Let
$ H$
be a real Hilbert space and
$ X$
be a convex bounded subset of
$ H$
. If
$\varphi $
is a real convex function on X, we will denote by
${\theta }(\varphi , X)$
the oscillation on
$ X$
of the convex function
$x\mapsto \left \Vert x\right \Vert {}^2 + \varphi (x)$
, so
$$ \begin{align*}{\theta}(\varphi, X) = \sup_{x\in X} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 +\varphi(x) \big)- \inf_{x\in X} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 +\varphi(x)\big) ,\end{align*} $$
hence
$\delta _X= \inf _{\varphi \in \Gamma _X} {\theta }(\varphi ,X)$
. Ricceri shows in [Reference Ricceri2] in a rather indirect way that
$\delta _X>0$
for any convex subset of H containing more than one point. He notices also that Zalinescu gave him in a private communication another indirect (and not quantitative) proof of this result based on quite hard arguments of convex analysis.
Recall that if
$ C$
is a bounded subset of some normed space
$ E$
, the Chebyshev radius (for short radius) of
$ C$
is the infimum
$\rho $
of all
$r>0$
such that
$ C$
is contained in some ball
$ B(x,r)$
for
$x\in E$
. So,
$$ \begin{align*}\rho = \inf_{a\in E} \nolimits\,\sup_{x\in C}\nolimits \left\Vert x-a\right\Vert \kern-1.2pt. \end{align*} $$
And a Chebyshev center (for short center) of
$ C$
is any point
$ a\in E$
such that
${C\subset B( a, \rho )}$
. Of course, there are examples, where such a center does not exist or is not unique. One can find in [Reference Alimov and Tsar’kov1] what is well known about radius and center of a bounded set and Jung inequalities. In particular, we will use in the sequel the following three theorems.
Theorem 1 If the Banach space
$ E$
is uniformly convex (in particular, if it is a Hilbert space), any nonempty bounded subset has a unique center.
Theorem 2 Let
$ H$
be a Hilbert space and
$ X$
be a bounded subset of
$ H$
of center a and radius
$ \rho $
. Then, for any
$ \varepsilon>0$
,
$ a$
belongs to the closed convex hull of the set of those points of
$ X$
such that
$ \left \Vert x-a\right \Vert \geq \rho -\varepsilon $
. In particular,
$a\in \overline{\mathrm{conv}}( X)$
.
Theorem 3 Let
$ H$
be a Hilbert space and
$ X$
be a bounded subset of
$ H$
of center a and radius
$ \rho $
. Then,
$\rho \sqrt 2\leq \operatorname {\mathrm {{diam}}}(X) \leq 2\rho $
. Moreover, if
$ H$
has finite dimension d, one has even
$\rho \leq \operatorname {\mathrm {{diam}}}(X) \sqrt {\dfrac d{2(d+1)}}$
.
We now prove our main theorem.
Theorem 4 If
$ X$
is a bounded convex subset of
$ H$
having radius
$\rho $
,
$ \delta _X$
is equal to
$\rho ^2$
. In particular,
$\delta _X\geq \dfrac {\operatorname {\mathrm {{diam}}}(X)^2}4$
.
For the proof of this statement we need the next lemmas.
Lemma 5 Let X be a convex bounded subset of H,
$a\in H$
and
$\in \mathbb R$
. One has
$\delta _X = \delta _{X+a}$
and
$ \delta _{ \lambda X} = \lambda ^2 \delta _X$
. Moreover, if
$ X\subset Y$
, one has
$\delta _Y\geq \delta _X$
.
Proof If
$\varphi $
is convex on
$ X$
the function
$\varphi _a : x\mapsto \varphi (x-a)$
is convex on
$X+a$
, and so is the function
$\varphi ^{\prime }_a = x\mapsto \varphi _a(x) + \left \Vert a\right \Vert {}^2 - 2{\langle }a, x{\rangle }$
. Then, for
$ y= x+a\in X+a$
:
$$ \begin{align*}\left\Vert y\right\Vert^2 +\varphi^{\prime}_a(y) &= \left\Vert x+a\right\Vert^2 + \varphi_a(y) - \left\Vert a\right\Vert^2 + 2{\langle}a, y{\rangle} \\&= \left\Vert x\right\Vert^2 + \left\Vert a\right\Vert^2 + 2{\langle}a,x{\rangle} +\varphi( x) +\left\Vert a\right\Vert^2 - 2{\langle} a, x+a{\rangle}=\varphi(x), \end{align*} $$
whence we deduce that
$\delta _{X+a} \geq \delta _X$
, and
$ \delta _{X+a} \leq \delta _X$
by replacing a by
$-a$
.
In the same way, if
$ \lambda \in \mathbb R^*$
, and if
$ \varphi $
is convex on X, the function
$\varphi _\lambda : y\mapsto \lambda ^2 \varphi ( \dfrac y\lambda )$
is convex on
$ \lambda X$
and for
$ y=\lambda x\in \lambda X$
, we get
$\left \Vert y\right \Vert {}^2 + \varphi _\lambda (y)= \lambda ^2 \left \Vert x\right \Vert + \lambda ^2\varphi (x)$
, whence
$ \delta _{ \lambda X}\geq \lambda ^2 \delta _X$
, and the equality
$\delta _{ \lambda X} = \lambda ^2 \delta _X$
by replacing
$ \lambda $
by
$ 1/\lambda $
.
If
$ Y\supset X$
and if
$\varphi $
is convex on Y, the function
$\psi = \varphi _{|X}$
is convex on
$ X$
, and we get
$$ \begin{align*}\sup_{y\in Y} \left\Vert y\right\Vert{}^2+\varphi(y) \geq \sup_X \left\Vert x\right\Vert{}^2 + \psi(x) \text{ and } \inf_{y\in Y}\big(\kern-1.2pt\left\Vert y\right\Vert{}^2+\varphi(y)\Big) \leq \inf_X \left\Vert x\right\Vert{}^2 + \psi(x) .\end{align*} $$
Thus,
$\delta _X\leq {\theta }(\psi , X) \leq {\theta }(\varphi ,Y)$
, whence we deduce that
$ \delta _Y\geq \delta _X$
.
Lemma 6 Let X be a convex bounded subset of H. If
$\rho $
is the radius of
$ X$
one has
$ \delta _X\leq \rho ^2$
.
Proof If a is the center of X and
$ \varphi _a$
the affine function
$ x\mapsto \left \Vert a\right \Vert {}^2-2{\langle } a,x{\rangle } $
, one has
$\rho = \sup _{x\in X} \left \Vert x-a\right \Vert $
and
$\left \Vert x\right \Vert {}^2 +\varphi _a(x) = \left \Vert x-a\right \Vert {}^2$
, hence
${\sup _x \big (\kern-1.2pt\left \Vert x\right \Vert {}^2 +\varphi _a(x)\big )=\rho ^2}$
and
$\inf _x \big (\kern-1.2pt\left \Vert x\right \Vert {}^2 +\varphi _a(x)\big )=\inf _{x\in X} \left \Vert x-a\right \Vert {}^2=0$
, since
$ a\in \bar X$
by Theorem 2. Thus,
$\delta _X\leq {\theta }(\varphi _a, X) =\rho ^2$
.
Lemma 7 Let
$ X$
be a bounded convex subset of
$ H$
. If
$ \varphi $
belongs to
$\Gamma ( X)$
, there exists a convex lower semi-continuous function
$\bar \varphi $
on
$ \overline {X}$
such that
${\theta }( \bar \varphi , \overline {X}) \leq {\theta }( \varphi , X)$
.
Proof If
$\varphi $
is not bounded from below on
$ X$
, one has
${\theta }(\varphi , X) = +\infty $
. Then, it is enough to take
$\bar \varphi =0$
for getting
$ {\theta }( \bar \varphi , \overline {X}) \leq \sup _{x\in X} \left \Vert x\right \Vert {}^2<{\theta }(\varphi , X) = +\infty $
.
On the contrary, if
$ \varphi $
is bounded from below take
$\bar \varphi : \overline {X}\to \mathbb R$
the lower semi-continuous envelope of
$ \varphi $
, that is,
$$ \begin{align*}\bar\varphi(x) = \liminf_{y\in X, y\to x} \varphi(y) = \sup\{ \ell(x) : \ell \text{ affine continuous on } H, \ \ell\leq \varphi \text { on } X\}, \end{align*} $$
which is convex on
$ \overline {X}$
. We, clearly, have
$ \bar \varphi \leq \varphi $
on
$ X$
, hence
$$ \begin{align*}\sup_{x\in X} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 +\varphi(x) \big)\geq \sup_{x\in X} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 +\bar\varphi(x)\big) =\sup_{x\in \bar X} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 +\bar\varphi(x)\big) .\end{align*} $$
And letting,
$\mu =\inf _{x\in X} \big (\kern-1.2pt\left \Vert x\right \Vert {}^2 +\varphi (x)\big )$
, we have for all
$ y\in X$
:
$\varphi (y) \geq \mu -\left \Vert y\right \Vert {}^2$
hence
$$ \begin{align*}\bar\varphi(x) =\liminf_{y\to x} \varphi(y) \geq \liminf_{y\to x} \mu-\left\Vert y\right\Vert{}^2= \mu-\left\Vert x\right\Vert{}^2\end{align*} $$
and
$\mu =\inf _{x\in X} \big (\kern-1.2pt\left \Vert x\right \Vert {}^2 +\varphi (x)\big )\geq \inf _{x\in \bar X}\big (\kern-1.2pt\left \Vert x\right \Vert {}^2 +\bar \varphi (x)\big ) \geq \mu $
. We conclude that
$ {\theta }(\bar \varphi , X)\leq {\theta }(\varphi , X)$
.
The convex sets
$ X$
and
$ \overline {X}$
have same center and same radius. We have already proved that
$ \delta _X\leq \rho ^2$
. We can reduce by translation the proof to the case, where the center a of
$ X$
is 0. Let
$ \varphi $
be a convex function on
$ X$
. Following Lemma 7, there exists an lower semi-continuous (l.s.c.) convex function
$\bar \varphi $
on
$ \overline {X}$
such that
${\theta }( \bar \varphi , \overline {X}) \leq {\theta }( \varphi , X)$
. For proving that
$ {\theta }( \varphi , X)\geq \rho ^2$
, we can thus assume moreover that
$ X$
is closed and that
$\varphi $
is l.s.c.
For
$ \varepsilon>0$
, define
$ X_\varepsilon = \{ x\in X : \left \Vert x\right \Vert \geq \rho -\varepsilon \}$
and
$ K_\varepsilon = \overline { \operatorname {\mathrm {{conv}}}(X_\varepsilon )}$
then put
${\mu _\varepsilon = \sup _{x\in X_\varepsilon } \varphi (x)}$
. For all
$x\in X_\varepsilon $
one has:
$$ \begin{align*}\left\Vert x\right\Vert{}^2 + \varphi(x)\geq (\rho-\varepsilon)^2 + \varphi(x) ,\end{align*} $$
hence
$$ \begin{align*}\sup_{x\in X_\varepsilon}\nolimits \big(\kern-1.2pt\left\Vert x\right\Vert^2 +\varphi(x)\big)\geq (\rho-\varepsilon)^2 + \sup_{x\in X_\varepsilon}\nolimits \varphi(x)=(\rho-\varepsilon)^2 + \mu_\varepsilon, \end{align*} $$
and
$$ \begin{align*}\sup_{x\in X} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 + \varphi(x)\big) \geq \sup_{x\in X_\varepsilon} \big(\kern-1.2pt\left\Vert x\right\Vert{}^2 +\varphi(x)\big)\geq (\rho-\varepsilon)^2 + \mu_\varepsilon .\end{align*} $$
Since the set C of those x in
$ X$
such that
$\varphi (x) \leq \mu _\varepsilon $
is a closed convex set containing
$X_\varepsilon $
, hence containing
$ K_\varepsilon $
, it follows from Theorem 2 that
$a=0\in C$
, thus that
${\varphi (0)\leq \mu _\varepsilon }$
. Thus, we get
$$ \begin{align*}{\theta}( \varphi, X) \geq \big((\rho-\varepsilon)^2 + \mu_\varepsilon\big) -\varphi(0) \geq \big((\rho-\varepsilon)^2 + \mu_\varepsilon\big) -{\mu}_\varepsilon=(\rho-\varepsilon)^2 .\end{align*} $$
Since
$\varepsilon>0$
is arbitrary, we deduce that
${\theta }( \varphi , X) \geq \rho ^2$
, hence that
$$ \begin{align*}\delta_X = \inf_{\varphi\in \Gamma_X}{\theta}(\varphi,X)\geq \rho^2\ge\delta_X .\end{align*} $$
By Theorem 3, we deduce that we necessarily have
$ \operatorname {\mathrm {{diam}}}(X) \leq 2\rho $
and thus
${\delta _X=\rho ^2\geq \dfrac {\operatorname {\mathrm {{diam}}}(X)^2}4}$
.
Using Theorem 1.2 from [Reference Ricceri2] and Theorem 4 above allows us to state the following.
Theorem Let H be a real Hilbert space,
$\Omega \subset H $
a convex open set,
$\Phi : \Omega \to H$
an operator of class
$ \mathcal C^1$
, with L-Lipschitz derivative and
$V\subseteq \Omega $
a set such that
${\eta := \inf _{x\in V} \left \Vert \Phi (x)\right \Vert>0}$
.
Then, for all bounded convex set
$X\subseteq V$
with radius smaller than
$2\eta .L$
, we have
${0\notin \operatorname {\mathrm {{conv}}}( \Phi (X))}$
.
We can also rewrite this result in the following form, denoting by
$\mathcal A_L$
the set of
$\mathcal C^1$
functions with
$ L$
-Lipschitz derivative from
$ X$
into a Hilbert space and
$ \rho $
the radius of the bounded convex open set
$ X$
:
$$ \begin{align*}\sup \Big\{\sqrt{\dfrac{ 2 d( 0, \Phi( X))} L } : \Phi\in \mathcal A_L, \ 0\in \overline{\operatorname{\mathrm{{conv}}}}\big(\Phi(X)\big) \Big\} \leq \rho. \end{align*} $$
We now prove that this result is optimal.
Theorem 8 Let X be a bounded convex nonempty open subset of the Hilbert space H and
$\rho $
be its radius. Then, there exists a function
$ \Phi $
of class
$ \mathcal C^1$
with
$ L$
-Lipschitz derivative from X to the Hilbert space
$ H\oplus \mathbb R$
satisfying
$\rho ^2 =\dfrac { 2 d( 0, \Phi (X))}L$
and
$ 0\in \overline{\operatorname {\mathrm {{conv}}}}\big (\Phi (X)\big )$
.
Denote by a the center of X. Define
$\lambda = \dfrac 1{\rho \sqrt 2}$
, then the continuous functions
${f : X\to \mathbb R}$
and
$ \Phi : X\to H\times \mathbb R$
by,
$$ \begin{align*}\left\{ \begin{aligned} f(x)\, &= \lambda \big(\rho^2-\left\Vert x-a\right\Vert{}^2\big) ,\\ \Phi(x) &=\big ( x-a, f(x)\big)\,. \end{aligned} \right. \end{align*} $$
Since
$ \left \Vert x-a\right \Vert <\rho $
for all
$ x\in X$
, we have
$ f>0$
on X.
Lemma 9 One has
$\left \Vert \Phi (x)\right \Vert \geq \left \Vert \Phi (a)\right \Vert = \dfrac {\rho }{\sqrt 2}$
for all
$x\in X$
.
Proof Indeed:
$$ \begin{align*} \left\Vert \Phi(x)\right\Vert{}^2 &= \left\Vert x-a\right\Vert{}^2 + \left\vert f(x)\right\vert{}^2 = \left\Vert x-a\right\Vert{}^2 + \lambda^2\big( \rho^2-\left\Vert x-a\right\Vert{}^2\big)^2\\ &=\lambda^2\rho^4 + \lambda^2 \left\Vert x-a\right\Vert{}^4 + (1- 2\lambda^2\rho^2) \left\Vert x-a\right\Vert{}^2=\lambda^2\rho^4 + \lambda^2 \left\Vert x-a\right\Vert{}^4\\ &\geq \lambda^2\rho^4=\dfrac{\rho^2}2, \end{align*} $$
thus
$\left \Vert \Phi (x)\right \Vert \geq \dfrac {\rho }{\sqrt 2}= \left \Vert \Phi (a)\right \Vert $
for all
$ x\in X$
.
Lemma 10 The function
$\Phi $
is of class
$ \mathcal C^1$
with L-Lipschitz derivative, where
$ L$
is equal to
$2\lambda $
.
Proof For all
$x\in X$
and
$ h\in H,$
we have:
$$ \begin{align*}\Phi'(x).h= \big( h, f'(x).h\big) = \big( h, -2\lambda {\langle} x-a, h{\rangle}\big), \end{align*} $$
what shows that
$ \Phi $
is of class
$ \mathcal C^1$
(and even of class
$ \mathcal C^\infty $
). Moreover, if x and
$ y$
belong to
$ X$
, we have
$$ \begin{align*}\left\Vert\Phi'(x).h - \Phi'(y).h\right\Vert= \left\Vert\big( 0, 2\lambda{\langle}y-x, h{\rangle}\big)\right\Vert\leq 2\lambda \left\Vert x-y\right\Vert.\left\Vert h\right\Vert\kern-1.2pt, \end{align*} $$
whence
$\left \Vert \Phi '(x)-\Phi '(y)\right \Vert \leq 2\lambda \left \Vert x-y\right \Vert $
and the fact that
$ \Phi '$
is
$2\lambda $
-Lipschitz.
Lemma 11 The origin of
$ H\oplus \mathbb R$
belongs to the closed convex hull of
$\Phi (X)$
.
Proof It follows from Theorem 2 that for all
$ \varepsilon>0$
there exist points
$( x_i)_{i\leq n} $
of
$ X\setminus B( a, \rho -\varepsilon )$
and non-negative real numbers
$ (\alpha _i)_{i\leq n}$
with sum 1 such that
$ \left \Vert a- \sum _i \alpha _i x_i\right \Vert \leq \varepsilon $
. Then, we have
$$ \begin{align*}z=\sum_i \alpha_i \Phi( x_i) = \Big( \sum_i \alpha_i x_i -a, \ \sum_i \alpha_i f( x_i) \Big). \end{align*} $$
Since
$\left \Vert x_i-a\right \Vert>\rho -\varepsilon $
, one has
$0<f(x_i) <\lambda \big ( \rho ^2-(\rho -\varepsilon )^2\big )<2\lambda \rho \varepsilon =\varepsilon \sqrt 2$
, hence
${0\leq \sum _i \alpha _i f(x_i) < \varepsilon \sqrt 2}$
and finally,
$$ \begin{align*}\left\Vert z\right\Vert^2 = \left\Vert \sum_i\nolimits \alpha_i x_i -a\right\Vert^2 + \left\vert \sum_i\nolimits \alpha_i f(x_i)\right\vert^2 \leq \varepsilon^2 +2\varepsilon^2 =3\varepsilon^2 , \end{align*} $$
whence we deduce that
$z\in \operatorname {\mathrm {{conv}}}( \Phi (X))\cap B\big ( (0,0), \varepsilon \sqrt 3\big )$
, thus that every neighborhood of
$(0,0)$
in
$ H\oplus \mathbb R$
meets the convex hull of
$\Phi (x)$
and finally that the origin
$(0,0)$
of
$ H\oplus \mathbb R$
belongs to
$\overline{\operatorname {\mathrm {{conv}}}}\big (\Phi (X)\big )$
.
This completes the proof of Theorem 8.
Corollary 12 Let X be a bounded convex nonempty open subset of the Hilbert space H and
$\rho $
be its radius. Then,
$$ \begin{align*}\sup \Big\{\sqrt{\dfrac{ 2 d( 0, \Phi( X))} L } : \Phi\in \mathcal A_L, \ 0\in \overline{\operatorname{\mathrm{{conv}}}}\big(\Phi(X)\big) \Big\} = \rho. \end{align*} $$
