Fact 2.1 Let $B\subset X$ be a convex set. If $y_0\in \partial B$ and $\emptyset \neq J\subset \mathrm {int}(B)$ then

$$ \begin{align*}[y_0-\overline{\mathrm{cone}}(J-y_0)]\cap\mathrm{int} (B)=\emptyset\qquad \text{and}\qquad [y_0-\mathrm{cone}(J-y_0)]\cap B=\emptyset.\end{align*} $$

Fact 2.2 Let $C\subset X$ be a nonempty closed convex set and $v\in X\setminus \{0\}$ . Then:

1. (i) v is a recession direction for C if and only if there exists $c_0\in C$ such that the half-line $c_0+\mathbb {R}^+v$ is contained in C;

2. (ii) if v is a recession direction for C and there exist $c_0\in C$ and $\lambda>0$ such that $c_0+\lambda v\in \partial C$ , then the half-line $c_0+\mathbb {R}^+v$ is contained in $\partial C$ .

Proof If $[0,x_0]\subset \partial C$ then $\delta =0$ and the statement is quite clear. So let us assume that $[0,x_0]$ is not contained in $\partial C$ , and hence $(0,x_0)\subset \mathrm {int}(C)$ and $\delta>0$ . Let $z_0\in \partial C$ be such that $\|\frac 12x_0-z_0\|=\delta $ . Then (3.1) easily implies that $x_0$ and $z_0$ are linearly independent. Hence, via an appropriate linear homeomorphism, we can (and do) assume that X coincides with $\mathbb {R}^2$ , equipped with a certain norm $\|\cdot \|$ , and

$$ \begin{align*}x_0=(2,0),\qquad \qquad z_0=(1,1).\end{align*} $$

Let us define

$$ \begin{align*}v_0=z_0-\frac12x_0=(0,1),\qquad w_0=z_0-x_0=(-1,1),\end{align*} $$

$$ \begin{align*}H=\{(u,v)\in X:\, v\geq |u|\},\qquad K=\{(u,v)\in X:\, u\geq |v|\},\end{align*} $$

$$ \begin{align*}H^\circ=\{(u,v)\in X:\, v> |u|\},\qquad K^\circ=\{(u,v)\in X:\, u> |v|\}.\end{align*} $$

We claim that:

1. (i) $(-K^\circ )\cap C=\emptyset $ and $(-K)\cap \mathrm {int}(C)=\emptyset $ ;

2. (ii) $(z_0+H^\circ )\cap C=\emptyset $ and $(z_0+H)\cap \mathrm {int} (C)=\emptyset $ .

Let us show (i). By (3.1), we have $\frac 12x_0+\delta B_X\subset C$ . In particular, the segment ${J=(z_0,x_0-z_0)}$ is contained in $\mathrm {int} (C)$ (indeed, $\frac 12 x_0$ is its midpoint). Since clearly ${K^\circ =\mathrm {cone}(J) }$ , (i) follows by Fact 2.1. The proof of (ii) follows similarly, taking into account that $z_0\in \partial C$ and that the segment $(0,x_0)$ is contained in $\mathrm {int} (C)$ .

Some simple cases. If $\partial C$ contains a half-line from $0$ , we are immediately done by picking x from that half-line. Now, if $w_0$ is a recession direction for C then the half-line $\mathbb {R}^+w_0$ is contained in $(-K)\cap C$ , and hence in $\partial C$ by (i) above. If $z_0$ is a recession direction for C, the half-line $\mathbb {R}^+ z_0$ contains $z_0\in \partial C$ , and hence $\mathbb {R}^+ z_0\subset \partial C$ by Fact 2.2,(ii). So, in both these cases we are done.

The main case. Assume now that $w_0$ and $z_0$ are not recession directions for C.

Let us denote $W=\{(u,v)\in X:\, v\geq 0\}$ and notice that, by our previous observations, we can consider the simple curve $\gamma $ describing the part of $\partial C$ contained in $W\setminus (x_0+ K^\circ )$ , with endpoints $0$ and $q_0$ (see Figure 1).

Figure 1

The curve $\gamma$ .

We claim that $\|q_0\|\geq \|x_0\|$ . Indeed, if we consider $B=\|x_0\|B_X$ , we clearly have ${x_0\in \partial B}$ ; moreover, since by (3.1) we have

$$ \begin{align*} \|2z_0-x_0\|=\|x_0-2z_0\|=2\delta<\|x_0\|, \end{align*} $$

the interval $I=(2z_0-x_0, x_0-2z_0)$ is contained in $\mathrm {int}(B)$ . Now, by Fact 2.1, we have that the set

$$ \begin{align*} [x_0-\overline{\mathrm{cone}}(I-x_0)]=x_0+K \end{align*} $$

does not intersect $\mathrm {int} (B)$ . Since

$$ \begin{align*} q_0\in x_0+\mathbb{R}^+z_0\subset x_0+K, \end{align*} $$

$q_0\notin \mathrm {int}(B)$ and the claim is proved.

Now, by continuity, if $\varepsilon \in \left (0,\|x_0\|\right ]$ then there exists $x\in \gamma $ such that $\|x\|=\varepsilon $ . To conclude the proof it is sufficient to show that $d\left (\frac 12x,\partial C\right )\leq \delta $ . Since the image of $\gamma $ is contained in $\partial C\cap [W\setminus (x_0+ K^\circ )]$ and since $(-K^\circ )\cap C=\emptyset $ , we have that

$$ \begin{align*} \textstyle \frac12x\in W\setminus[(\frac12x_0+ K^\circ)\cup(-K^\circ)]=\{(u,v)\in X:\, v\geq 0,\, v\geq u-1,\, v\geq-u\}. \end{align*} $$

Notice that the set $W\setminus [(\frac 12x_0+ K^\circ )\cup (-K^\circ )]$ is the union of the following sets (see Figure 2):

• • $T_0=\{\frac \alpha 2x_0+\frac \beta 2z_0:\, \alpha>0,\, \beta \geq 0,\, \alpha +\beta <1\} $ ;

• • $T_1= \frac 12x_0+ H$ ;

• • $T_2=\{(u,v)\in X:\, -u\leq v< -u+1,\, u<0\}$ ;

• • $T_3=\{\frac \alpha 2z_0+\beta v_0:\, \alpha \geq 0,\, \beta \geq 0,\, \alpha +\beta <1\}$ .

Figure 2

Proof of Proposition 3.1.

Let us observe that, since the segment $(0,x_0)$ is contained in $\mathrm {int}(C)$ , we have that

$$ \begin{align*}2T_0=\mathrm{conv}\bigl((0,x_0)\cup\{z_0\}\bigr)\setminus\{z_0\}\subset \mathrm{int}(C).\end{align*} $$

Since $x\in \partial C$ , it follows that $x\not \in 2T_0$ and hence that $\frac 12x\not \in T_0$ .

Now, consider the case in which $\frac 12 x\in T_1$ . Then necessarily

$$ \begin{align*}\textstyle \frac12x+v_0\in z_0+H\subset X\setminus\mathrm{int}(C),\end{align*} $$

and hence

$$ \begin{align*}\textstyle d\left(\frac12x,\partial C\right)\leq\|v_0\|=\|\frac12x_0-z_0\|=\delta.\end{align*} $$

Similarly, if $\frac 12 x\in T_2$ , then necessarily the segment $[\frac 12x,\frac 12x-v_0]$ intersects $-K$ , and hence we get $d\left (\frac 12x,\partial C\right )\leq \|\frac 12x-(\frac 12x-v_0)\|=\|v_0\| =\delta .$

It remains to consider the case in which $\frac 12x\in T_3$ . In that case, by Observation 3.2,(i) the ray $x+\mathbb {R}^+w_0$ does not intersect $\mathrm {int}(C)$ . Let us denote $\frac 12x=(\overline {u},\overline {v})$ and define

$$ \begin{align*}z=x+(-\overline{u},\overline{u})=x+\overline{u} w_0\in x+\mathbb{R}^+w_0\in X\setminus\mathrm{int}(C).\end{align*} $$

Observe that

$$ \begin{align*}\textstyle \|z-\frac12x\|=\|\frac12x+\overline{u} w_0\|=\| (0, \overline{v}+\overline{u})\|\leq\|(0,1)\|=\|v_0\|=\delta,\end{align*} $$

where the inequality holds since $0\leq \overline {v}+\overline {u}\leq 1$ . This clearly implies that ${d\left (\frac 12x,\partial C\right )\leq \delta }$ . The proof of Proposition 3.1 is complete.

Proof By (2.1), it is sufficient to prove that $\delta ^{\prime }_C(x,\varepsilon ) \leq \widetilde {\delta }_C(x,\varepsilon )$ , whenever $0\le \varepsilon <\Delta _x(C)$ . If $\widetilde {\delta }_C(x,\varepsilon )=\frac {\varepsilon }2$ or $\delta ^{\prime }_C(x,\varepsilon )=0$ there is nothing to prove, hence we can suppose that $\widetilde {\delta }_C(x,\varepsilon )<\frac {\varepsilon }2$ and $\delta ^{\prime }_C(x,\varepsilon )>0$ . Moreover, without any loss of generality, we can (and do) suppose that $x=0$ .

Let us consider a real number $\eta $ such that $ \widetilde {\delta }_C(x,\varepsilon )<\eta <\frac {\varepsilon }2 $ . Then there exist $x_0\in C$ and $w_0\in \partial C$ such that $\|x_0\|\geq \varepsilon $ and $\|\frac {x_0}2-w_0\|<\eta $ . It is easy to see that, since $\delta ^{\prime }_C(x,\varepsilon )>0$ , one must have $(0,x_0)\subset \mathrm {int}(C)$ . Hence, since $\eta <\frac {\varepsilon }{2}\leq \frac {\|x_0\|}{2}$ , the vectors $x_0,w_0$ are linearly independent. Let Y be the two-dimensional subspace of X defined as $Y=\mathrm {span}\{x_0,w_0\}$ . By compactness, there exists $z_0\in Y\cap \partial C$ such that:

$$ \begin{align*}\textstyle \delta:=\|\frac{x_0}2-z_0\|=d\left(\frac{x_0}{2},Y\cap\partial C\right)\leq \|\frac{x_0}2-w_0\|<\eta<\frac{\varepsilon}2\leq\frac{\|x_0\|}2.\end{align*} $$

By Proposition 3.1, there exists $x'\in Y\cap \partial C$ such that

$$ \begin{align*}\textstyle\|x'\|=\varepsilon\qquad \text{ and}\qquad d\left(\frac12 x',Y\cap \partial C\right)\leq\delta.\end{align*} $$

In particular, we have

$$ \begin{align*}\textstyle \delta^{\prime}_C(x,\varepsilon)\leq d\left(\frac{x+x'}2 , \partial C\right)\leq d\left(\frac{x+x'}2,Y\cap \partial C\right)\leq\delta<\eta.\end{align*} $$

By the arbitrariness of $\eta \in \left ( \widetilde {\delta }_C(x,\varepsilon ),\frac {\varepsilon }2\right )$ , we get $\delta ^{\prime }_C(x,\varepsilon ) \leq \widetilde {\delta }_C(x,\varepsilon )$ and the proof is done.

Observation 3.4 (see [Reference De Bernardi and Veselý9, Observations 2.1 and 3.3])

Let $C\subset X$ be a convex body.

1. (i) If $\partial C$ contains no line, then ${C}$ contains no line either.

2. (ii) The distance-function $d(\cdot ,\partial C)$ is concave on ${C}$ .

3. (iii) $\widetilde {\delta }^{\prime }_C(\varepsilon )=\widetilde {\delta }_C(\varepsilon )\le \delta _C(\varepsilon )\le \delta ^{\prime }_C(\varepsilon )\le \varepsilon /2$ for each $\varepsilon \in [0,\mathrm {diam}(C))$ .

4. (iv) At $\varepsilon =0$ , all four moduli are null and right-continuous.

5. (v) If $\partial C$ contains a segment of length $0<\ell <\mathrm {diam}(C)$ then $\delta ^{\prime }_C(\varepsilon )=0$ for each $\varepsilon \in [0,\ell ]$ .

Proof First, we proceed as at the beginning of the proof of [Reference De Bernardi and Veselý9, Theorem 3.6]. By Observation 3.4(c),(d), it suffices to show that $\delta ^{\prime }_C(\varepsilon )\le \widetilde {\delta }^{\prime }_C(\varepsilon )$ whenever $0< \varepsilon <\mathrm {diam}\,C$ . Without any loss of generality, we can assume that

$$ \begin{align*}0< \varepsilon<\mathrm{diam}\,C \,, \quad \delta^{\prime}_C(\varepsilon)>0\,, \quad \widetilde{\delta}^{\prime}_C(\varepsilon)<\varepsilon/2\, \end{align*} $$

and hence, by Observation 3.4(a),(e), that C does not contain lines.

Fix an arbitrary $0<\eta <(\varepsilon /2) - \widetilde {\delta }^{\prime }_C(\varepsilon )$ , the proof is concluded if we prove that $\delta ^{\prime }_C(x)\leq \widetilde \delta ^{\prime }_C(\varepsilon )+\eta $ . By the definition of $\delta ^{\prime }_C(\varepsilon )$ , there exist points $x,y\in C$ such that $\|x-y\|=\varepsilon $ and $d(\frac {x+y}2,\partial C)<\widetilde {\delta }^{\prime }_C(\varepsilon ) + \eta $ . By Observation 3.4(b) and since C does not contain lines, we can shift the segment $[x,y]$ along the line $\mathrm {aff}\{x,y\}$ in one of the two directions till the boundary in such a way that the new segment $[\hat {x},\hat {y}]$ satisfies $d(\frac {\hat {x}+\hat {y}}2,\partial C) \le d(\frac {x+y}2,\partial C)$ . Hence we can (and do) assume that $x\in \partial C$ . If also $y\in \partial C$ then we immediately get that $\delta ^{\prime }_C(\varepsilon )\le \widetilde {\delta }^{\prime }_C(\varepsilon ) + \eta $ ; so let us assume that $y\in \mathrm {int}\,C$ . Notice that, in particular, $\Delta _x(C)>\varepsilon $ and $\widetilde \delta ^{\prime }_C(x,\varepsilon )<\widetilde \delta ^{\prime }_C(\varepsilon )+\eta $ . Now, an application of Theorem 3.3 yields

$$ \begin{align*}\delta^{\prime}_C(\varepsilon)\leq\delta^{\prime}_C(x,\varepsilon)=\widetilde\delta^{\prime}_C(x,\varepsilon)<\widetilde\delta^{\prime}_C(\varepsilon)+\eta,\end{align*} $$

and the proof is done since $\eta $ can be arbitrarily small.