A question on the directional derivative in a real Banach space

Question

Let $K$ be a convex and compact set in the real Banach space $(\mathbb{R}^n, |\cdot|_{\infty})$. Here, the max-norm is defined as $|v|_{\infty}=\max_{i}|v_{i}|$ and the distance function $d_{K}(\cdot)$ is defined to be $d_{K}(x):=\inf_{y \in K}|x-y|_{\infty}$. Let us use the notation $D_{K}(x;v)$ to denote the directional derivative of $d_{K}(x)$ with respect to the direction $v \in \mathbb{R}^{n}$, i.e., $$D_{K}(x;v):=\lim_{h \to 0^{+}}\frac{d_{K}(x+hv)-d_{K}(x)}{h}$$

Then the question that I want to ask is:

If $x$ is a point in $\mathbb{R}^{n}-K$ with $D_{K}(x;v) \leq 0$ for some $v \in \mathbb{R}^{n}$, then is it possible that there exists a sequence $\{ x_{n} \}$ in $\mathbb{R}^{n}-K$ and $\epsilon>0$ such that $x_{n}$ converges to $x$ and $D_{K}(x_{n};v) \geq \epsilon$ for all $n \in \mathbb{N}$?

I provide some remarks and my opinions related to this question.

• If the ambient space was just a Euclidean space, i.e., $(\mathbb{R}^{n},|\cdot|_{2})$, then the answer of this question is obviously NO. The reason is that the directional derivative $D_{K}(x;v)$ is continuous with respect to both $x$ and $v$ since $d_{K}(\cdot)$ is continuously differentiable ($C^{1}$).
• I checked that this question also fails if $K$ is convex polytope. So, I guess that this statement is false for general convex and compact set $K$ in the real Banach space $(\mathbb{R}^{n}, |\cdot|_{\infty})$ but I don't have an idea to derive a contradiction for such a general set.

I really appreciate that you read and answer my question.

Answer 1

The statement is also false for general $K$ in arbitrary Banach spaces with arbitrary norms.

Let $h>0$ be given. We choose $n$ such that $\|x_n-x\|\leq \varepsilon h /4$. Using convexity of $d_K$, one can obtain $d_K(x_n+hv)-d_K(x_n)\ge h \varepsilon$ from $D_K(x_n;v)\ge \varepsilon$. Then we can use the Lipschitz continuity (with Lipschitz constant $1$) of $d_K$ to estimate $$ \frac1h(d_K(x+hv)-d_K(x)) \geq \frac1h(d_K(x_n+hv)-d_K(x_n) - 2\|x_n-x\|) \geq \frac1h h \varepsilon - \frac2h \|x_n-x\| \geq \varepsilon - \varepsilon/2 = \varepsilon/2. $$ Taking the limit $h\to 0^+$ yields $D_K(x;v)\geq \varepsilon/2$, which makes $D_K(x;v)\leq0$ impossible.