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

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

Let $G\subset{\mathbb{R}}^{n}$ be an open convex set which is either bounded
or contains a translation of a convex cone with nonempty interior. It is known
that then, for every modulus $\omega$, every function on $G$ which is both
semiconvex and semiconcave with modulus $\omega$ is (globally)
$C^{1,\omega}$-smooth. We show that this result is optimal in the sense that
the assumption on $G$ cannot be relaxed. We also present direct short proofs
of the above mentioned result and of some its quantitative versions. Our
results have immediate consequences concerning (i) a first-order quantitative
converse Taylor theorem and (ii) the problem whether $f\in C^{1,\omega}(G)$
whenever $f$ is continuous and smooth in a corresponding sense on all lines.
We hope that these consequences are of an independent interest.

###### Key words and phrases:

$\omega$-semiconvex functions, $\omega$-semiconcave functions,
$C^{1,\omega}$-smooth functions, smoothness on all lines, converse Taylor
theorem, strongly $\alpha(\cdot)$-paraconvex functions

###### 1991 Mathematics Subject Classification:

Primary: 26B25; Secondary: 26B35

The research was supported by GAČR 18-11058S

##  1\. Introduction

First we recall a classical reasult on smoothness of functions which are both
semiconvex and semiconcave (i.e., semiconvex and semiconcave with linear
modulus).

######  Definition 1.1.

Let $X$ be a Hilbert space, $G\subset X$ an open convex set and $f$ a function
on $G$. We say that $f$ is semiconvex, if there exists $C\geq 0$ and a
continuous convex function $g$ on $G$ such that

$f(x)=g(x)-\frac{C}{2}\cdot\|x\|^{2},\ \ \ x\in G.$

Then we say that $f$ is semiconvex with constant $C$. We say that $f$ is
semiconcave on $G$ (with constant $C$) if $-f$ is semiconvex on $G$ (with
constant $C$).

Let $X$, $G$ and $f$ be as in Definition 1.1. Then it is very easy to show
(see Remark 2.2) that

(1.1) $f^{\prime}\ \text{exists and is Lipschitz with constant}\ C\\\
\Longrightarrow\ \ \ f\ \text{ is both semiconvex and semiconcave with
constant}\ C.$

Rather suprisingly, also the converse of (1.1) holds:

(1.2) $f\ \text{ is both semiconvex and semiconcave with constant}\ C\\\
\Longrightarrow\ \ \ f^{\prime}\ \text{exists and is Lipschitz with constant}\
C.$

Some bibliographic notes concerning this well-known interesting nontrivial
result are contained in Section 3, where also its short elementary proof is
presented.

Consequently, we have that

(1.3) $f\ \text{ is both semiconvex and semiconcave with constant}\ C\\\
\Longleftrightarrow\ \ \ f^{\prime}\ \text{exists and is Lipschitz with
constant}\ C.$

In particular,

(1.4) $f\ \text{is both semiconvex and semiconcave}\ \Longleftrightarrow\ f\in
C^{1,1}(G).$

In the present article we study analogues of (1.2) and (1.4) for functions
which are both $\omega$-semiconvex and $\omega$-semiconcave. We use the
following terminology.

######  Definition 1.2.

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

######  Definition 1.3.

Let $X$ be a normed linear space, $G\subset X$ an open convex set and
$\omega\in{\cal M}$. We say that a continuous $f:G\to{\mathbb{R}}$ 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 $\omega$-semiconcave if $-f$ is $\omega$-semiconvex.

######  Remark 1.4.

If $X$ is a Hilbert space, it is easy to show (cf. [CaSi, p. 30]) that $f$ is
semiconvex on $G\subset X$ with constant $C\geq 0$ if and only if $f$ is
$\omega$-semiconvex with modulus $\omega(t)=\frac{C}{2}t,\ t\geq 0$.

######  Remark 1.5.

The notion of $\omega$-semiconvex functions is very close to Rolewicz’s notion
of strongly $\alpha(\cdot)$-paraconvex functions (which has in [Rol3] slightly
different sense than in [Rol2]). Namely (see e.g. [DZ1, Remark 2.11]) if
$\omega\in{\cal M}$ and $\alpha(t)=t\omega(t),\ t\geq 0$, then (by the
definition from [Rol3]),

$f$ is $\omega$-semiconvex $\Longrightarrow$ $f$ is strongly
$\alpha(\cdot)$-paraconvex $\Longrightarrow$ $f$ is $(2\omega)$-semiconvex.

Thus most results conserning $\omega$-semiconvex functions have obvious
consequences for strongly $\alpha(\cdot)$-paraconvex functions (and vice
versa). We will not formulate such consequences of our results.

######  Definition 1.6.

If $G$ is an open subset of a normed linear space and $\omega\in{\cal M}$,
then we denote by $C^{1,\omega}_{*}(G)$ the set of all Fréchet differentiable
$f:G\rightarrow{\mathbb{R}}$ such that $f^{\prime}$ is uniformly continuous
with modulus $\omega$. We set
$C^{1,\omega}(G):=\bigcup_{K>0}C^{1,K\omega}_{*}(G)$.

Note that so defined $C^{1,\omega}$-smoothness, which coincides with (global)
$C^{1,1}$-smoothness in the case of a linear $\omega$, is common in the
literature. We will sometimes use also (stronger)
$C_{*}^{1,\omega}$-smoothness (which is somewhere, e.g. in [Jo] and [JZ],
called “$C^{1,\omega}$-smoothness”).

Let us first mention that a (weaker) version of (1.1) for general moduli is
also easy (see Remark 2.2). It asserts that, if $X$, $G$ and $\omega$ are as
in Definition 1.3, then

(1.5) $f\in C^{1,\omega}_{*}(G)\Longrightarrow\ \text{$f$ is both
$\omega$-semiconvex and $\omega$-semiconcave on}\ G.$

However (if $X$, $G$ and $\omega$ are as in Definition 1.3) the implication

(1.6) $f\ \text{is both $\omega$-semiconvex and $\omega$-semiconcave on $G$}\
\Longrightarrow\ f\in C^{1,\omega}(G)$

does not hold in general. The validity of (1.6) (or of its local versions) was
studied e.g. in [CaSi, Theorem 3.3.7], [JTZ] and [Kr] (and essentially already
in [Rol1] for $\omega(t)=t^{\alpha}$, $0<\alpha\leq 1$). As far as we know,
the strongest and the most general versions of (1.6) follow from versions of
Converse Taylor Theorem proved in [Jo] and [HJ]. In particular (see [Kr,
Theorem 2.6 and Remark 2.7]), [HJ, Corollary 126] easily implies the following
result.

######  Theorem 1.7.

[Kr] Let $X$ be a normed linear space and $G\subset X$ an open convex set.
Suppose that $G$ is either bounded or has the property that there are $a\in
X$, $r>0$ and $\\{u_{n}\\}_{n\in{\mathbb{N}}}\subset X$, $\|u_{n}\|=n$, such
that $B(a+u_{n},nr)\subset G$ for each $n\in{\mathbb{N}}$. Then, for each
modulus $\omega\in{\cal M}$ and for each function $f$ on $G$,

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

######  Remark 1.8.

Using (1.5), we easily see that the assertion of Theorem 1.7 implies the
following analogue of (1.4):

(1.7) $f\ \text{is both $(K\omega)$-semiconvex and $(K\omega)$-semiconcave}\
\text{for some}\ K>0\ \Longleftrightarrow\ f\in C^{1,\omega}(G).$

For a local version of (1.7) see [JTZ, Theorem 6.1]. See also Proposition 6.8
for reformulations of this characterization of $C^{1,\omega}$-smoothness on
some convex sets.

Let us note that an easy compactness argument shows that an unbounded open
convex $G\subset{\mathbb{R}}^{n}$ satisfies the assumption of Theorem 1.7 if
and only if it contains a translation of a convex cone with nonempty interior.

The main result of the present article is the following theorem which shows
that, for $X={\mathbb{R}}^{n}$, Theorem 1.7 is optimal in the sense that the
assumption on $G$ cannot be relaxed. It follows immediately from Proposition
5.13 (proved in Section 5) via Remark 5.7.

######  Theorem 1.9.

Let $G\subset{\mathbb{R}}^{n}$ ($n\geq 2$) be an unbounded open convex set
which does not contain a translation of a convex cone with nonempty interior.
Then there exists a concave modulus $\omega\in{\cal M}$ with
$\lim_{t\to\infty}\omega(t)=\infty$ and a function $f$ which is both
$\omega$-semiconvex and $\omega$-semiconcave on $G$ but $f\notin
C^{1,\omega}(G)$.

Since [HJ, Corollary 126] is a rather general result (working with classes
$C^{k,\omega}(G)$ for each $k\in{\mathbb{N}}$), we find useful to present (in
Section 4) direct proofs (which partly follow arguments from [HJ]) of Theorem
1.7 and of some its quantative versions (sharper then those which directly
follow from the proof of [HJ, Corollary 126]). Namely, we prove the following
result.

######  Theorem 1.10.

Let $X$ be a normed linear space and $\emptyset\neq G\subset X$ an open convex
set. Let $\omega\in{\cal M}$ and let $f$ be a function on $G$ which is both
$\omega$-semiconvex and $\omega$-semiconcave. Then the following assertions
hold.

  1. (i)

If $G$ is bounded, then $f^{\prime}$ is uniformly continuous with modulus
$6e_{G}\omega$, where
$e_{G}:=\frac{\operatorname{diam}G}{\sup\\{r:B(a,r)\subset G\\}}$.

  2. (ii)

If $a\in X$, $r>0$ and $\\{u_{n}\\}_{n\in{\mathbb{N}}}\subset X$ with
$\|u_{n}\|=n$ are such that $B(a+u_{n},nr)\subset G$ for each
$n\in{\mathbb{N}}$, then $f^{\prime}$ is uniformly continuous with modulus
$12\,(1+1/r)\omega$.

######  Remark 1.11.

  1. (i)

The proof of [HJ, Corollary 126] gives the multiplicative constants 15 and 30
instead of our constants 6 and 12.

  2. (ii)

In the special cases when $G=X$ or $\omega$ is linear, we can assert that $f$
is $C_{*}^{1,4\omega}$-smooth (see Corollary 4.3 and Remark 4.5).

In the last Section 6 we show that our results have immediate consequences (or
reformulations) concerning

  1. (i)

a first-order quantitative converse Taylor theorem and

  2. (ii)

the problem whether $f\in C^{1,\omega}(G)$ whenever $f$ is continuous and
$C^{1,\omega}_{*}$-smooth on all lines.

We hope that these versions of our new results can be interesting also for
readers which are not interested in $\omega$-semiconvex functions.

At the end of Section 6, a natural open question is formulated.

##  2\. Preliminaries

###  2.1. Basic definitions

In the following, $X$ will be always a real normed linear space and $X^{*}$
its dual space. The norm and the origin in any normed linear space will be
denoted by $\|\cdot\|$ and $0$, respectively. We consider real Hilbert spaces
which can be finite-dimensional. By the derivative $f^{\prime}$ of a function
$f$ defined on a subset of $X$ we always mean the Fréchet derivative and the
notion of $C^{1}$-smoothness has the standard sense. The symbol $B(x,r)$
($\overline{B}(x,r)$) denotes the open (closed) ball with centre $x$ and
radius $r$ in $X$. For $A\subset{\mathbb{R}}^{n}$, the symbols $\overline{A}$
and $\operatorname{span}A$ denote the closure and the linear span of $A$,
respectively. 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
$\lambda>0$. A ray in ${\mathbb{R}}^{n}$ is a set of the form $\rho=\\{a+tv:\
t\geq 0\\}$, where $a\in{\mathbb{R}}^{n}$ and $0\neq v\in{\mathbb{R}}^{n}$. By
the dimension $\dim(C)$ of a convex set $C\subset{\mathbb{R}}^{n}$ we mean the
dimension of the affine hull of $C$ (see [Roc, p. 12]).

###  2.2. $\omega$-semiconvexity and $\omega$-semiconcavity

We will frequently use the obvious fact that if $f$ is $\omega$-semiconvex
(resp. $\omega$-semiconcave) on $G$ and $c>0$, then the function $cf$ is
$(c\omega)$-semiconvex (resp. $(c\omega)$-semiconcave).

Further, it follows easily from Definition 1.3 that $\omega$-semiconvexity is
equivalent to “$\omega$-semiconvexity on all lines”. More precisely, the
following statement holds.

######  Lemma 2.1.

Let $X$ be a normed linear space, $G\subset X$ an open convex set,
$\omega\in{\cal M}$, and let $f$ be a continuous function on $G$. Then the
following conditions are equivalent.

  1. (i)

$f$ is $\omega$-semiconvex on $G$.

  2. (ii)

For every $a\in G$ and $v\in X$, $\|v\|=1$, the function $f_{a,v}:t\mapsto
f(a+tv)$ defined on the open interval $\\{t\in{\mathbb{R}}:\ a+tv\in G\\}$ is
$\omega$-semiconvex.

######  Remark 2.2.

  1. (i)

Using Lemma 2.1, it is easy to see that, to prove implications (1.1) and
(1.5), it is sufficient to prove them for $X={\mathbb{R}}$. The validity of
(1.1) for $X={\mathbb{R}}$ is almost obvious. Indeed, the assumption of (1.1)
implies that $g(x):=f(x)+\frac{C}{2}x^{2}$ is convex on $G$, since
$g^{\prime}(x)=f^{\prime}(x)+Cx$ is clearly nondecreasing on the interval $G$.
And the validity of (1.5) for $X={\mathbb{R}}$ immediately follows e.g. from
[CaSi, Proposition 2.1.2].

  2. (ii)

Using Lemma 2.1, Remark 1.4 and (1.1), we obtain that if $\omega$ is linear,
then we can assert in (1.5) even that $f$ is both $\tilde{\omega}$-semiconvex
and $\tilde{\omega}$-semiconcave, where $\tilde{\omega}:=\frac{1}{2}\omega$.
However, for general modulus it is not true even for $X={\mathbb{R}}$ and
$\tilde{\omega}:=C\omega$, where $C<1$ is any absolute constant. Indeed, if
$0<C<1$ is given, we choose $p\in(1,2)$ such that $C<1/p$ and set
$\omega(t):=t^{p-1}$, $t\geq 0$, and $f(x):=x^{p}/p,\ x\in(0,1)$. Then
$f^{\prime}(x)=\omega(x),\ x\in(0,1),$ and consequently the concavity of
$\omega$ implies that $f$ is $C_{*}^{1,\omega}$-smooth. However (as observed
in [Kr2, Remark 2.10 (ii)]), it is not difficult to show that $f$ is not
$(C\omega)$-semiconcave.

We will need the following fact which is an easy consequence of (1.5) and
Lemma 2.1.

######  Lemma 2.3.

Let $X$ be a normed linear space, $G\subset X$ an open convex set,
$\omega\in{\cal M}$ and let $f$ be a Fréchet differentiable function on $G$.
Suppose that

(2.1) $|f^{\prime}(y)(y-x)-f^{\prime}(x)(y-x)|\leq\|y-x\|\omega(\|y-x\|),\quad
x,y\in G.$

Then $f$ is both $\omega$-semiconvex and $\omega$-semiconcave on $G$.

###### Proof.

We will first observe that if $f_{a,v}$ is as any function as in Lemma 2.1
(ii), then

(2.2) $f_{a,v}$ is $C^{1,\omega}_{*}$-smooth.

Indeed, for each $t_{1}\neq t_{2}$ from the domain of $f_{a,v}$ we can apply
(2.1) to $x:=a+t_{1}v$ and $y:=a+t_{2}v$, and obtain

$|(f_{a,v})^{\prime}(t_{2})-(f_{a,v})^{\prime}(t_{1})|=|f^{\prime}(a+t_{2}v)(v)-f^{\prime}(a+t_{1}v)(v)|\\\
=\frac{1}{|t_{2}-t_{1}|}|f^{\prime}(a+t_{2}v)((t_{2}-t_{1})v)-f^{\prime}(a+t_{1}v)((t_{2}-t_{1})v)|\leq\omega(|t_{2}-t_{1}|),$

Therefore each $f_{a,v}$ is $\omega$-semiconvex by (1.5) and so $f$ is
$\omega$-semiconvex by Lemma 2.1. Using the same argument to $-f$, we obtain
that $f$ is also $\omega$-semiconcave. ∎

An other basic tool for us is the following essentially well-known result.

######  Lemma 2.4.

Let $X$ be a normed linear space, $G\subset X$ an open convex set and
$\omega\in{\cal M}$. Let a function $f$ be both $\omega$-semiconvex and
$\omega$-semiconcave on $G$. Then $f$ is Fréchet differentiable on $G$ and

(2.3) $|f(y)-f(x)-f^{\prime}(x)(y-x)|\leq\|y-x\|\,\omega(\|y-x\|),\ \ \ x,y\in
G.$

###### Proof.

A proof in the case $X={\mathbb{R}}^{n}$ can be found in [CaSi] (see the proof
of [CaSi, (3.23), p. 60]). The proof of the general case is factually
contained in the proof of [Kr, Theorem 2.6]. For the convenience of the
reader, we shortly repeat the argument from [Kr]. It is based on [Rol2,
Theorem 3] (cf. also [Rol3, Proposition 2]) which easily implies (cf. [Kr] or
[DZ1, Corollary 4.5]) that $\omega$-semiconvexity of $f$ implies that for each
$x\in G$ there exists $\Phi_{x}\in X^{*}$ such that

$f(x+h)-f(x)-\Phi_{x}(h)\geq-\|h\|\omega(\|h\|)\ \ \text{whenever}\ \ \ x+h\in
G.$

Since $-f$ is also $\omega$-semiconvex, for each $x\in G$ there exists
$\Psi_{x}\in X^{*}$ such that

$-f(x+h)+f(x)-\Psi_{x}(h)\geq-\|h\|\omega(\|h\|)\ \ \text{whenever}\ \ \
x+h\in G.$

Adding these two inequalities, we easily obtain that $\Psi_{x}=-\Phi_{x}$ and

$|f(x+h)-f(x)-\Phi_{x}(h)|\leq\|h\|\omega(\|h\|)\ \ \text{whenever}\ \ \
x+h\in G,$

which implies that $f^{\prime}(x)=\Phi_{x}$ and so (2.3) holds. ∎

Lemma 2.4 and Remark 1.4 immediately imply the following well-known result.

######  Corollary 2.5.

Let $X$ be a Hilbert space, $G\subset X$ an open convex set, and let $f$ be a
function which is both semiconvex and semiconcave with constant $C>0$ on $G$.
Then $f$ is Fréchet differentiable on $G$ and

(2.4) $|f(y)-f(x)-f^{\prime}(x)(y-x)|\leq\frac{C}{2}\|y-x\|^{2},\ \ \ x,y\in
G.$

The following result (which clearly implies that the converse of Corollary 2.5
holds) is also essentially well-known, but we prefer to present a simple
proof.

######  Lemma 2.6.

If $X$, $G$, $\omega$ are as in Lemma 2.4 and $f$ satisfies (2.3), then $f$ is
both $(2\omega)$-semiconvex and $(2\omega)$-semiconcave. Moreover, if $\omega$
is linear, then $f$ is both $\omega$-semiconvex and $\omega$-semiconcave.

###### Proof.

For each $x\in G$, $y\in G$, we can apply (2.3) for $x:=y$ and $y:=x$ and
obtain

$|f(x)-f(y)-f^{\prime}(y)(x-y)|\leq\|x-y\|\,\omega(\|x-y\|).$

Adding this inequality with (2.3) and using the triangle inequality, we obtain

$|f^{\prime}(y)(y-x)-f^{\prime}(x)(y-x)|\leq 2\|y-x\|\omega(\|y-x\|).$

Since we have proved that (2.1) implies (2.2), we obtain that each function
$f_{a,v}$ is $C_{*}^{1,2\omega}$-smooth and consequently both
$(2\omega)$-semiconvex and $(2\omega)$-semiconcave by (1.5) (and even both
$\omega$-semiconvex and $\omega$-semiconcave by Remark 2.2 (ii) if $\omega$ is
linear). So the assertions of the lemma folow by Lemma 2.1. ∎

We will need also the following special case of [DZ2, Lemma 2.7].

######  Lemma 2.7.

Let $\omega\in{\cal M}$ and $n\geq k\geq 2$. Let $\emptyset\neq
G\subset{\mathbb{R}}^{n}$ be an open convex set, and let
$L:{\mathbb{R}}^{n}\to{\mathbb{R}}^{k}$ be a linear surjection. Let $f$ be a
function on $L(G)$ which is both $\omega$-semiconvex and $\omega$-semiconcave
on $L(G)$. Then there exists $C>0$ such that the function $f\circ L$ is both
$(C\omega)$-semiconvex and $(C\omega)$-semiconcave on $G$.

(Note that by well-known facts, $L(G)$ is an open convex set.)

###  2.3. Recession cones

In the proof of our main result, we will use some properties of recession
cones (called sometimes also asymptotic cones) of unbounded convex sets
$A\subset{\mathbb{R}}^{n}$. We refer to the exposition in [Roc], however we
denote the recession cone of $A$ by ${\rm rec}\,(A)$ (and not $0^{+}A$ as in
[Roc]).

######  Definition 2.8.

Let $\emptyset\neq A\subset{\mathbb{R}}^{n}$ be a convex set. By the recession
cone of $A$ we mean the set ${\rm rec}\,(A)$ of all $y\in{\mathbb{R}}^{n}$
such that $x+\lambda y\in A$ for each $x\in A$ and $\lambda\geq 0$.

We will need the following two lemmas on open convex sets which easily follow
from results proved in [Roc] for closed convex sets.

######  Lemma 2.9.

Let $\emptyset\neq G\subset{\mathbb{R}}^{n}$ be an open convex set. Then the
following assertions hold.

  1. (i)

${\rm rec}\,(G)={\rm rec}\,(\overline{G})$.

  2. (ii)

${\rm rec}\,(G)$ is a closed convex cone in ${\mathbb{R}}^{n}$ and $0\in{\rm
rec}\,(G)$.

  3. (iii)

${\rm rec}\,(G)=\\{y\in{\mathbb{R}}^{n}:\ \exists x\in G\ \forall\lambda\geq
0:\ x+\lambda y\in G\\}$.

  4. (iv)

If $G$ is unbounded then ${\rm rec}\,(G)\setminus\\{0\\}\neq\emptyset$.

  5. (v)

$\dim({\rm rec}\,(G))=\dim(\operatorname{span}({\rm rec}\,(G)))$.

  6. (vi)

$\dim({\rm rec}\,(G))=n$ if and only if $G$ contains a translation of a cone
with nonempty interior.

###### Proof.

Property (i) follows from [Roc, Corollary 8.3.1] and (ii) from [Roc, Theorem
8.2] and (i). Theorem [Roc, Theorem 8.3] and (i) imply (iii). Property (iv)
follows from [Roc, Theorem 8.4] and (i). Property (v) is obvious since
$0\in{\rm rec}\,(G)$. Using (iii) and (v), it is easy to prove (vi). ∎

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

######  Lemma 2.10.

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
$L^{-1}(\\{0\\})\cap{\rm rec}\,(G)=\\{0\\}$. Then ${\rm rec}\,(L(G))=L({\rm
rec}\,(G))$.

###### Proof.

By Lemma 2.9 (i) we have ${\rm rec}\,(G)={\rm rec}\,(\overline{G})$. So
$L^{-1}(\\{0\\})\cap{\rm rec}\,(\overline{G})=\\{0\\}$ and thus [Roc, Theorem
9.1] (applied to $C:=\overline{G}$) implies that $L(\overline{G})$ is closed
and ${\rm rec}\,(L(\overline{G}))=L({\rm rec}\,(\overline{G}))$. It is an easy
well-known fact that $L(G)$ is an open convex set. Continuity of $L$ implies
$L(\overline{G})\subset\overline{L(G)}$ and so
$L(\overline{G})=\overline{L(G)}$ since $L(\overline{G})$ is closed. Therefore
${\rm rec}\,(L(\overline{G}))={\rm rec}\,(L(G))$ by Lemma 2.9 (i) and the
equality ${\rm rec}\,(L(G))=L({\rm rec}\,(G))$ follows. ∎

##  3\. Linear modulus

The present short section is devoted to result (1.2) which is rewritten below
as Theorem 3.1. We do not know the origin of this interesting useful result
which is well-known for a long time.

It is stated (for $G=X$) in [LL, p. 265](1986), where it is written (without a
reference) that the case $G=X={\mathbb{R}}^{n}$ is well-known and it is shown
how the general case easily follows from the finite-dimensional case. The case
$G=X$ is completely proved in [HP](1989). Let us note that a weaker version of
Theorem 3.1 which asserts that $f\in C^{1,1}(G)$ only, is factually proved in
[Rol1](1979) (for $G=X$ and Lipschitz $f$) and very easily follows (for $G=X$)
from [Mo, Proposition 10.b](1965) which asserts that if $g$ and $h$ are
continuous convex functions such that $g(x)+h(x)=(1/2)\|x\|^{2},\ x\in X$,
then $g^{\prime}$ is Lipschitz with constant $1$. The equality $G=X$ is vital
in [Mo] and [HP].

Several proofs of Theorem 3.1 work with second derivatives of $f$. For
example, [CaSi, Corollary 3.8] and [BC, Exercise 3.8, p. 183] use
distributional derivatives, and [BPP, Lemma 2.7] and [BBEM, Theorem 2.3] use
Aleksandrov’s theorem on a.e. second differentiability of convex functions.

We present below a short elementary proof using the procedure (based on (3.2))
which will be used also in the proof of Lemma 4.2. Our proof is similar to
elementary proofs given in [Iv, Theorem 1.1] and [BAJN, Corollary 5.1] and we
guess that a similar proof is (essentially) contained also in a much older
article.

######  Theorem 3.1 (Well-known theorem).

Let $X$ be a Hilbert space, $G\subset X$ an open convex set and let a function
$f$ on $G$ be both semiconvex and semiconcave with constant $C$. Then $f$ is
differentiable on $G$ and $f^{\prime}$ is Lipschitz with constant $C$.

###### Proof.

By Corollary 2.5 we know that $f^{\prime}$ exists on $G$ and (2.4) holds.
First we will show that if $u,v\in G$, $u\neq v$, then

(3.1) $\overline{B}\bigg{(}\frac{u+v}{2},\frac{\|v-u\|}{2}\bigg{)}\subset G\
\Longrightarrow\ \|f^{\prime}(v)-f^{\prime}(u)\|\leq C\|v-u\|.$

To this end, suppose that the assumption of (3.1) holds. Without any loss of
generality we can suppose that

(3.2)
$f(u)-f(v)+\frac{1}{2}f^{\prime}(u)(v-u)-\frac{1}{2}f^{\prime}(v)(u-v)\leq 0,$

since otherwise we could work with $\tilde{f}:=-f$ instead of $f$. Now
consider an arbitrary $w\in X$, $\|w\|=\|v-u\|$. Then we have
$\frac{1}{2}(u+v+w)\in G$ and therefore, using (2.4) and (3.2), we obtain

$\displaystyle f^{\prime}(v)$
$\displaystyle\bigg{(}\frac{1}{2}w\bigg{)}-f^{\prime}(u)\bigg{(}\frac{1}{2}w\bigg{)}$
$\displaystyle=f\bigg{(}\frac{1}{2}(u+v+w)\bigg{)}-f(u)-f^{\prime}(u)\bigg{(}\frac{1}{2}(w+v-u)\bigg{)}$
$\displaystyle+f(u)-f(v)+\frac{1}{2}f^{\prime}(u)(v-u)-\frac{1}{2}f^{\prime}(v)(u-v)$
$\displaystyle-\bigg{(}f\bigg{(}\frac{1}{2}(u+v+w)\bigg{)}-f(v)-f^{\prime}(v)\bigg{(}\frac{1}{2}(w-v+u)\bigg{)}\bigg{)}$
$\displaystyle\leq\frac{C}{8}\|w+(v-u)\|^{2}+\frac{C}{8}\|w-(v-u)\|^{2}$
$\displaystyle=\frac{C}{4}\|w\|^{2}+\frac{C}{4}\|v-u\|^{2}=\frac{C}{2}\|v-u\|^{2}$

and hence

$\displaystyle\|f^{\prime}(v)-f^{\prime}(u)\|=\sup_{w\in
X,\|w\|=\|u-v\|}\frac{f^{\prime}(v)(w)-f^{\prime}(u)(w)}{\|w\|}\leq C\|v-u\|.$

Now consider arbitrary $x,y\in G$, $x\neq y$. We can clearly choose
$n\in{\mathbb{N}}$ so large that, for each $1\leq i\leq n$,

$\overline{B}\bigg{(}\frac{z_{i-1}+z_{i}}{2},\frac{\|z_{i}-z_{i-1}\|}{2}\bigg{)}\subset
G,\ \text{where}\ z_{i}:=x+\frac{i}{n}(y-x),\quad i\in\\{0,\dots,n\\}.$

Then (3.1) implies that $\|f^{\prime}(z_{i})-f^{\prime}(z_{i-1})\|\leq
C\|z_{i}-z_{i-1}\|$ for each $1\leq i\leq n$, and consequently

$\displaystyle\|f^{\prime}(y)-f^{\prime}(x)\|$
$\displaystyle\leq\sum_{i=1}^{n}\|f^{\prime}(z_{i})-f^{\prime}(z_{i-1})\|\leq\sum_{i=1}^{n}C\|z_{i}-z_{i-1}\|=C\|y-x\|.$

∎

##  4\. Results for a general modulus

In this section, we prove Theorem 1.10 and interesting Corollary 4.3. Other
more special versions of implication (1.6) (for $X={\mathbb{R}}$ or for
special $\omega$) can be found in Remark 4.5 and Remark 5.3.

######  Remark 4.1.

Note that in the proof of Lemma 4.2 (and so also in the proofs of Corollary
4.3 and Theorem 1.10) the assumption that $f$ is both $\omega$-semiconvex and
$\omega$-semiconcave is applied via property (2.3) only.

Proceeding similarly as in the proof of Theorem 3.1, we obtain the following
estimate.

######  Lemma 4.2.

Let $X$ be a normed linear space, $G\subset X$ an open convex set,
$\omega\in{\cal M}$, and let $f$ be a function on $G$ which is both
$\omega$-semiconvex and $\omega$-semiconcave. If $y,z\in G$, $r>0$ and
$\overline{B}(y,r)\subset G$, then

(4.1) $\|f^{\prime}(z)-f^{\prime}(y)\|\leq
2\bigg{(}1+\frac{\|z-y\|}{r}\bigg{)}\omega\bigg{(}\frac{1}{2}r+\frac{1}{2}\|z-y\|\bigg{)}.$

###### Proof.

As in the proof of Theorem 3.1, we may suppose that

$\displaystyle
f(y)-f(z)+\frac{1}{2}f^{\prime}(y)(z-y)-\frac{1}{2}f^{\prime}(z)(y-z)\leq 0.$

Then for every $w\in X$, $\|w\|=r$, we obtain, using $\frac{1}{2}(z+y+w)\in G$
and (2.3),

$\displaystyle f^{\prime}$
$\displaystyle(z)\bigg{(}\frac{1}{2}w\bigg{)}-f^{\prime}(y)\bigg{(}\frac{1}{2}w\bigg{)}$
$\displaystyle=f\bigg{(}\frac{1}{2}(z+y+w)\bigg{)}-f(y)-f^{\prime}(y)\bigg{(}\frac{1}{2}(w+z-y)\bigg{)}$
$\displaystyle+f(y)-f(z)+\frac{1}{2}f^{\prime}(y)(z-y)-\frac{1}{2}f^{\prime}(z)(y-z)$
$\displaystyle-\bigg{(}f\bigg{(}\frac{1}{2}(z+y+w)\bigg{)}-f(z)-f^{\prime}(z)\bigg{(}\frac{1}{2}(w-z+y)\bigg{)}\bigg{)}$
$\displaystyle\leq\frac{1}{2}\|w+(z-y)\|\omega\bigg{(}\frac{1}{2}\|w+(z-y)\|\bigg{)}+\frac{1}{2}\|w-(z-y)\|\omega\bigg{(}\frac{1}{2}\|w-(z-y)\|\bigg{)}$
$\displaystyle\leq(r+\|z-y\|)\|\omega\bigg{(}\frac{1}{2}(r+\|z-y\|)\bigg{)}$

and hence

$\displaystyle\|f^{\prime}(z)-f^{\prime}(y)\|$ $\displaystyle=\sup_{w\in
X,\|w\|=r}\frac{f^{\prime}(z)(w)-f^{\prime}(y)(w)}{\|w\|}$ $\displaystyle\leq
2\bigg{(}1+\frac{\|z-y\|}{r}\bigg{)}\omega\bigg{(}\frac{1}{2}r+\frac{1}{2}\|z-y\|\bigg{)}.$

∎

Using (4.1) for $z\neq y$ and $r:=\|z-y\|$, we immediately obtain the
following result.

######  Corollary 4.3.

Let $X$ be a normed linear space and $\omega\in{\cal M}$. Suppose that a
function $f$ on $X$ is both $\omega$-semiconvex and $\omega$-semiconcave. Then
$f^{\prime}$ is uniformly continuous with modulus $4\omega$.

######  Remark 4.4.

Corollary 4.3 is optimal in the sense that the constant $4$ cannot be
diminished, even for linear modulus. It is factually shown in [BAJN, Example
5], where the authors consider the example of the function
$f(x,y)=x^{2}-y^{2}$ on the space $X=({\mathbb{R}}^{2},\|\cdot\|_{\infty})$.
They (factually) show that then (2.3) holds with $G=X$ and $\omega(t)=t$,
which implies by Lemma 2.6 that $f$ is both $\omega$-semiconvex and
$\omega$-semiconcave on $X$. Observing that
$\|f^{\prime}((1,1))-f^{\prime}((0,0))\|\geq 4\|(1,1)\|_{\infty}=4$ (see
[BAJN, Example 5]), our assertion follows.

Using Lemma 4.2, we give also a relatively short proof of Theorem 1.10
formulated in the introduction.

######  Proof of Theorem 1.10.

(i) Suppose that $G$ is bounded.

Let $x,x+h\in G$ and $\varepsilon>0$. We find $y_{0}\in G$ and $r_{0}>0$ such
that $\overline{B}(y_{0},r_{0})\subset G$ and $\operatorname{diam}G/r_{0}\leq
e_{G}+\varepsilon$. Set $\lambda:=\|h\|/\operatorname{diam}G$,
$s:=x+\frac{1}{2}h$, $y:=\lambda y_{0}+(1-\lambda)s$ and $r:=\lambda r_{0}$.
Then

$\overline{B}(y,r)=\lambda\overline{B}(y_{0},r_{0})+(1-\lambda)\\{s\\}\subset
G\ \ \ \text{and}\ \ \ \frac{\|h\|}{e_{G}+\varepsilon}\leq
r\leq\frac{\|h\|}{e_{G}}.$

Clearly $\|s-y_{0}\|\leq\operatorname{diam}G-r_{0}$ and so

$\|s-y\|=\|\lambda(s-y_{0})\|=\frac{\|h\|}{\operatorname{diam}G}\|s-y_{0}\|\leq\|h\|\left(1-\frac{r_{0}}{\operatorname{diam}G}\right)\leq\|h\|-\frac{1}{e_{G}+\varepsilon}\|h\|.$

Consequently, if $z=x$ or $z=x+h$, we have
$\|z-y\|\leq\frac{3}{2}\|h\|-\frac{1}{e_{G}+\varepsilon}\|h\|$. Using this
inequality, estimates of $r$ and the obvious inequality $e_{G}\geq 2$, we
obtain

$2\left(1+\frac{\|z-y\|}{r}\right)\leq
2\left(1+\frac{3}{2}(e_{G}+\varepsilon)-1\right)=3(e_{G}+\varepsilon),\
\text{and}$
$\frac{1}{2}r+\frac{1}{2}\|z-y\|\leq\frac{1}{2}\left(\frac{\|h\|}{e_{G}}+\frac{3}{2}\,\|h\|-\frac{1}{e_{G}+\varepsilon}\,\|h\|\right)<\|h\|.$

Therefore (4.1) gives $\|f^{\prime}(z)-f^{\prime}(y)\|\leq
3(e_{G}+\varepsilon)\omega(\|h\|)$ and

$\|f^{\prime}(x+h)-f^{\prime}(x)\|\leq\|f^{\prime}(x+h)-f^{\prime}(y)\|+\|f^{\prime}(x)-f^{\prime}(y)\|\leq
6(e_{G}+\varepsilon)\omega(\|h\|)$

which implies the assertion of (i).

(ii) Suppose that $a\in X$, $r>0$ and $\\{u_{n}\\}_{n\in{\mathbb{N}}}$ are
such that $\|u_{n}\|=n$ and $B(a+u_{n},nr)\subset G$ for each
$n\in{\mathbb{N}}$.

Let $x,x+h\in G$. Then there exists $n\in{\mathbb{N}}$ such that
$x,x+h\in\overline{B}(a,n)$. Set $B:=G\cap B(a,(1+r)n)$. Then
$x,x+h,a+u_{n}\in B$. Since $B(a+u_{n},rn)\subset B$, we have

$\displaystyle
e_{B}\leq\frac{\operatorname{diam}B}{rn}\leq\frac{2(1+r)n}{rn}=2\left(1+\frac{1}{r}\right).$

Thus $\|f^{\prime}(x+h)-f^{\prime}(x)\|\leq 12(1+1/r)\omega(\|h\|)$ by the
case (i). $\square$

######  Remark 4.5.

Let $X$ be a normed linear space, $G\subset X$ an open convex set,
$\omega\in{\cal M}$, and let $f$ be both $\omega$-semiconvex and
$\omega$-semiconcave on $G$.

  1. (i)

If $X$ is a Hilbert space and $\omega$ is linear, then $f\in
C_{*}^{1,2\omega}(G)$.

  2. (ii)

If $X={\mathbb{R}}$, then $f\in C_{*}^{1,2\omega}(G)$ too.

  3. (iii)

If $X$ is general and $\omega$ is linear, then $f\in C_{*}^{1,4\omega}(G)$.

Indeed, (i) is a reformulation of (1.2) by Remark 1.4. Easy assertion (ii)
follows e.g. from [DZ2, Proposition 2.8 (i)]. To prove (iii), we can proceed
similarly as in the second part of the proof of Theorem 3.1, choosing for
given $x$, $y$ the number $n$ so large that the points $z_{i}$ have the
property that $\overline{B}(z_{i},\|z_{i}-z_{i-1}\|)\subset G$, $i=1,\dots,n$.
Then, applying (4.1) to $y:=z_{i}$, $z:=z_{i-1}$ and $r:=\|z_{i}-z_{i-1}\|$,
the inequality $\|f^{\prime}(y)-f^{\prime}(x)\|\leq 4\omega(\|y-x\|)$ easily
follows.

##  5\. Optimality in finite-dimansional spaces

By Remark 4.5 (iii), we know that if $\omega$ is linear, then the implication
(1.6) holds for each open convex subset $G$ of an arbitrary normed linear
space. This assertion holds also for some nonlinear $\omega\in{\cal M}$ (cf.
Remark 5.3 below), but the following example shows that for the most
interesting nonlinear moduli this assertion does not hold.

######  Example 5.1.

Let $X={\mathbb{R}}^{2}$ and $G:={\mathbb{R}}\times(0,1)$. Let $\omega\in{\cal
M}$ be such that

(5.1) $\liminf_{t\to 0+}\frac{\omega(t)}{t}>0\ \ \text{and}\ \
\liminf_{t\to\infty}\frac{\omega(t)}{t}=0.$

Then the implication (1.6) does not hold.

###### Proof.

Set

$g(x):=x_{1}x_{2},\ \ x=(x_{1},x_{2})\in G.$

By (5.1) we can choose $1>\delta>0$ and $c>0$ such that $ct\leq\omega(t)$ for
each $t\in(0,\delta]$. Then the monotonicity of $\omega$ implies that
$c\delta\leq\omega(t)$ for each $t\in(\delta,\infty)$. We will show that

(5.2) $g$ is both $\left(\frac{2}{c\delta}\omega\right)$-semiconvex and
$\left(\frac{2}{c\delta}\omega\right)$-semiconcave on $G$.

To this end consider $x=(x_{1},x_{2})\in G$ and
$h=(h_{1},h_{2})\in{\mathbb{R}}^{2}$ with $x+h\in G$. Note that $|h_{2}|<1$.
We have

$|g(x+h)-g(x)-g^{\prime}(x)(h)|=|(x_{1}+h_{1})(x_{2}+h_{2})-x_{1}x_{2}-(x_{2}h_{1}+x_{1}h_{2})|=|h_{1}h_{2}|.$

Now we distinguish two cases. If $\|h\|\leq\delta$, then

$|g(x+h)-g(x)-g^{\prime}(x)(h)|=|h_{1}h_{2}|\leq\|h\|\cdot\|h\|\leq\|h\|\,\frac{1}{c}\omega(\|h\|).$

If $\|h\|>\delta$, we have

$|g(x+h)-g(x)-g^{\prime}(x)(h)|=|h_{1}h_{2}|\leq|h_{1}|\leq\|h\|\,\frac{1}{c\delta}\omega(\|h\|).$

So Lemma 2.6 implies (5.2), and consequently the function
$f:=\frac{c\delta}{2}g$ is both $\omega$-semiconvex and $\omega$-semiconcave
on $G$.

However, we will show that $f\notin C^{1,\omega}(G)$. Suppose to the contrary
that $f^{\prime}$ is uniformly continuous with modulus $D\omega$ on $G$ for
some $D>0$. Then also $h(x_{1},x_{2}):=\frac{\partial g}{\partial
x_{2}}(x_{1},x_{2})=x_{1}$ is uniformly continuous with modulus $E\omega$ on
$G$ for some $E>0$. By (5.1) there exists $t_{0}>0$ such that
$E\omega(t_{0})<t_{0}$ and therefore

$t_{0}=h\bigg{(}t_{0},\frac{1}{2}\bigg{)}=h\bigg{(}t_{0},\frac{1}{2}\bigg{)}-h\bigg{(}0,\frac{1}{2}\bigg{)}\leq
E\omega(t_{0})<t_{0},$

which is a contradiction. ∎

######  Remark 5.2.

We have proved that if $\omega\in{\cal M}$ has properties (5.1), then there
exists a $C^{1}$ function $f$ on the strip ${\mathbb{R}}\times(0,1)$ which is
both $\omega$-semiconvex and $\omega$-semiconcave, and for which $f^{\prime}$
is uniformly continuous on the ray $[0,\infty)\times\\{\frac{1}{2}\\}$ with
modulus $D\omega$ for no $D>0$.

######  Remark 5.3.

  1. (i)

If $\liminf_{t\to 0+}\frac{\omega(t)}{t}=0$, then (for arbitrary $X$ and $G$)
$f$ is $\omega$-semiconvex (resp. $\omega$-semiconcave) on $G$ if and only if
$f$ is continuous and convex (resp. concave) on $G$ (see e.g. [Kr, p. 3,
(2)]). Consequently (1.6) clearly holds for arbitrary $X$ and $G$.

  2. (ii)

If $\liminf_{t\to\infty}\frac{\omega(t)}{t}>0$, it is not difficult to prove
that (1.6) holds for $X={\mathbb{R}}^{2}$ and an arbitrary $G\subset X$.

It follows that condition (5.1) in Example 5.1 cannot be relaxed.

The proof of Theorem 1.9 for $n=2$ is based on the constructions of the
following two lemmas. (The case of general $n$ needs a nontrivial inductive
argument.)

######  Lemma 5.4.

Let $\eta:[0,\infty)\to[0,\infty)$ be a continuous nondecreasing nonconstant
concave function with $\eta(0)=0$ and $\lim_{t\to\infty}\eta(t)/t=0$. Then
there exists a concave modulus $\omega$ such that

  1. (i)

$\omega(t)\geq\min(1,t),\ \ t\in[0,\infty),$

  2. (ii)

the function $g(t):=\frac{\eta(t)\cdot\log t}{t\cdot\omega(t)},\
t\in[1,\infty),$ is bounded,

  3. (iii)

$\lim_{t\to\infty}\omega(t)/\log t=0$,

  4. (iv)

$\lim_{t\to\infty}\omega(t)=\infty$.

###### Proof.

First we will construct a concave modulus $\tilde{\omega}\in{\cal M}$, for
which conditions (i)-(iii) hold. Set $\tilde{\omega}(t):=t,\ \ t\in[0,1]$, and

$\tilde{\omega}(t):=1+\frac{1}{\eta(1)}\cdot\int_{1}^{t}\frac{\eta(u)}{u^{2}}\
du,\ \ t\geq 1.$

Then clearly $\tilde{\omega}\in{\cal M}$ and (i) holds for $\tilde{\omega}$.

Note that the function $u\mapsto\frac{\eta(u)}{u}$ is nonincreasing on
$[1,\infty)$ by concavity of $\eta$. So the function
$u\mapsto\frac{\eta(u)}{u^{2}}$ is nonincreasing on $[1,\infty)$ too.
Consequently we obtain that $\tilde{\omega}$ is concave on $[1,\infty)$ and
since $\tilde{\omega}_{+}^{\prime}(1)=1$, we obtain that $\tilde{\omega}$ is
concave on $[0,\infty)$.

Since $u\mapsto\frac{\eta(u)}{u}$ is nonincreasing on $[1,\infty)$, we obtain
that, for $1<u\leq t$, we have

$\frac{\eta(u)}{u^{2}}\geq\frac{\eta(t)}{t}\cdot\frac{1}{u},$

consequently

$\tilde{\omega}(t)\geq\frac{1}{\eta(1)}\cdot\frac{\eta(t)}{t}\cdot\int_{1}^{t}\frac{1}{u}\
du=\frac{1}{\eta(1)}\cdot\frac{\eta(t)}{t}\cdot\log t,\ \ t\geq 1,$

and so (ii) for $\tilde{\omega}$ follows.

Finally, (iii) for $\tilde{\omega}$ follows by L’Hospital’s rule, since

$\lim_{t\to\infty}\frac{\int_{1}^{t}\frac{\eta(u)}{u^{2}}\ du}{\log
t}=\lim_{t\to\infty}\frac{\frac{\eta(t)}{t^{2}}}{\frac{1}{t}}=0.$

Now it is easy to see that the modulus

$\omega(t):=\tilde{\omega}(t)+\sqrt{\log(1+t)},\ \ t\in[0,\infty],$

has all demanded properties. ∎

######  Lemma 5.5.

Let $\eta:[0,\infty)\to[0,\infty)$ be a nondecreasing nonconstant continuous
concave function such that $\eta(0)=0$ and $\lim_{t\to\infty}\eta(t)/t=0$. Set

$H:=\\{(x_{1},x_{2})\in{\mathbb{R}}^{2}:x_{1}>1,\;|x_{2}|<\eta(x_{1})\\}$

and

$\displaystyle f(x_{1},x_{2}):=\log(x_{1})x_{2},\quad(x_{1},x_{2})\in H.$

Then there exists a concave $\omega\in{\cal M}$ such that

(5.3) $\lim_{t\to\infty}\omega(t)=\infty,\ \ \ \
\lim_{t\to\infty}\omega(t)/\log t=0,$

and

(5.4) $f$ is both $(C\omega)$-semiconvex and $(C\omega)$-semiconcave for some
$C>0$.

###### Proof.

By Lemma 5.4 there exists a concave $\omega\in{\cal M}$ such that (5.3) and
conditions (i), (ii) of Lemma 5.4 hold. So we can find $D>0$ such that

(5.5) $\eta(t)\log(t)\leq Dt\omega(t)\ \ \ \text{and}\ \ \ \eta(t)\leq
Dt,\quad t\geq 1.$

To prove (5.4), we will use Lemma 2.3.

So we consider arbitrary $x=(x_{1},x_{2})\in H$, $y=(y_{1},y_{2})\in H$ and
set $h=(h_{1},h_{2}):=y-x$. By the symmetry, we can suppose that $h_{1}\geq
0$. Now we will estimate from above

$V:=|f^{\prime}(y)(y-x)-f^{\prime}(x)(y-x)|=|f^{\prime}(x+h)(h)-f^{\prime}(x)(h)|.$

An elementary computation gives

$\displaystyle V$ $\displaystyle=\left|\frac{x_{2}+h_{2}}{x_{1}+h_{1}}\cdot
h_{1}+\log(x_{1}+h_{1})h_{2}-\frac{x_{2}}{x_{1}}\,h_{1}-\log(x_{1})h_{2}\right|$
$\displaystyle\leq
h_{1}\,\left|\frac{x_{2}+h_{2}}{x_{1}+h_{1}}-\frac{x_{2}}{x_{1}}\right|+|h_{2}|\,(\log(x_{1}+h_{1})-\log(x_{1}))=:A+B.$

We have

$A=h_{1}\,\left|\frac{h_{2}x_{1}-h_{1}x_{2}}{(x_{1}+h_{1})x_{1}}\right|\leq\frac{h_{1}|h_{2}|}{x_{1}+h_{1}}+\frac{(h_{1})^{2}|x_{2}|}{(x_{1}+h_{1})x_{1}}.$

Using Lemma 5.4 (i) and $x_{1}>1$, we obtain

$\frac{h_{1}|h_{2}|}{x_{1}+h_{1}}\leq\min(\|h\|,\|h\|^{2})=\|h\|\min(1,\|h\|)\leq\|h\|\omega(\|h\|)$

and, using also $|x_{2}|\leq\eta(x_{1})\leq Dx_{1}$,

$\frac{(h_{1})^{2}|x_{2}|}{(x_{1}+h_{1})x_{1}}\leq
D\,\frac{(h_{1})^{2}}{x_{1}+h_{1}}\leq D\,\min(\|h\|,\|h\|^{2})\leq
D\,\|h\|\omega(\|h\|),$

and consequently $A\leq(D+1)\,\|h\|\omega(\|h\|)$.

If $h_{1}\leq 1$, then $h_{1}\leq\omega(h_{1})$ by Lemma 5.4 (i), and
therefore

$\displaystyle
B=|h_{2}|\log\bigg{(}1+\frac{h_{1}}{x_{1}}\bigg{)}\leq|h_{2}|\frac{h_{1}}{x_{1}}\leq|h_{2}|h_{1}\leq|h_{2}|\omega(h_{1})\leq\|h\|\omega(\|h\|).$

If $1\leq h_{1}\leq x_{1}$, note that Lemma 5.4 (i) implies
$1\leq\omega(h_{1})$ and obviously $|h_{2}|\leq 2\eta(x_{1}+h_{1})$. Therefore
we obtain, using also (5.5),

$\displaystyle B$
$\displaystyle=|h_{2}|\log\bigg{(}1+\frac{h_{1}}{x_{1}}\bigg{)}\leq
2\eta(x_{1}+h_{1})\frac{h_{1}}{x_{1}}\leq 2\frac{\eta(2x_{1})}{x_{1}}h_{1}$
$\displaystyle\leq 4Dh_{1}\leq 4Dh_{1}\omega(h_{1})\leq 4D\|h\|\omega(\|h\|).$

If $x_{1}\leq h_{1}$, then we obtain, using $|h_{2}|\leq 2\eta(x_{1}+h_{1})$,
(5.5) and concavity of $\omega$,

$\displaystyle B$ $\displaystyle=|h_{2}|(\log(x_{1}+h_{1})-\log(x_{1}))$
$\displaystyle\leq 2\eta(x_{1}+h_{1})\log(x_{1}+h_{1})\leq
2D(x_{1}+h_{1})\omega(x_{1}+h_{1})$ $\displaystyle\leq
4Dh_{1}\omega(2h_{1})\leq 8Dh_{1}\omega(h_{1})\leq 8D\|h\|\omega(\|h\|).$

Using the above estimates, we obtain $V=A+B\leq(9D+1)\,\|h\|\omega(\|h\|)$. So
Lemma 2.3 implies that (5.4) holds for $C:=9D+1$. ∎

The proof of our main result in general ${\mathbb{R}}^{n}$ use an inductive
argument which needs the following notion.

######  Definition 5.6.

Let $n\in{\mathbb{N}}$ and $\omega\in\cal M$. Then we denote by $\cal
G_{n}(\omega)$ the set of all open convex sets $G\subset{\mathbb{R}}^{n}$ for
which there exist a ray $\rho\subset G$, $C>0$ and a function $f\in C^{1}(G)$
which is both $(C\omega)$-semiconvex and $(C\omega)$-semiconcave and for which
$f^{\prime}$ is uniformly continuous on $\rho$ with modulus $D\omega$ for no
$D>0$.

######  Remark 5.7.

If $G\in\cal G_{n}(\omega)$, then the implication (1.6) does not hold. Indeed,
then the function $f^{*}:=(1/C)f$ (where $C$ and $f$ are as in Definition 5.6)
is both $\omega$-semiconvex and $\omega$-semiconcave but clearly $f^{*}\notin
C^{1,\omega}(G)$.

######  Remark 5.8.

Note that Remark 5.2 immediately implies that, if $\omega\in{\cal M}$ has
properties (5.1), then ${\mathbb{R}}\times(0,1)\in\cal G_{2}(\omega)$.

The following result on further unbounded open convex subsets in
${\mathbb{R}}^{2}$ easily follows from Lemma 5.5.

######  Lemma 5.9.

Let $u$ be a positive nondecreasing concave function on $(1,\infty)$ and let
$l$ be a negative nonincreasing convex function on $(1,\infty)$ such that
$\lim_{x\to\infty}u(x)/x=0$ and $\lim_{x\to\infty}l(x)/x=0$. Then there exists
a concave modulus $\omega\in{\cal M}$ such that
$\lim_{t\to\infty}\omega(t)=\infty$ and

(5.6) $G:=\\{(x,y):\ x>1,\ l(x)<y<u(x)\\}\in\cal G_{2}(\omega).$

###### Proof.

Set $h(x):=u(x)-l(x),\ x>1$, $a:=\max(h(2),h^{\prime}_{+}(2))$ and
$b:=2a-h(2)$. Clearly $a>0$, $b>0$. Let $\eta(x):=ax$ for $x\in[0,2]$ and
$\eta(x):=h(x)+b$ for $x\in(2,\infty)$. Then $\eta$ is clearly a nonnegative
nonconstant nondecreasing continuous concave function on $[0,\infty)$ such
that $\eta(0)=0$, $\lim_{x\to\infty}\eta(x)/x=0$ and

$G\subset
H:=\\{(x_{1},x_{2})\in{\mathbb{R}}^{2}:x_{1}>1,\;|x_{2}|<\eta(x_{1})\\}.$

Let $f(x_{1},x_{2}):=\log(x_{1})x_{2},\ (x_{1},x_{2})\in H$, and let
$\omega\in{\cal M}$ be a modulus which corresponds to $\eta$ and $f$ by Lemma
5.5. Suppose to the contrary $G\notin\cal G_{2}(\omega)$. Since $G\subset H$
and (5.4) implies that $f$ is both $(C\omega)$-semiconvex and
$(C\omega)$-semiconcave on $G$ for some $C>0$, we obtain that $f^{\prime}$ is
uniformly continuous with modulus $D\omega$ on the ray
$\rho:=[2,\infty)\times\\{0\\}\subset G$ for some $D>0$. Consequently

$\bigg{|}\frac{\partial f}{\partial x_{2}}(t,0)-\frac{\partial f}{\partial
x_{2}}(2,0)\bigg{|}=|\log t-\log 2|\leq D\omega(t-2),\ \ t>2,$

which clearly contradicts (5.3). ∎

The following lemma is geometrically obvious.

######  Lemma 5.10.

Let $V\subset{\mathbb{R}}^{n}$ be a two-dimensional Hilbert space and
$C\subset V$ a two-dimensional closed convex cone such that $\\{v,-v\\}\subset
C$ for no $v\neq 0$. Then there exists $0\neq w\in V$ such that

$\operatorname{span}(\\{w\\})\cap C=\\{0\\}.$

######  Lemma 5.11.

Let $\omega\in{\cal M}$ be concave. Let $n\geq k\geq 2$, let
$G\subset{\mathbb{R}}^{n}$ be an open convex set, and let
$L:{\mathbb{R}}^{n}\to{\mathbb{R}}^{k}$ be a linear surjection. Suppose that
$H:=L(G)\in\cal G_{k}(\omega)$ and ${\rm rec}\,(H)\subset L({\rm rec}\,(G))$.
Then $G\in\cal G_{n}(\omega)$.

###### Proof.

Since $H\in\cal G_{k}(\omega)$, there exist a ray $\rho\subset H$ and a
function $f\in C^{1}(H)$ which is both $\omega$-semiconvex and
$\omega$-semiconcave and for which $f^{\prime}$ is uniformly continuous on
$\rho$ with modulus $D\omega$ for no $D>0$. Set

$\tilde{f}(y):=f(L(y)),\ \ y\in G.$

Then $\tilde{f}\in C^{1}(G)$ and there exists $C>0$ such that $\tilde{f}$ is
both $(C\omega)$-semiconvex and $(C\omega)$-semiconcave on $G$ by Lemma 2.7.

We can write $\rho=\\{a+t\,v:\ t\in[0,\infty)\\}$, where $a\in H$ and
$v\in{\mathbb{R}}^{k}\setminus\\{0\\}$. Then $v\in{\rm rec}\,(H)$ by Lemma 2.9
(iii) and so, by the assumptions, there exists $\tilde{v}\in{\rm rec}\,(G)$
with $L(\tilde{v})=v$. Choose $\tilde{a}\in G$ with $L(\tilde{a})=a$. Then
$\tilde{\rho}:=\\{\tilde{a}+t\,\tilde{v}:\ t\in[0,\infty)\\}\subset G$. It is
easy to see that $L(\tilde{\rho})=\rho$ and, for
$\alpha:=\|\tilde{v}\|/\|v\|$,

(5.7) $\alpha\cdot\|L(\tilde{x})-L(\tilde{y})\|=\|\tilde{x}-\tilde{y}\|\ \
\text{for each}\ \ \tilde{x},\tilde{y}\in\tilde{\rho}.$

Now suppose, to the contrary, that $G\notin\cal G_{n}(\omega)$. Then there
exists $\tilde{D}>0$, such that

(5.8)
$\|(\tilde{f})^{\prime}(\tilde{x})-(\tilde{f})^{\prime}(\tilde{y})\|\leq\tilde{D}\,\omega(\|\tilde{x}-\tilde{y}\|),\
\ \ \tilde{x},\tilde{y}\in\tilde{\rho}.$

Consider the dual mapping
$L^{*}:({\mathbb{R}}^{k})^{*}\to({\mathbb{R}}^{n})^{*}$ defined by
$L^{*}(\varphi):=\varphi\circ L$, $\varphi\in({\mathbb{R}}^{k})^{*}$. Since
$L$ is surjective, $L^{*}$ is injective and so
$L^{*}:({\mathbb{R}}^{k})^{*}\to L^{*}(({\mathbb{R}}^{k})^{*})$ is a linear
isomorphism. So there exists $\beta>0$ such that

(5.9) $\beta\|\varphi\circ L\|\geq\|\varphi\|\ \ \text{for each}\ \
\varphi\in({\mathbb{R}}^{k})^{*}.$

Now consider arbitrary $x,y\in\rho$ and find
$\tilde{x},\tilde{y}\in\tilde{\rho}$ such that $L(\tilde{x})=x$,
$L(\tilde{y})=y$. Using (5.9) for $\varphi:=f^{\prime}(x)-f^{\prime}(y)$,
(5.8), (5.7) and the concavity of $\omega$, we obtain

$\displaystyle\|f^{\prime}(x)-f^{\prime}(y)\|$
$\displaystyle\leq\beta\|(f^{\prime}(x)-f^{\prime}(y))\circ
L\|=\beta\|(f^{\prime}(L(\tilde{x}))-f^{\prime}(L(\tilde{y}))\circ L\|$
$\displaystyle=\beta\|(\tilde{f})^{\prime}(\tilde{x})-(\tilde{f})^{\prime}(\tilde{y})\|\leq\beta\tilde{D}\omega(\|\tilde{x}-\tilde{y}\|)$
$\displaystyle=\beta\tilde{D}\omega(\alpha\|x-y\|)\leq\beta\tilde{D}\max(1,\alpha)\omega(\|x-y\|).$

So $f^{\prime}$ is uniformly continuous on $\rho$ with modulus $D\omega$ for
$D:=\beta\tilde{D}\max(1,\alpha)$, which is a contradiction. ∎

As an easy consequence we obtain the following fact.

######  Lemma 5.12.

Let $\omega\in{\cal M}$ be concave. Let $n\geq 2$, $G\in\cal G_{n}(\omega)$,
and let $A:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n}$ be an affine bijection. Then
$A(G)\in\cal G_{n}(\omega)$.

###### Proof.

If $A$ is linear, then the assertion easily follows from Lemma 5.11. If $A$ is
a translation, then the proof is obvious. So the lemma clearly follows. ∎

######  Proposition 5.13.

Let $n\geq 2$ and let $G\subset{\mathbb{R}}^{n}$ be an unbounded open convex
set which contains no translation of a convex cone with nonempty interior.
Then there exists a concave modulus $\omega\in\cal M$ with
$\lim_{t\to\infty}\omega(t)=\infty$ such that $G\in\cal G_{n}(\omega)$.

###### Proof.

We will proceed by induction on $n$.

a) Let $n=2$. Then our assumptions and Lemma 2.9 (iv), (vi) imply that
$\dim({\rm rec}\,(G))=1$. On account of Lemma 5.12 we can suppose
$(1,0)\in{\rm rec}\,(G)$. Now we will distinguish two cases.

If also $-(1,0)\in{\rm rec}\,(G)$, then clearly $G={\mathbb{R}}\times H$ for
an open convex set $H\subset{\mathbb{R}}$. Since $\dim({\rm rec}\,(G))=1$, $H$
is a bounded open interval. By Lemma 5.12 we can suppose $H=(0,1)$ and so
$G\in\cal G_{2}(\omega)$ e.g. for $\omega(t)=\sqrt{t}$ by Remark 5.8.

If $-(1,0)\notin{\rm rec}\,(G)$, then we can suppose (by Lemma 5.12) that
$\\{x\in{\mathbb{R}}:(x,0)\in G\\}=(1,\infty)$. We can also suppose that a
support line to $\overline{G}$ at $(1,0)$ is parallel to the $y$-axis. For
$x>1$, set $u(x):=\sup\\{y:(x,y)\in G\\}$ and $l(x):=\inf\\{y:(x,y)\in G\\}$.
It is easy to show that $u$ is a positive nondecreasing concave function on
$(1,\infty)$, $l$ is a negative nonincreasing convex function on $(1,\infty)$
and

$G=\\{(x,y):\ x>1,\ l(x)<y<u(x)\\}.$

Then $A:=\lim_{x\to\infty}u(x)/x=0$, since otherwise $(1,A/2)\in{\rm
rec}\,(G)$ and thus $\dim({\rm rec}\,(G))=2$. Similarly we obtain
$\lim_{x\to\infty}l(x)/x=0$. So our assertion follows from Lemma 5.9.

b) Now suppose that $n\geq 3$ is given and that the proposition “holds for
$n-1$”. We will distinguish two cases.

b1) There exists $v\neq 0$ such that $\\{v,-v\\}\subset{\rm rec}\,(G)$ (i.e.,
$G$ contains a line).

By Lemma 5.12, we can suppose that $v=(0,\dots,0,1)$ and thus there exists an
open convex set $H\subset{\mathbb{R}}^{n-1}$ such that
$G=H\times{\mathbb{R}}$. Then clearly $\dim({\rm rec}\,(H))<n-1$ and so the
inductive assumption gives that $H\in\cal G_{n-1}(\omega)$ for some concave
$\omega\in\cal M$ with $\lim_{t\to\infty}\omega(t)=\infty$. Let
$L(x_{1},\dots,x_{n}):=(x_{1},\dots,x_{n-1})$. Then clearly $L(G)=H$ and ${\rm
rec}\,(H)\subset L({\rm rec}\,(G))$. Therefore $G\in\cal G_{n}(\omega)$ by
Lemma 5.11.

b2) There is no $v\neq 0$ such that $\\{v,-v\\}\subset{\rm rec}\,(G)$.

We will show that there exists a linear surjection
$L:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n-1}$ such that

(5.10) $L^{-1}(\\{0\\})\cap{\rm rec}\,(G)=\\{0\\}\ \ \text{and}$ (5.11) $\ \
\dim(L({\rm rec}\,(G)))<n-1.$

If $\dim({\rm rec}\,(G))<n-1$, it is easy (using Lemma 2.9 (v)) to find $L$
satisfying (5.10) and observe that (5.11) holds.

In the opposite case Lemma 2.9 (v),(vi) imply that $\dim({\rm rec}\,(G))=n-1$
and there exist linearly independent vectors $v_{1},\dots,v_{n-1}\in{\rm
rec}\,(G)$. Now we apply Lemma 5.10 to the space
$V:=\operatorname{span}\\{v_{1},v_{2}\\}$ and the cone $C:=V\cap\,{\rm
rec}\,(G)$ (which is closed by Lemma 2.9 (ii)) and obtain a corresponding
vector $w$. Then any linear surjection
$L:{\mathbb{R}}^{n}\to{\mathbb{R}}^{n-1}$ with $L(w)=0$ clearly satisfies
(5.10). Moreover, $L(w)=0$ implies that the vectors $L(v_{1})$ and $L(v_{2})$
are linearly dependent and consequently $\dim(L({\rm rec}\,(G)))<n-1$.

So there exists $L$ for which (5.10) and (5.11) hold. Using (5.10) and Lemma
2.10 we obtain ${\rm rec}\,(L(G))=L({\rm rec}\,(G))$. So $\dim({\rm
rec}\,(L(G)))<n-1$ and consequently $L(G)\in\cal G_{n-1}(\omega)$ for some
concave $\omega\in\cal M$ with $\lim_{t\to\infty}\omega(t)=\infty$ by the
induction assumption. Therefore we obtain $G\in\cal G_{n}(\omega)$ by Lemma
5.11. ∎

##  6\. Consequences and open questions

As already mentioned, our results are very closely related to a first-order
quantitative converse Taylor theorem and to the problem whether $f\in
C^{1,\omega}(G)$ whenever $f$ is continuous and smooth in a corresponding
sense on all lines.

To describe precisely these relations, first recall [HJ, Definition 121] in
the special case of real-valued functions and $p=1$ (which deal with
approximation of a function by polynomials of degree $p\leq 1$).

######  Definition 6.1.

Let $X$ be a normed linear space and $G\subset X$ an open set. We say that a
function $f:G\to{\mathbb{R}}$ is $UT^{1}$-smooth on $G$ with modulus
$\omega\in{\cal M}$ (or shortly $UT^{1}_{\omega}$-smooth) if for each $x\in G$
there exist $b\in{\mathbb{R}}$ and $x^{*}\in X^{*}$ such that

(6.1) $|f(x+h)-(b+x^{*}(h))|\leq\|h\|\,\omega(\|h\|)\ \ \ \text{whenever}\ \ \
x+h\in G.$

######  Remark 6.2.

Note that if (6.1) holds then clearly $b=f(x)$ and $x^{*}=f^{\prime}(x)$. So
$f$ is $UT^{1}_{\omega}$-smooth on $G$ if and only if (2.3) holds.

Further, we will use the following terminology.

######  Definition 6.3.

Let $X$ be a normed linear space, $G\subset X$ an open convex set,
$\omega\in{\cal M}$, and let $f$ be a function on $G$. We will say that $f$ is
$C^{1,\omega}_{*}$-smooth on all lines if, for every $a\in G$ and $v\in X$,
$\|v\|=1$, the function $f_{a,v}:t\mapsto f(a+tv)$ defined on the open
interval $\\{t\in{\mathbb{R}}:\ a+tv\in G\\}$ is $C^{1,\omega}_{*}$-smooth.

Using this notation, we can reformulate some already mentioned facts as
follows.

######  Proposition 6.4.

Let $X$ be a normed linear space, $G\subset X$ an open convex set,
$\omega\in{\cal M}$, and let $f$ be a continuous function on $G$. Then the
following implications hold.

  1. (i)

If $f$ is both $\omega$-semiconvex and $\omega$-semiconcave, then $f$ is
$UT^{1}_{\omega}$-smooth.

  2. (ii)

If $f$ is $UT^{1}_{\omega}$-smooth, then $f$ is both $(2\omega)$-semiconvex
and $(2\omega)$-semiconcave.

  3. (iii)

If $f$ is $C^{1,\omega}_{*}$-smooth on all lines, then $f$ is both
$\omega$-semiconvex and $\omega$-semiconcave.

  4. (iv)

If $f$ is both $\omega$-semiconvex and $\omega$-semiconcave, then $f$ is
$C^{1,2\omega}_{*}$-smooth on all lines.

###### Proof.

Assertions (i) and (ii) are reformulations of Lemma 2.4 and Lemma 2.6,
respectively. Using (1.5) and Lemma 2.1 we obtain (iii). Lemma 2.1 together
with Remark 4.5 (ii) give (iv). ∎

######  Remark 6.5.

If $\omega$ in Proposition 6.4 is linear, then the converse of (i) holds by
Lemma 2.6 and the converse of (iv) holds by Remark 2.2 (ii) and Lemma 2.1.

Using Proposition 6.4, Remark 4.1 and Remark 6.2, we easily see that

Theorem 1.10 and Corollary 4.3 remain true if we replace the property “$f$ is
both $\omega$-semiconvex and $\omega$-semiconcave” by the property that “$f$
is $UT^{1}$-smooth function with modulus $\omega$” or by the property that
“$f$ is continuous and $C^{1,\omega}_{*}$-smooth on all lines”.

So we obtain in the very special case $p=1$, $Y={\mathbb{R}}$ (and finite
modulus $\omega$) a more precise qualitative versions of converse Taylor
theorem [HJ, Corollary 126] and also of [JZ, Proposition 12] and [HJ, Theorem
127] (which deal with functions which are “$C^{p,\omega}_{*}$-smooth on all
lines”).

In particular, we obtain the following version of Corollary 4.3.

######  Corollary 6.6.

Let $X$ be a normed linear space and $\omega\in{\cal M}$. Suppose that a
continuous function $f$ on $X$ is $C_{*}^{1,\omega}$-smooth on all lines. Then
$f$ is $C_{*}^{1,4\omega}$-smooth on $X$.

Further, Proposition 6.4 shows that our basic result Theorem 1.9 can be
reformulated in the following form.

######  Theorem 6.7.

Let $G\subset{\mathbb{R}}^{n}$ ($n\geq 2$) be an unbounded open convex set
which does not contain a translation of a convex cone with nonempty interior.
Then there exists a concave modulus $\omega\in{\cal M}$ with
$\lim_{t\to\infty}\omega(t)=\infty$ and a continuous function $f$ such that

  1. (i)

$f$ is $C^{1,\omega}_{*}$-smooth on all lines,

  2. (ii)

$f$ is $UT^{1}$-smooth function with modulus $\omega$ and

  3. (iii)

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

In particular, the assumptions on $G$ of [HJ, Corollary 126] cannot be relaxed
in the case $X={\mathbb{R}}^{n}$.

Finally observe that, by Proposition 6.4, equivalence (1.7) can be extended as
follows.

######  Proposition 6.8.

Let $X$ and $G$ be as in Theorem 1.7. Then, for each $\omega\in{\cal M}$ and
for each function $f$ on $G$, the following conditions are equivalent.

  1. (i)

$f\in C^{1,\omega}(G).$

  2. (ii)

$f\ \text{is both $(K\omega)$-semiconvex and $(K\omega)$-semiconcave}\
\text{for some}\ K>0.$

  3. (iii)

$f$ is $UT^{1}_{K\omega}$-smooth for some $K>0$.

  4. (iv)

$f$ is continuous and $C_{*}^{1,K\omega}$-smooth on all lines for some $K>0$.

The following natural (cf. Remark 5.3) question remains open.

######  Question 6.9.

Let $\omega\in{\cal M}$, $n\geq 2$ and let $G\subset{\mathbb{R}}^{n}$ be an
open convex set which contains no translation of a convex cone with nonempty
interior. Does there exist a function $f$ on $G$ which is both
$\omega$-semiconvex and $\omega$-semiconcave and does not belong to
$C^{1,\omega}(G)$, if

  1. (i)

$\liminf_{t\to 0+}\frac{\omega(t)}{t}>0$ and
$\liminf_{t\to\infty}\frac{\omega(t)}{t}=0$?

  2. (ii)

$\omega$ is concave, $\liminf_{t\to 0+}\frac{\omega(t)}{t}>0$ and
$\liminf_{t\to\infty}\frac{\omega(t)}{t}=0$?

  3. (iii)

$\omega(t)=t^{\alpha},\ t\geq 0,$ for some $0<\alpha<1$?

  4. (iv)

$\omega(t)=\sqrt{t},\ t\geq 0$?

## References

  * [BAJN] G.O. Berger, P.A. Absil, R.M. Jungers, Y. Nesterov: On the quality of first-order approximation of functions with Hölder continuous gradient, J. Optim. Theory Appl. 185 (2020) 17–33.
  * [BBEM] M. Bačák, J.M. Borwein, A. Eberhard, B.S. Mordukhovich: Infimal convolutions and Lipschitzian properties of subdifferentials for prox-regular functions in Hilbert spaces, J. Convex Anal. 17 (2010) 737–763.
  * [BC] M. Bardi, I. Capuzzo-Dolcetta: Optimal control and viscosity solutions of Hamilton-Jacobi-Bellman equations, Birkhäuser Boston, Boston, MA (1997).
  * [BPP] M. Bougeard, J.-P. Penot, A. Pommellet: Towards minimal assumptions for the infimal convolution regularization, J. Approx. Theory 64 (1991) 245–270.
  * [CaSi] P. Cannarsa, C. Sinestrari: Semiconcave functions, Hamilton-Jacobi equations, and optimal control, Progress in Nonlinear Differential Equations and their Applications, 58. Birkhäuser Boston, Inc., Boston, MA (2004).
  * [DZ1] J. Duda, L. Zajíček: Semiconvex functions: representations as suprema of smooth functions and extensions, J. Convex Analysis 16 (2009) 239–260.
  * [DZ2] J. Duda, L. Zajíček: Smallness of singular sets of semiconvex functions in separable Banach spaces, J. Convex Anal. 20 (2013) 573–598.
  * [HJ] P. Hájek, M. Johanis: Smooth Analysis in Banach Spaces, Walter de Gruyter, Berlin (2014).
  * [HP] J.-B. Hiriart-Urruty, P. Plazanet: Moreau’s decomposition theorem revisited, Ann. Inst. H. Poincaré Anal. Non Linéaire 6 (1989), suppl. 325–338.
  * [Iv] G.E. Ivanov: Continuity of optimal controls in differential games, and some properties of weakly and strongly convex functions (Russian), Mat. Zametki 66 (1999) 816–839; translation in Math. Notes 66 (1999) 675–693.
  * [Jo] M. Johanis: A quantitative version of the Converse Taylor theorem: $C^{k,\omega}$-smoothness, Colloq. Math. 136 (2014) 57–64.
  * [JTZ] A. Jourani, L. Thibault, D. Zagrodny: $C^{1,\omega(\cdot)}$-regularity and Lipschitz-like properties of subdifferential, Proc. Lond. Math. Soc. (3) 105 (2012) 189–223.
  * [JZ] M. Johanis, L. Zajíček: Smoothness via directional smoothness and Marchaud’s theorem in Banach spaces, J. Math. Anal. Appl. 423 (2015) 594–607.
  * [Kr] V. Kryštof: Generalized versions of Ilmanen lemma: Insertion of $C^{1,\omega}$ or $C^{1,\omega}_{{\rm loc}\,}$ functions, Comment. Math. Univ. Carolin. 59 (2018) 223–231.
  * [Kr2] V. Kryštof: Semiconvex functions and their differences, Bachelor thesis, Charles University, Prague (2013) (Czech).
  * [LL] J.M. Lasry, P.L. Lions: A remark on regularization in Hilbert spaces, Israel J. Math. 55 (1986) 257–266.
  * [Mo] J.-J. Moreau: Proximité et dualité dans un espace hilbertien, Bull. Soc. Math. France 93 (1965) 273–299.
  * [Roc] R.T. Rockafellar: Convex analysis, Princeton Mathematical Series, No. 28, Princeton University Press, Princeton, N.J. (1970).
  * [Rol1] S. Rolewicz: On $\gamma$-paraconvex multifunctions, Math. Japon. 24 (1979/80) 293–300.
  * [Rol2] S. Rolewicz: On the coincidence of some subdifferentials in the class of $\alpha(\cdot)$-paraconvex functions, Optimization 50 (2001) 353–360.
  * [Rol3] S. Rolewicz: An extension of Mazur’s theorem on Gâteaux differentiability to the class of strongly $\alpha(\cdot)$-paraconvex functions, Studia Math. 172 (2006) 243–248.
  * [Za] L. Zajíček: Differentiability of approximately convex, semiconcave and strongly paraconvex functions, J. Convex Analysis 15 (2008) 1–15.