About the definition of directional derivative in the clousure of an open set

Question

Let be $X$ a normed space and thus we prove that if $x_0\in\overset{\circ}Y\subseteq X$ then for any $\vec v\in X$ such that $\lVert\vec v\lVert=1$ the set $$ V:=\{h\in\Bbb R:(x_0+h\vec v)\in\overset{\circ}Y\} $$ is a not empty open set that has $0$ as a cluster point. So first of all we observe that if $x_0\in\overset{\circ}Y$ then $0\in V$ and thus it is not empty. Anyway if $x_0\in\overset{\circ}Y$ then there exists $\delta>0$ such that $B(x_0,\delta)\subseteq\overset{\circ} Y$ and thus $(x_0+h\vec v)\in \overset{\circ}Y$ for any $h\in(-\delta,\delta)$ so that $0$ is a cluster point of $V$. Then if $(x_0+h\vec v)\in\overset{\circ}Y$ there exists $\delta>0$ such that $B(x_0+h\vec v,\delta)\subseteq\overset{\circ}Y$ and thus $(x_0+t\vec v)\in\overset{\circ}Y$ for any $t\in(h-\delta,h+\delta)$ so that $(h-\delta,h+\delta)\subseteq V$ and thus this it is open. Now if $\overset{\circ}Y$ is not empty and so that for any $x_0\in\overset{\circ}Y$ there exists $\delta>0$ such that $B(x_0,\delta)\subseteq\overset{\circ}Y$ then $V\neq\emptyset$ for any $\vec v\in V$ such that $\lVert\vec v\lVert=1$ (indeed an open ball is a convex set) and so there exist at least a net in $V$ converging to $0$ so that if $f:Y\rightarrow\Bbb R^n$ is a function then the limit $$ \lim_{h\rightarrow 0}\frac{f(x_0+h\vec v)-f(x)}h $$ is well defined and we call it the deriviative of $f$ in the direction $\vec v$. Anyway if $x_0\notin\overset{\circ}Y$ it could happen that one of the limits $$ \lim_{h\rightarrow 0^+}\frac{f(x_0+h\vec v)-f(x)}h\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\lim_{h\rightarrow 0^-}\frac{f(x_0+h\vec v)-f(x)}h $$ could be defined and we can think to exted the notion of directional derivative: e.g. the upper half space $H^k:=\{x\in\Bbb R^k:x_k\ge0\}$ is closed and if $x_0\in\partial H^k=\{x\in\Bbb R^k:x_k=0\}$ the limit $$ \lim_{h\rightarrow 0^+}\frac{f(x_0+h\vec e_k)-f(x)}h $$ is well defined for any function $f$ whose domain is the upper half space. So I ask if $x_0$ is a cluster point of $\overset{\circ}Y$ $\Biggl($and thus $x_0\in\overline{\overset{\circ\,}Y}\Biggl)$ then at least for one $\vec v\in X$ such that $\lVert\vec v\lVert=1$ the set $V$ as above defined is a not empty open set that has $0$ as cluster point so that it is possible to exted the notion of directional derivative in the clousure of an open set. So could someone help me, please?

Answer 1

It is not true that $V$ will always be non-empty.

We should note that the vector $\vec{v}$ was chosen first, then $V$ is a subset of the real numbers such that $x_0+\vec{v}V \subset \mathring{Y}$. It is a linear crossection of a particular direction of $Y$, particularly in the direction of $\vec{v}$.

To construct a counterexample for any $\vec{v}$ consider the set of points in $\mathbb{R}^2$ satisfying $$(x, y) \in \mathring{Y} \iff x>0 \wedge 0<y<x^2.$$ (Convince yourself that this is an open set.)

The limit point we will consider is $(0, 0)$. Take any vector $\vec{v}$. The set $V$ is either empty or a half infinite open interval that does not contain $0$. It is clear that $0$ is not a cluster point of any $V$.