorname{rec}(\overline{G}))$
and thus [R, Theorem 9.1] (applied to $C:=\overline{G}$) implies that
$L(\overline{G})$ is closed and
$\operatorname{rec}(L(\overline{G}))=L(\operatorname{rec}(\overline{G}))$.
Hence we easily obtain that $L(\overline{G})=\overline{L(G)}$. Therefore,
$\operatorname{rec}(L(\overline{G}))=\operatorname{rec}(L(G))$ by Lemma 2.12
(i) and the equality $\operatorname{rec}(L(G))=L(\operatorname{rec}(G))$
follows. ∎

##  3\. Modulus $\omega_{\eta}$

One of the main ingredients of this article is the introduction of modulus
$\omega_{\eta}$:

######  Definition 3.1.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\omega\in\mathcal{M}$. Then for every
$h>0$ we set

(3)
$\displaystyle\omega_{\eta}(h):=\inf\bigg{\\{}\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})\bigg{\\}},$

where the infimum is taken over all partitions $0=x_{0}<x_{1}<\cdots<x_{n}=h$
of $[0,h]$. Further we set $\omega_{\eta}(0):=0$.

The fact that $\omega_{\eta}$ is indeed a modulus will be proved in Lemma 3.4
(i) below.

######  Remark 3.2.

The original motivation for $\omega_{\eta}$ was that the following statement
holds:

  * •

Let $\eta:(0,\infty)\to(0,\infty)$ be concave, let $\omega\in\mathcal{M}$ and
set

$\displaystyle G:=\\{(x,y)\in\mathbb{R}^{2}:x>0,\;|y|<\eta(x)\\}.$

Let $f:G\to\mathbb{R}$ be both $\omega$-semiconvex and $\omega$-semiconcave.
Then $f\in C^{1,\omega_{\eta}}(G)$.

However, we do not need this statement and therefore not prove it. Note also
that $\omega_{\eta}$ is “minimal” in the sense of Proposition 4.4 below which
we use substantially at the end of the article.

######  Remark 3.3.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\omega\in\mathcal{M}$. Then

$\displaystyle\omega_{\eta}(h)\leq\max\bigg{\\{}1,\frac{h}{\eta(h)}\bigg{\\}}\omega(h),\quad
h>0.$

Indeed, taking $n=1$ in the sum in (3), we immediately obtain the estimate.

######  Lemma 3.4.

Let $\eta:(0,\infty)\to(0,\infty)$ be concave and let $\omega\in\mathcal{M}$.
Then the following hold:

  * (i)

$\omega_{\eta}\in\mathcal{M}$.

  * (ii)

For every $x,h>0$ we have
$\omega_{\eta}(x+h)\leq\omega_{\eta}(x)+\omega_{\eta}(h)$.

###### Proof.

(i): Obviously $\omega_{\eta}(h)\geq 0$ for every $h\geq 0$. By Remark 3.3 and
Remark 2.7 (ii), we have

$\displaystyle\omega_{\eta}(h)\leq\max\bigg{\\{}1,\frac{h}{\eta(h)}\bigg{\\}}\omega(h)\leq\max\bigg{\\{}1,\frac{1}{\eta(1)}\bigg{\\}}\omega(h),\quad
h\in(0,1),$

and thus $\lim_{h\to 0+}\omega_{\eta}(h)=\omega_{\eta}(0)=0$. To prove that
$\omega_{\eta}$ is non-decreasing, it is sufficient to show

(4) $\displaystyle\omega_{\eta}(x)\leq\omega_{\eta}(x+h),\quad x,h>0.$

Let $x,h>0$ and let $0=x_{0}<x_{1}<\cdots<x_{n}=x+h$ be a partition of
$[0,x+h]$. Then there exists $m\in\mathbb{N}$ such that $m\leq n$ and
$x_{m-1}<x\leq x_{m}$. According to Remark 2.7 (i), (ii), we have

$\displaystyle\frac{x-x_{m-1}}{\eta(x)}=\frac{x}{\eta(x)}-\frac{x_{m-1}}{\eta(x)}\leq\frac{x_{m}}{\eta(x_{m})}-\frac{x_{m-1}}{\eta(x_{m})}=\frac{x_{m}-x_{m-1}}{\eta(x_{m})}$

and thus

$\displaystyle\omega_{\eta}(x)$
$\displaystyle\leq\max\bigg{\\{}1,\frac{x-x_{m-1}}{\eta(x)}\bigg{\\}}\omega(x-x_{m-1})$
$\displaystyle+\sum_{j=1}^{m-1}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle\leq\sum_{j=1}^{m}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle\leq\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1}).$

Hence (4) holds.

(ii): Let $x,h>0$ and $\varepsilon>0$. Then there exists a partition
$0=x_{0}<x_{1}<\cdots<x_{n}=x$ of $[0,x]$ such that

$\displaystyle\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})\leq\omega_{\eta}(x)+\frac{\varepsilon}{2},$

and a partition $0=y_{0}<y_{1}<\cdots<y_{m}=h$ of $[0,h]$ such that

$\displaystyle\sum_{i=1}^{m}\max\bigg{\\{}1,\frac{y_{i}-y_{i-1}}{\eta(y_{i})}\bigg{\\}}\omega(y_{i}-y_{i-1})\leq\omega_{\eta}(h)+\frac{\varepsilon}{2}.$

Hence, for $x_{j}:=x+y_{j-n}$ ($j=n+1,\dots,n+m$) we obtain using Remark 2.7
(i)

$\displaystyle\omega_{\eta}(x+h)\leq\sum_{j=1}^{n+m}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle=\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})+\sum_{i=1}^{m}\max\bigg{\\{}1,\frac{y_{i}-y_{i-1}}{\eta(x+y_{i})}\bigg{\\}}\omega(y_{i}-y_{i-1})$
$\displaystyle\leq\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})+\sum_{i=1}^{m}\max\bigg{\\{}1,\frac{y_{i}-y_{i-1}}{\eta(y_{i})}\bigg{\\}}\omega(y_{i}-y_{i-1})$
$\displaystyle\leq\omega_{\eta}(x)+\omega_{\eta}(h)+\varepsilon.$

Thus (ii) holds. ∎

In the rest of this section we will prove several properties of
$\omega_{\eta}$ that we will need further.

######  Lemma 3.5.

Let $\eta:(0,\infty)\to(0,\infty)$ be non-decreasing and let
$\omega\in\mathcal{M}$ be concave. Then

(5) $\displaystyle\frac{h\omega(\eta(h))}{\eta(h)}\leq\omega_{\eta}(h),\quad
h>0.$

###### Proof.

Let $h>0$ and $\varepsilon>0$. Then there exists a partition
$0=x_{0}<x_{1}<\cdots<x_{n}=h$ of $[0,h]$ such that

$\displaystyle\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})\leq\omega_{\eta}(h)+\varepsilon.$

Set

$\displaystyle J_{1}:=\\{j\in\\{1,\dots,n\\}:\eta(x_{j})\leq
x_{j}-x_{j-1}\\},$ $\displaystyle
J_{2}:=\\{j\in\\{1,\dots,n\\}:x_{j}-x_{j-1}<\eta(x_{j})\\}.$

Using Remark 2.7 (ii) we obtain

$\displaystyle(x_{j}-x_{j-1})\frac{\omega(\eta(h))}{\eta(h)}$
$\displaystyle\leq(x_{j}-x_{j-1})\frac{\omega(\eta(x_{j}))}{\eta(x_{j})}$
$\displaystyle\leq\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\omega(x_{j}-x_{j-1})$
$\displaystyle\leq\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1}),\quad
j\in J_{1},$

and

$\displaystyle(x_{j}-x_{j-1})\frac{\omega(\eta(h))}{\eta(h)}$
$\displaystyle\leq(x_{j}-x_{j-1})\frac{\omega(x_{j}-x_{j-1})}{x_{j}-x_{j-1}}$
$\displaystyle\leq\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1}),\quad
j\in J_{2}.$

Consequently

$\displaystyle\frac{h\omega(\eta(h))}{\eta(h)}$ $\displaystyle=\sum_{j\in
J_{1}}(x_{j}-x_{j-1})\frac{\omega(\eta(h))}{\eta(h)}+\sum_{j\in
J_{2}}(x_{j}-x_{j-1})\frac{\omega(\eta(h))}{\eta(h)}$
$\displaystyle\leq\sum_{j\in
J_{1}}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle+\sum_{j\in
J_{2}}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle=\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})\leq\omega_{\eta}(h)+\varepsilon.$

Hence (5) holds. ∎

######  Lemma 3.6.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\omega\in\mathcal{M}$. Then

(6)
$\displaystyle\omega_{\eta}(x+h)\leq\omega_{\eta}(x)+\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h),\quad
x,h>0.$

###### Proof.

Let $x,h>0$ and $\varepsilon>0$. Then there exists a partition
$0=x_{0}<x_{1}<\cdots<x_{n}=x$ of $[0,x]$ such that

$\displaystyle\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})\leq\omega_{\eta}(x)+\varepsilon.$

Hence, for $x_{n+1}:=x+h$ we have

$\displaystyle\omega_{\eta}(x+h)$
$\displaystyle\leq\sum_{j=1}^{n+1}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle=\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h)+\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle\leq\omega_{\eta}(x)+\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h)+\varepsilon.$

Consequently (6) holds. ∎

######  Lemma 3.7.

Let $\eta:(0,\infty)\to(0,\infty)$ be non-decreasing and let
$\omega\in\mathcal{M}$. Then

(7)
$\displaystyle\omega_{\eta}(x+h)\leq\omega_{\eta}(x)+2\frac{h\omega(\eta(x))}{\eta(x)},\quad
x,h>0,h\geq\frac{\eta(x)}{2}.$

###### Proof.

Let $x,h>0$, $h\geq\eta(x)/2$, and $\varepsilon>0$. Then there exists a
partition $0=x_{0}<x_{1}<\cdots<x_{n}=x$ of $[0,x]$ such that

$\displaystyle\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})\leq\omega_{\eta}(x)+\varepsilon.$

Denote by $m$ the upper integer part of $h/\eta(x)$ and set
$x_{j}:=x+(j-n)h/m$ for each $j=n+1,\dots,n+m$. Then

$\displaystyle\frac{m}{2}\leq\frac{h}{\eta(x)}\leq m$

and $h/m\leq\eta(x)\leq\eta(x_{j})$ for each $j=n+1,\dots,n+m$. Thus

$\displaystyle\omega_{\eta}(x+h)$
$\displaystyle\leq\sum_{j=1}^{n+m}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle=\sum_{j=1}^{n}\max\bigg{\\{}1,\frac{x_{j}-x_{j-1}}{\eta(x_{j})}\bigg{\\}}\omega(x_{j}-x_{j-1})$
$\displaystyle+\sum_{j=n+1}^{n+m}\max\bigg{\\{}1,\frac{\frac{h}{m}}{\eta(x_{j})}\bigg{\\}}\omega\bigg{(}\frac{h}{m}\bigg{)}$
$\displaystyle\leq\omega_{\eta}(x)+\varepsilon+\sum_{j=1}^{m}\omega\bigg{(}\frac{h}{m}\bigg{)}\leq\omega_{\eta}(x)+\varepsilon+\sum_{j=1}^{m}\omega(\eta(x))$
$\displaystyle=\omega_{\eta}(x)+m\omega(\eta(x))+\varepsilon\leq\omega_{\eta}(x)+2\frac{h\omega(\eta(x))}{\eta(x)}+\varepsilon.$

Hence (7) holds. ∎

##  4\. The case $\mathbb{R}^{2}$

We begin with three lemmas that we need to prove important Proposition 4.4
below.

######  Lemma 4.1.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\omega\in\mathcal{M}$ be both concave,
let $g:(0,\infty)\to\mathbb{R}$ be differentiable and let $C>0$. Suppose that

(8) $\displaystyle|g(x+h)-g(x)|$ $\displaystyle\leq
C\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h),\quad x,h>0,$ (9)
$\displaystyle\eta(x)|g^{\prime}(x)|$ $\displaystyle\leq
C\omega(\eta(x)),\quad x>0,$ (10)
$\displaystyle\eta(x)|g^{\prime}(x)-g^{\prime}(x+h)|$ $\displaystyle\leq
C\omega(h),\quad x,h>0,$

and set

$\displaystyle G$
$\displaystyle:=\\{(x,y)\in\mathbb{R}^{2}:x>0,\;|y|<\eta(x)\\},$
$\displaystyle f(x,y)$ $\displaystyle:=g(x)y,\quad(x,y)\in G.$

Then the function $f$ is both $(5C\omega)$-semiconvex and
$(5C\omega)$-semiconcave.

###### Proof.

By [KZ, Lemma 2.3] it is sufficient to show that

$|f^{\prime}(x+h,y+k)[h,k]-f^{\prime}(x,y)[h,k]|\\\ \leq
5C|(h,k)|\omega(|(h,k)|),\quad(x,y),(x+h,y+k)\in G,h\geq 0.$

Let $(x,y),(x+h,y+k)\in G$, $h\geq 0$. Then, according to Remark 2.7 (i), we
have

(11) $\displaystyle|k|\leq|y+k|+|y|\leq\eta(x+h)+\eta(x)\leq 2\eta(x+h)$

and thus by (9) and Remark 2.7 (ii)

$\displaystyle|k||g^{\prime}(x+h)|$
$\displaystyle=|k|\frac{\eta(x+h)|g^{\prime}(x+h)|}{\eta(x+h)}\leq
C|k|\frac{\omega(\eta(x+h))}{\eta(x+h)}$ $\displaystyle\leq
C|k|\frac{\omega\big{(}\frac{|k|}{2}\big{)}}{\frac{|k|}{2}}\leq
2C\omega(|k|).$

Further, by (10) we have

$\displaystyle|y||g^{\prime}(x+h)-g^{\prime}(x)|\leq\eta(x)|g^{\prime}(x+h)-g^{\prime}(x)|\leq
C\omega(h).$

Consequently

(12)
$|g^{\prime}(x+h)(y+k)-g^{\prime}(x)y|\leq|y||g^{\prime}(x+h)-g^{\prime}(x)|+|k||g^{\prime}(x+h)|\\\
\leq C\omega(h)+2C\omega(|k|)\leq 3C\omega(|(h,k)|).$

Since by (11) we have

$\displaystyle|k|\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}=\max\bigg{\\{}|k|,\frac{|k|}{\eta(x+h)}h\bigg{\\}}\leq
2\max\\{|k|,h\\},$

we obtain by (8)

(13) $|k||g(x+h)-g(x)|\leq
C|k|\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h)\\\ \leq
2C\max\\{|k|,h\\}\omega(h)\leq 2C|(h,k)|\omega(|(h,k)|).$

Hence (12) and (13) give

$\displaystyle|f^{\prime}(x+h,y+k)[h,k]-f^{\prime}(x,y)[h,k]|$
$\displaystyle\leq|h|\bigg{|}\frac{\partial f}{\partial
x}(x+h,y+k)-\frac{\partial f}{\partial
x}(x,y)\bigg{|}+|k|\bigg{|}\frac{\partial f}{\partial
y}(x+h,y+k)-\frac{\partial f}{\partial y}(x,y)\bigg{|}$
$\displaystyle=|h||g^{\prime}(x+h)(y+k)-g^{\prime}(x)y|+|k||g(x+h)-g(x)|$
$\displaystyle\leq
3C|(h,k)|\omega(|(h,k)|)+2C|(h,k)|\omega(|(h,k)|)=5C|(h,k)|\omega(|(h,k)|).$

∎

######  Lemma 4.2.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\omega\in\mathcal{M}$ be both concave.
Define the function $\psi$ by

$\psi(x):=\sup\\{\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x_{2}):x_{1},x_{2}\geq
0,\lambda\in[0,1],\\\ x=\lambda x_{1}+(1-\lambda)x_{2}\\},\quad x\geq 0.$

Then $\psi\in\mathcal{M}$, $\psi$ is concave, $\omega_{\eta}\leq\psi\leq
2\omega_{\eta}$ and

(14) $\displaystyle\psi(t+\eta(t))\leq\psi(t)+2\omega(\eta(t)),\quad
t>0,\eta(t)\leq t.$

###### Proof.

By Remark 2.6 (ii) and Lemma 3.4, there exists a concave
$\varphi\in\mathcal{M}$ such that $0\leq\omega_{\eta}\leq\varphi\leq
2\omega_{\eta}$. Note that $\psi$ is the upper concave envelope of
$\omega_{\eta}$ (see [R, Corollary 17.1.5]). In particular, $\psi$ is concave
and $0\leq\omega_{\eta}\leq\psi\leq\varphi$. Thus $\psi\leq 2\omega_{\eta}$,
$\lim_{x\to 0+}\psi(x)=0$ and it follows from Remark 2.7 (i) that $\psi$ is
non-decreasing. Therefore, $\psi\in\mathcal{M}$.

Let $t>0$, $x_{1},x_{2}\geq 0$ and $\lambda\in[0,1]$ be such that $\eta(t)\leq
t$, $x_{1}\leq x_{2}$ and $\lambda x_{1}+(1-\lambda)x_{2}=t+\eta(t)$. Then to
prove (14) it is sufficient to show

$\displaystyle\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x_{2})\leq\psi(t)+2\omega(\eta(t)).$

Note also that $\eta$ is non-decreasing by Remark 2.7 (i).

If $t\leq x_{1}$, then $\eta(t)\leq t\leq x_{1}\leq x_{2}$ and thus by Lemma
3.6 (with $h=\eta(t)$ and $x=x_{1}-\eta(t)$ or $x=x_{2}-\eta(t)$)

$\displaystyle\omega_{\eta}(x_{i})$
$\displaystyle\leq\omega_{\eta}(x_{i}-\eta(t))+\max\bigg{\\{}1,\frac{\eta(t)}{\eta(x_{i})}\bigg{\\}}\omega(\eta(t))$
$\displaystyle=\omega_{\eta}(x_{i}-\eta(t))+\omega(\eta(t)),\quad i=1,2.$

Hence

$\displaystyle\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x_{2})$
$\displaystyle\leq\lambda\omega_{\eta}(x_{1}-\eta(t))+(1-\lambda)\omega_{\eta}(x_{2}-\eta(t))+\omega(\eta(t))$
$\displaystyle\leq\lambda\psi(x_{1}-\eta(t))+(1-\lambda)\psi(x_{2}-\eta(t))+\omega(\eta(t))$
$\displaystyle\leq\psi(\lambda(x_{1}-\eta(t))+(1-\lambda)(x_{2}-\eta(t)))+\omega(\eta(t))$
$\displaystyle=\psi(t)+\omega(\eta(t))\leq\psi(t)+2\omega(\eta(t)).$

Now suppose that $x_{1}\leq t$. Then $\lambda<1$ and therefore we may set

$\displaystyle
h:=\frac{\eta(t)}{1-\lambda},\;x:=x_{2}-\frac{\eta(t)}{1-\lambda}.$

Then $(1-\lambda)t=t-\lambda t\leq t-\lambda
x_{1}=(1-\lambda)x_{2}-\eta(t)=(1-\lambda)x$ and thus

(15) $\displaystyle t\leq x.$

In particular, $0<x$. If $h\leq\eta(x+h)$, then by Lemma 3.6 and Remark 2.7
(iii)

$\displaystyle\omega_{\eta}(x_{2})$
$\displaystyle=\omega_{\eta}(x+h)\leq\omega_{\eta}(x)+\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h)$
$\displaystyle=\omega_{\eta}(x)+\omega\bigg{(}\frac{\eta(t)}{1-\lambda}\bigg{)}\leq\omega_{\eta}(x)+\frac{\omega(\eta(t))}{1-\lambda}$

and thus

$\displaystyle\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x_{2})$
$\displaystyle\leq\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x)+\omega(\eta(t))$
$\displaystyle\leq\lambda\psi(x_{1})+(1-\lambda)\psi(x)+\omega(\eta(t))$
$\displaystyle\leq\psi(\lambda x_{1}+(1-\lambda)x)+\omega(\eta(t))$
$\displaystyle=\psi(t)+\omega(\eta(t))\leq\psi(t)+2\omega(\eta(t)).$

If $\eta(x+h)\leq h$, then by Lemma 3.7, Remark 2.7 (ii) and (15) we have

$\displaystyle\omega_{\eta}(x_{2})$
$\displaystyle=\omega_{\eta}(x+h)\leq\omega_{\eta}(x)+2h\frac{\omega(\eta(x))}{\eta(x)}$
$\displaystyle\leq\omega_{\eta}(x)+2h\frac{\omega(\eta(t))}{\eta(t)}=\omega_{\eta}(x)+2\frac{\omega(\eta(t))}{1-\lambda}$

and therefore

$\displaystyle\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x_{2})$
$\displaystyle\leq\lambda\omega_{\eta}(x_{1})+(1-\lambda)\omega_{\eta}(x)+2\omega(\eta(t))$
$\displaystyle\leq\lambda\psi(x_{1})+(1-\lambda)\psi(x)+2\omega(\eta(t))$
$\displaystyle\leq\psi(\lambda x_{1}+(1-\lambda)x)+2\omega(\eta(t))$
$\displaystyle=\psi(t)+2\omega(\eta(t)).$

∎

######  Lemma 4.3.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\psi\in\mathcal{M}$ be both concave,
let $\omega\in\mathcal{M}$ and let $C>0$. Suppose that

(16) $\displaystyle\psi(t+\eta(t))-\psi(t)$ $\displaystyle\leq
C\omega(\eta(t)),\quad t>0,\;\eta(t)\leq t,$ (17) $\displaystyle\psi(h)$
$\displaystyle\leq C\max\bigg{\\{}1,\frac{h}{\eta(h)}\bigg{\\}}\omega(h),\quad
h>0.$

Then

$\displaystyle\psi(x+h)-\psi(x)$ $\displaystyle\leq
2C\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h),\quad x,h>0.$

###### Proof.

Let $x,h>0$. We will distinguish four cases.

(a) $x\leq h$: Then $\eta(x+h)\leq\eta(2h)\leq 2\eta(h)$ by Remark 2.7 (i),
(iii). And so, using Remark 2.6 (i) and (17), we obtain

$\displaystyle\psi(x+h)-\psi(x)\leq\psi(h)\leq
C\max\bigg{\\{}1,\frac{h}{\eta(h)}\bigg{\\}}\omega(h)$ $\displaystyle\leq
C\max\bigg{\\{}1,\frac{2h}{\eta(x+h)}\bigg{\\}}\omega(h)\leq
2C\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h).$

(b) $h\leq\eta(h)$: Then, according to Remark 2.6 (i) and (17), we have

$\displaystyle\psi(x+h)-\psi(x)$ $\displaystyle\leq\psi(h)\leq
C\max\bigg{\\{}1,\frac{h}{\eta(h)}\bigg{\\}}\omega(h)=C\omega(h)$
$\displaystyle\leq 2C\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h).$

(c) $\eta(x)\leq h\leq x$: Then $\eta(x+h)\leq\eta(2x)\leq 2\eta(x)$ by Remark
2.7 (i), (iii). Thus, using the concavity of $\psi$ and (16), we obtain

$\displaystyle\psi(x+h)-\psi(x)=h\frac{\psi(x+h)-\psi(x)}{h}\leq
h\frac{\psi(x+\eta(x))-\psi(x)}{\eta(x)}$ $\displaystyle\leq
C\frac{h}{\eta(x)}\omega(\eta(x))\leq 2C\frac{h}{\eta(x+h)}\omega(h)\leq
2C\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h).$

(d) Suppose that none of cases (a), (b) or (c) holds: Then clearly $h\leq x$
and $\eta(h)\leq h\leq\eta(x)$. Since $\eta$ is continuous, there exists
$t\in[h,x]$ such that $\eta(t)=h$. Hence, using the concavity of $\psi$ and
(16), we obtain

$\displaystyle\psi(x+h)-\psi(x)$
$\displaystyle=\eta(t)\frac{\psi(x+\eta(t))-\psi(x)}{\eta(t)}\leq\eta(t)\frac{\psi(t+\eta(t))-\psi(t)}{\eta(t)}$
$\displaystyle\leq C\omega(\eta(t))\leq
2C\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h).$

∎

Now we can prove the crucial proposition of this article, which we will need
to prove Theorem 5.5 below. However, we believe that it is also interesting in
itself.

######  Proposition 4.4.

Let $\eta:(0,\infty)\to(0,\infty)$ and $\omega\in\mathcal{M}$ be both concave.
Suppose that $\omega(x)>0$ for all $x>0$, and set

$\displaystyle G:=\\{(x,y)\in\mathbb{R}^{2}:x>0,\;|y|<\eta(x)\\}.$

Then there exists a concave and strictly positive $g\in C^{1}((0,\infty))$
such that the function $f$ defined by

$\displaystyle f(x,y)=g(x)y,\quad(x,y)\in G,$

has the following properties:

  * (i)

$f$ is both $\omega$-semiconvex and $\omega$-semiconcave.

  * (ii)

If $\varphi\in\mathcal{M}$ satisfies
$\liminf_{x\to\infty}\varphi(x)/\omega_{\eta}(x)=0$, then there is no $C>0$
such that

(18) $\displaystyle\bigg{|}\frac{\partial f}{\partial y}(x,0)-\frac{\partial
f}{\partial y}(1,0)\bigg{|}\leq C\varphi(|x-1|),\quad x>0,$

in particular, $f\notin C^{1,\varphi}(G)$.

###### Proof.

Firstly, note that $\eta$ is non-decreasing by Remark 2.7 (i). So it follows
from Lemma 3.5 that $\omega_{\eta}(x)>0$ for all $x>0$. Hence by Lemma 4.2
there exists a concave $\psi\in\mathcal{M}$ such that (16) holds with $C=2$
and

(19) $\displaystyle 0<\omega_{\eta}(x)\leq\psi(x)\leq 2\omega_{\eta}(x),\quad
x>0.$

Further, it follows from (19) and Remark 3.3 that (17) holds with $C=2$. Hence
by Lemma 4.3 we have

(20) $\displaystyle\psi(x+h)-\psi(x)$ $\displaystyle\leq
4\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h),\quad x,h>0.$

Note that since $\psi$ is concave, we have

$\displaystyle\psi(x)=\int_{0}^{x}\psi^{\prime}_{+}(s)ds,\quad x\geq 0.$

Since $\psi\in\mathcal{M}$ and $\psi(x)>0$ for every $x>0$, there exists $a>0$
such that

(21) $\displaystyle\psi^{\prime}_{+}(4a)>0.$

By Remark 2.7 (ii) we have

(22) $\displaystyle\frac{\eta(x)}{x}\leq\frac{\eta(a)}{a}\leq
q:=\max\bigg{\\{}1,\frac{\eta(a)}{a}\bigg{\\}},\quad x\geq a.$

Denote by $\delta$ the real function on $[a,\infty)$ that is affine on each
$[2^{n-1}a,2^{n}a]$ ($n\in\mathbb{N}$) and

$\displaystyle\delta(2^{n}a)=\psi^{\prime}_{+}(2^{n}a),\quad
n\in\mathbb{N}\cup\\{0\\}.$

Then $\delta$ is continuous and non-negative. Since $\psi^{\prime}_{+}$ is
non-increasing, $\delta$ is also non-increasing. Hence for every
$n\in\mathbb{N}$ and $x\in[2^{n-1}a,2^{n}a]$ we have

$\displaystyle\psi^{\prime}_{+}(2x)\leq\psi^{\prime}_{+}(2^{n}a)=\delta(2^{n}a)\leq\delta(x)\leq\delta(2^{n-1}a)=\psi^{\prime}_{+}(2^{n-1}a)\leq\psi^{\prime}_{+}\bigg{(}\frac{x}{2}\bigg{)}$

and thus

(23)
$\displaystyle\psi^{\prime}_{+}(2x)\leq\delta(x)\leq\psi^{\prime}_{+}\bigg{(}\frac{x}{2}\bigg{)},\quad
x\geq a.$

We set $b:=1280q^{2}$ and

$\displaystyle g(x):=\frac{1}{b}\int_{a}^{a+x}\delta(s)ds,\quad x>0.$

Then $g$ is concave and $g\in C^{1}((0,\infty))$. Furthermore, for all $x>0$
we obtain using (23) and (21)

$\displaystyle g(x)$
$\displaystyle\geq\frac{1}{b}\int_{a}^{\min\\{a+x,2a\\}}\delta(s)ds\geq\frac{1}{b}\int_{a}^{\min\\{a+x,2a\\}}\psi^{\prime}_{+}(2s)ds$
$\displaystyle\geq\frac{1}{b}\int_{a}^{\min\\{a+x,2a\\}}\psi^{\prime}_{+}(4a)ds>0.$

(i): Note that $g^{\prime}$ is non-increasing and, by Remark 2.7 (i), also
non-negative. Note also that $8/b\leq 8q/b\leq 1/5$. Hence by Lemma 4.1 (with
$C=1/5$) it is sufficient to show that

(24) $\displaystyle g(x+h)-g(x)$
$\displaystyle\leq\frac{8}{b}\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h),\quad
x,h>0,$ (25) $\displaystyle\eta(x)g^{\prime}(x)$
$\displaystyle\leq\frac{8q}{b}\omega(\eta(x)),\quad x>0,$ (26)
$\displaystyle\eta(x)(g^{\prime}(x)-g^{\prime}(x+h))$
$\displaystyle\leq\frac{1}{5}\omega(h),\quad x,h>0.$

Inequality (24): For all $x,h>0$ we have by (23), (20) and Remark 2.7 (iii)

$\displaystyle g(x+h)-g(x)$
$\displaystyle=\frac{1}{b}\int_{a+x}^{a+x+h}\delta(s)ds\leq\frac{1}{b}\int_{a+x}^{a+x+h}\psi^{\prime}_{+}\bigg{(}\frac{s}{2}\bigg{)}ds=\frac{2}{b}\int_{\frac{a+x}{2}}^{\frac{a+x+h}{2}}\psi^{\prime}_{+}(t)dt$
$\displaystyle=\frac{2}{b}\bigg{(}\psi\bigg{(}\frac{a+x+h}{2}\bigg{)}-\psi\bigg{(}\frac{a+x}{2}\bigg{)}\bigg{)}\leq\frac{8}{b}\max\bigg{\\{}1,\frac{\frac{h}{2}}{\eta(\frac{a+x+h}{2})}\bigg{\\}}\omega\bigg{(}\frac{h}{2}\bigg{)}$
$\displaystyle\leq\frac{8}{b}\max\bigg{\\{}1,\frac{\frac{h}{2}}{\frac{\eta(a+x+h)}{2}}\bigg{\\}}\omega\bigg{(}\frac{h}{2}\bigg{)}\leq\frac{8}{b}\max\bigg{\\{}1,\frac{h}{\eta(x+h)}\bigg{\\}}\omega(h).$

Inequality (25): Let $x>0$. Then we have by (22)

$\displaystyle t:=\frac{\eta(x)}{2q}\leq\frac{\eta(a+x)}{2q}\leq\frac{a+x}{2}$

and by Remark 2.7 (iii)

$\displaystyle\frac{t}{\eta(\frac{a+x}{2})}=\frac{\eta(x)}{2q\eta(\frac{a+x}{2})}\leq\frac{\eta(\frac{x}{2})}{q\eta(\frac{a+x}{2})}\leq\frac{1}{q}\leq
1.$

Consequently, using the concavity of $\psi$, (23) and (20), we obtain

$\displaystyle\eta(x)g^{\prime}(x)$
$\displaystyle=\frac{1}{b}\eta(x)\delta(a+x)\leq\frac{1}{b}\eta(x)\psi^{\prime}_{+}\bigg{(}\frac{a+x}{2}\bigg{)}$
$\displaystyle\leq\frac{\eta(x)}{b}\dfrac{\psi(\frac{a+x}{2})-\psi(\frac{a+x}{2}-t)}{t}=\frac{2q}{b}\bigg{(}\psi\bigg{(}\frac{a+x}{2}\bigg{)}-\psi\bigg{(}\frac{a+x}{2}-t\bigg{)}\bigg{)}$
$\displaystyle\leq\frac{8q}{b}\max\bigg{\\{}1,\frac{t}{\eta(\frac{a+x}{2})}\bigg{\\}}\omega(t)=\frac{8q}{b}\omega\bigg{(}\frac{\eta(x)}{2q}\bigg{)}\leq\frac{8q}{b}\omega(\eta(x)).$

Inequality (26): By (25) and Remark 2.7 (iii) we have

(27)
$\eta(u)(g^{\prime}(u)-g^{\prime}(v))\leq\eta(u)g^{\prime}(u)\leq\frac{8q}{b}\omega(\eta(u))\leq\frac{8q}{b}\omega(4q(v-u))\\\
\leq\frac{32q^{2}}{b}\omega(v-u)=\frac{1}{40}\omega(v-u),\quad
0<u<v,\eta(u)\leq 4q(v-u).$

Set

$\displaystyle x_{n}:=2^{n}a-a,\quad n\in\mathbb{N}\cup\\{0\\}.$

Then by Remark 2.7 (iii) we have

$\displaystyle\eta(x_{n})\leq 2\eta\bigg{(}\frac{x_{n}}{2}\bigg{)}\leq
2\eta\bigg{(}\frac{x_{n-1}+x_{n}}{2}\bigg{)},\quad n\in\mathbb{N},$

and by (22)

$\displaystyle\eta\bigg{(}\frac{x_{n-1}+x_{n}}{2}\bigg{)}\leq\eta(x_{n})\leq
qx_{n}\leq q\cdot 2^{n}a=4q\frac{x_{n}-x_{n-1}}{2},\quad n\in\mathbb{N}.$

Hence by (27) (for every $n\in\mathbb{N}$ applied with $u=(x_{n-1}+x_{n})/2$
and $v=x_{n}$) we obtain

(28)
$q_{n}:=\frac{g^{\prime}(\frac{x_{n-1}+x_{n}}{2})-g^{\prime}(x_{n})}{x_{n}-\frac{x_{n-1}+x_{n}}{2}}\leq\frac{1}{40\eta(\frac{x_{n-1}+x_{n}}{2})}\frac{\omega(\frac{x_{n}-x_{n-1}}{2})}{\frac{x_{n}-x_{n-1}}{2}}\\\
\leq\frac{1}{20\eta(\frac{x_{n-1}+x_{n}}{2})}\frac{\omega(x_{n}-x_{n-1})}{x_{n}-x_{n-1}}\leq\frac{1}{10\eta(x_{n})}\frac{\omega(x_{n}-x_{n-1})}{x_{n}-x_{n-1}},\quad
n\in\mathbb{N}.$

Since $g^{\prime}$ is affine on $(x_{0},x_{1}]$ and on each $[x_{n-1},x_{n}]$
($n\in\mathbb{N}$, $n>1$), we obtain by (28) and Remark 2.7 (ii)

(29)
$\eta(u)(g^{\prime}(u)-g^{\prime}(v))=\eta(u)q_{n}(v-u)\leq\frac{\eta(u)}{10\eta(x_{n})}\frac{\omega(x_{n}-x_{n-1})}{x_{n}-x_{n-1}}(v-u)\\\
\leq\frac{\eta(u)}{10\eta(x_{n})}\frac{\omega(v-u)}{v-u}(v-u)\leq\frac{1}{10}\omega(v-u),\quad
n\in\mathbb{N},u,v\in[x_{n-1},x_{n}],0<u<v.$

Let $x,h>0$. Then there exists $n\in\mathbb{N}$ such that $x_{n-1}\leq x\leq
x_{n}$. If $\eta(x)\leq 4qh$ or $x+h\leq x_{n}$, then (26) holds by (27) or
(29). Hence we may further suppose that $4qh\leq\eta(x)$ and $x_{n}\leq x+h$.
Then we have by (22)

$\displaystyle x+h\leq x+\frac{\eta(x)}{4q}\leq
x_{n}+\frac{\eta(x_{n})}{4q}\leq x_{n}+\frac{x_{n}}{4}\leq 2x_{n}\leq x_{n+1}$

and thus, using two-times (29), we obtain

$\displaystyle\eta(x)(g^{\prime}(x)-g^{\prime}(x+h))$
$\displaystyle=\eta(x)(g^{\prime}(x)-g^{\prime}(x_{n}))+\eta(x)(g^{\prime}(x_{n})-g^{\prime}(x+h))$
$\displaystyle\leq\eta(x)(g^{\prime}(x)-g^{\prime}(x_{n}))+\eta(x_{n})(g^{\prime}(x_{n})-g^{\prime}(x+h))$
$\displaystyle\leq\frac{1}{10}\omega(x_{n}-x)+\frac{1}{10}\omega(x+h-x_{n})\leq\frac{1}{5}\omega(h).$

(ii): Let $\varphi\in\mathcal{M}$ satisfy
$\liminf_{x\to\infty}\varphi(x)/\omega_{\eta}(x)=0$. Suppose to the contrary
that there exists $C>0$ such that (18) holds. Hence

(30) $\displaystyle|g(x)-g(1)|=\bigg{|}\frac{\partial f}{\partial
y}(x,0)-\frac{\partial f}{\partial y}(1,0)\bigg{|}\leq C\varphi(|x-1|),\quad
x>0.$

If $\varphi(x)=0$ for all $x>0$, then (30) implies that $g$ is constant, and
thus by (23)

$\displaystyle\psi^{\prime}_{+}(4a)\leq\delta(2a)=bg^{\prime}(a)=0$

which is a contradiction with (21). So we may further suppose that
$\varphi(x_{0})>0$ for some $x_{0}>0$. Then

$\displaystyle
0<\frac{\varphi(x_{0})}{\omega_{\eta}(x)}\leq\frac{\varphi(x)}{\omega_{\eta}(x)},\quad
x>x_{0},$

and thus $\liminf_{x\to\infty}\varphi(x_{0})/\omega_{\eta}(x)=0$. Since
$\omega_{\eta}$ is non-decreasing by Lemma 3.4 (i), it follows that
$\lim_{x\to\infty}\omega_{\eta}(x)=\infty$. Now by (23) and (19) we have

$\displaystyle 2b(g(x)-g(1))$ $\displaystyle=2\int_{a+1}^{a+x}\delta(s)ds\geq
2\int_{a+1}^{a+x}\psi^{\prime}_{+}(2s)ds=\int_{2(a+1)}^{2(a+x)}\psi^{\prime}_{+}(t)dt$
$\displaystyle=\psi(2(a+x))-\psi(2(a+1))\geq\psi(x)-\psi(2(a+1))$
$\displaystyle\geq\omega_{\eta}(x)-\psi(2(a+1)),\quad x>1,$

and thus by (30)

$\displaystyle\omega_{\eta}(x)-\psi(2(a+1))\leq 2b(g(x)-g(1))\leq
2bC\varphi(x-1)\leq 2bC\varphi(x),\quad x>1.$

Hence

$\displaystyle 1-\frac{\psi(2(a+1))}{\omega_{\eta}(x)}\leq
2bC\frac{\varphi(x)}{\omega_{\eta}(x)},\quad x>1,$

which is a contradiction with
$\liminf_{x\to\infty}\varphi(x)/\omega_{\eta}(x)=0$ and
$\lim_{x\to\infty}\omega_{\eta}(x)=\infty$. ∎

##  5\. The case $\mathbb{R}^{n}$

The main result (Theorem 5.5) of the article works with $\omega\in\mathcal{M}$
that satisfy

(*)
$\displaystyle\inf_{n\in\mathbb{N}}\bigg{(}\liminf_{h\to\infty}\frac{\omega(h)}{n\omega\big{(}\frac{h}{n}\big{)}}\bigg{)}=0.$

An immediate consequence of Theorem 5.5 is Theorem 1.1, since we have:

(31) $\textnormal{If }\alpha\in(0,1)\textnormal{ and
}\omega(h)=h^{\alpha}\textnormal{ for all }h\geq 0,\textnormal{ then
}\eqref{de_condition}\textnormal{ holds}.$

Indeed, if $\alpha\in(0,1)$ and $\omega(h)=h^{\alpha}$ for all $h\geq 0$, then

$\displaystyle\liminf_{h\to\infty}\frac{\omega(h)}{n\omega\big{(}\frac{h}{n}\big{)}}=\frac{1}{n^{1-\alpha}},\quad
n\in\mathbb{N},$

and thus (* ‣ 5) holds.

The following remark concerns the validity of (* ‣ 5) for some other
$\omega\in\mathcal{M}$.

######  Remark 5.1.

  * (i)

If $\alpha\in[0,1)$ and $\beta>0$, then there exist $C\geq 0$, $h_{0}>1$ and a
concave $\omega\in\mathcal{M}$ such that
$\omega(h)=h^{\alpha}\log^{\beta}(h)+C$ for all $h\geq h_{0}$ and (* ‣ 5)
holds.

  * (ii)

If $\omega(h)=h$ for all $h\geq 0$, then (* ‣ 5) doesn’t hold.

  * (iii)

If $\beta>0$, then there exist $C\geq 0$, $h_{0}>1$ and a concave
$\omega\in\mathcal{M}$ such that $\omega(h)=h/\log^{\beta}(h)+C$ for all
$h\geq h_{0}$ and (* ‣ 5) doesn’t hold.

All these facts can be easily proved.

The main property of concave $\omega\in\mathcal{M}$ satisfying (* ‣ 5) is that
assertion (iii) of the following lemma holds.

######  Lemma 5.2.

Let $\omega\in\mathcal{M}$ be concave and satisfy (* ‣ 5). Then the following
hold:

  * (i)

$\omega(h)>0$ for all $h>0$ and $\liminf_{h\to 0+}\omega(h)/h>0$.

  * (ii)

$\lim_{h\to\infty}\omega(h)/h=0$.

  * (iii)

If $\eta:(0,\infty)\to(0,\infty)$ is non-decreasing and
$\lim_{h\to\infty}\eta(h)/h=0$, then

(32) $\displaystyle\liminf_{h\to\infty}\frac{\omega(h)}{\omega_{\eta}(h)}=0.$

###### Proof.

(i) It follows easily from (* ‣ 5) and Remark 2.7 (ii).

(ii): Firstly, the limit exists by Remark 2.7 (ii). Further, using (i) and
Remark 2.7 (iii), we obtain

$\displaystyle
0\leq\lim_{h\to\infty}\frac{\omega(h)}{h}=\omega(1)\lim_{h\to\infty}\frac{n\omega(h)}{\frac{h}{n}\omega(1)}\leq\liminf_{h\to\infty}\frac{n\omega(h)}{\omega(\frac{h}{n})},\quad
n\in\mathbb{N},$

and thus $\lim_{h\to\infty}\omega(h)/h=0$ by (* ‣ 5).

(iii): Let $\varepsilon>0$. Then, by (* ‣ 5), there exists $n\in\mathbb{N}$
such that

$\displaystyle\liminf_{h\to\infty}\frac{\omega(h)}{n\omega\big{(}\frac{h}{n}\big{)}}<\varepsilon.$

Further, there exists $h_{0}>0$ such that $\eta(h)/h\leq 1/n$ for every $h\geq
h_{0}$. Hence by (i), Remark 2.7 (ii) and Lemma 3.5, we have

$\displaystyle
0<\frac{n\omega\big{(}\frac{h}{n}\big{)}}{\omega(h)}=\frac{h}{\omega(h)}\frac{\omega\big{(}\frac{h}{n}\big{)}}{\frac{h}{n}}\leq\frac{h}{\omega(h)}\frac{\omega(\eta(h))}{\eta(h)}\leq\frac{\omega_{\eta}(h)}{\omega(h)},\quad
h\geq h_{0},$

and thus

$\displaystyle
0\leq\liminf_{h\to\infty}\frac{\omega(h)}{\omega_{\eta}(h)}\leq\liminf_{h\to\infty}\frac{\omega(h)}{n\omega\big{(}\frac{h}{n}\big{)}}<\varepsilon.$

Hence (32) holds. ∎

We will also need the following lemma:

######  Lemma 5.3.

Let $n\in\mathbb{N}$ and let $G\subset\mathbb{R}^{n}$ be an open convex set
such that $1\leq\dim(\operatorname{span}(\operatorname{rec}(G)))<n$. Then
there exists a linear surjection $L:\mathbb{R}^{n}\to\mathbb{R}^{2}$ such that

(33)
$\displaystyle(1,0)\in\operatorname{rec}(L(G))=L(\operatorname{rec}(G))\subset\operatorname{span}(\\{(1,0)\\}).$

###### Proof.

We will proceed by induction on $n$. If $n=2$, then
$\dim(\operatorname{span}(\operatorname{rec}(G)))=1$ and thus there exists a
linear bijection $L:\mathbb{R}^{n}\to\mathbb{R}^{2}$ such that $(1,0)\in
L(\operatorname{rec}(G))\subset\operatorname{span}(\\{(1,0)\\})$. Hence (33)
holds by Lemma 2.14.

Now suppose that $n\geq 3$ is given and the assertion holds for $n-1$. It
follows easily from Lemma 2.12 (ii) that
$W:=\operatorname{rec}(G)\cap(-\operatorname{rec}(G))$ is a linear subspace of
$\operatorname{span}(\operatorname{rec}(G))$. Thus

$\displaystyle m:=\dim(W)\leq
k:=\dim(\operatorname{span}(\operatorname{rec}(G)))$

and there exist $b_{1},\dots,b_{k}\in\mathbb{R}^{n}$ such that
$\\{b_{1},\dots,b_{m}\\}$ is an algebraic basis of $W$ and
$\\{b_{1},\dots,b_{k}\\}$ an algebraic basis of
$\operatorname{span}(\operatorname{rec}(G))$. We will show that there exists a
linear surjection $L_{1}:\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ such that

(34) $\displaystyle\operatorname{rec}(G)\cap L_{1}^{-1}(\\{0\\})\subset
W,\;1\leq\dim(L_{1}(\operatorname{span}(\operatorname{rec}(G))))<n-1.$

We will distinguish three cases:

(a) $k<n-1$: Then it is easy to find a linear surjection
$L_{1}:\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ such that
$\operatorname{span}(\operatorname{rec}(G))\cap L_{1}^{-1}(\\{0\\})=\\{0\\}$.
Then

$\displaystyle
L_{1}(\operatorname{span}(\operatorname{rec}(G)))=\operatorname{span}(\\{L_{1}(b_{1}),\dots,L_{1}(b_{k})\\})$

and thus (34) follows.

(b) $m=0$ and $k=n-1$: Then it is easy (cf. [KZ, Lemma 5.10] or see [S,
Consequence 2]) to find
$w\in\operatorname{span}\\{b_{1},b_{2}\\}\setminus\\{0\\}$ such that
$\operatorname{rec}(G)\cap\operatorname{span}(\\{w\\})=\\{0\\}$. Then there
exists a linear surjection $L_{1}:\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ such that
$L_{1}^{-1}(\\{0\\})=\operatorname{span}(\\{w\\})$. Since $L_{1}(w)=0$ and
$w\in\operatorname{span}\\{b_{1},b_{2}\\}$, it follows that $L_{1}(b_{1})$ and
$L_{1}(b_{2})$ are linearly dependent and thus

(35) $\displaystyle
L_{1}(\operatorname{span}(\operatorname{rec}(G)))=\operatorname{span}(\\{L_{1}(b_{2}),\dots,L_{1}(b_{k})\\}).$

Hence (34) holds.

(c) $1\leq m\leq k=n-1$: Then we choose a linear surjection
$L_{1}:\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ such that
$L_{1}^{-1}(\\{0\\})=\operatorname{span}(\\{b_{1}\\})$. Then
$L_{1}^{-1}(\\{0\\})\subset W$ and (35) holds. Consequently (34) holds.

So there exists $L_{1}$ for which (34) holds. Then $L_{1}(G)$ is an open
convex set and

(36) $\displaystyle L_{1}(\operatorname{rec}(G))=\operatorname{rec}(L_{1}(G))$

by Lemma 2.14. Thus

$\displaystyle\operatorname{span}(\operatorname{rec}(L_{1}(G)))$
$\displaystyle=\operatorname{span}(L_{1}(\operatorname{rec}(G)))=L_{1}(\operatorname{span}(\operatorname{rec}(G)))$

and hence $1\leq\dim(\operatorname{span}(\operatorname{rec}(L_{1}(G))))<n-1$.
Now by induction assumption there exists a linear surjection
$L_{2}:\mathbb{R}^{n-1}\to\mathbb{R}^{2}$ such that

(37)
$\displaystyle(1,0)\in\operatorname{rec}(L_{2}(L_{1}(G)))=L_{2}(\operatorname{rec}(L_{1}(G)))\subset\operatorname{span}(\\{(1,0)\\}).$

Then $L:=L_{2}\circ L_{1}$ is a linear surjection and it follows from (36) and
(37) that (33) holds. ∎

######  Remark 5.4.

The proof of the preceding lemma is similar to the proof of [KZ, Proposition
5.13]. Let us note that the proof of [KZ, Proposition 5.13] is incomplete.
Namely, in case (b1) we forgot to consider the case when $H$ is bounded. This
can be easily corrected as follows: If $H$ is bounded, then clearly
$\dim(\operatorname{span}(\operatorname{rec}(G)))=1$ and thus it is easy to
find a linear surjection $L:\mathbb{R}^{n}\to\mathbb{R}^{n-1}$ that satisfies
[KZ, (24) and (25)], and so we can finish the proof as in case (b2).

Now we will prove the main theorem of the article. Recall that it implies
Theorem 1.1 by (31) and that it can also be applied to other interesting
moduli than power type moduli (see Remark 5.1 (i)).

######  Theorem 5.5.

Let $G\subset\mathbb{R}^{n}$ ($n\geq 2$) be an unbounded open convex set that
doesn’t contain a translation of a cone with non-empty interior. Let
$\omega\in\mathcal{M}$ be concave and satisfy (* ‣ 5). Then there exists
$f:G\to\mathbb{R}$ that is both $\omega$-semiconvex and $\omega$-semiconcave
and $f\notin C^{1,\omega}(G)$.

###### Proof.

By Lemma 2.12 (iv), (v) and Lemma 5.3 there exists a linear surjection
$L:\mathbb{R}^{n}\to\mathbb{R}^{2}$ such that

(38)
$\displaystyle(1,0)\in\operatorname{rec}(L(G))=L(\operatorname{rec}(G))\subset\operatorname{span}(\\{(1,0)\\}).$

Then $L(G)$ is an open convex set. We will further distinguish two cases.

(a) $-(1,0)\notin\operatorname{rec}(L(G))$: Then it follows (see the proof of
[KZ, Proposition 5.13]) from (38) that there exist $b\in\mathbb{R}^{2}$ and a
concave $\eta:(0,\infty)\to(0,\infty)$ such that
$\lim_{x\to\infty}\eta(x)/x=0$ and

(39) $\displaystyle\\{(x,0)\in\mathbb{R}^{2}:x>0\\}\subset L(G)+b\subset
G_{0}:=\\{(x,y)\in\mathbb{R}^{2}:x>0,\;|y|<\eta(x)\\}.$

By Proposition 4.4 and Lemma 5.2 (iii) there exists $f_{0}\in C^{1}(G_{0})$
that is both $\omega$-semiconvex and $\omega$-semiconcave and such that
$f^{\prime}_{0}$ is uniformly continuous on $\\{(x,0)\in\mathbb{R}^{2}:x>0\\}$
with modulus $C\omega$ for no $C>0$. Thus it follows from (39) that
$L(G)+b\in\mathcal{G}_{2}(\omega)$. Then $L(G)\in\mathcal{G}_{2}(\omega)$ by
Remark 2.9 (iii), and hence $G\in\mathcal{G}_{n}(\omega)$ by Lemma 2.13 and
(38). Now the assertion follows from Definition 2.8.

(b) $-(1,0)\in\operatorname{rec}(L(G))$: Then it follows from Lemma 2.12 (ii)
and (38) that $\operatorname{rec}(L(G))=\operatorname{span}(\\{(1,0)\\})$.
Hence it is easy to find a bounded open interval $I\subset\mathbb{R}$ such
that $L(G)=\mathbb{R}\times I$. Then we find a linear bijection
$Q:\mathbb{R}^{2}\to\mathbb{R}^{2}$ and $b\in\mathbb{R}^{2}$ such that

$\displaystyle Q(L(G))+b=\mathbb{R}\times(0,1).$

Then $Q(L(G))+b\in\mathcal{G}_{2}(\omega)$ by Lemma 2.10 and Lemma 5.2 (i),
(ii). Hence $Q(L(G))\in\mathcal{G}_{2}(\omega)$ by Remark 2.9 (iii). Now by
Lemma 2.14 (applied to $Q$) and (38) we have

$\displaystyle\operatorname{rec}(Q(L(G)))=Q(\operatorname{rec}(L(G)))=Q(L(\operatorname{rec}(G)))$

and thus $G\in\mathcal{G}_{n}(\omega)$ by Lemma 2.13 (applied to $Q\circ L$).
Then the assertion follows from Definition 2.8. ∎

######  Acknowledgment.

I thank Luděk Zajíček for many helpful suggestions which improved this
article.

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