Let $K, L$ be two closed convex subsets of the normed space $X$. If $$ 0 \in \operatorname{Int}(K-L) $$
Prove that $$ \forall x \in K \cap L \Rightarrow T_{L \cap K}(x)=T_{K}(x) \cap T_{L}(x) $$ In which $T_{A}(x)$ is the Bouligand cone to $A$ at $x \in A$.
Let's recall that $y \in T_{A}(x)$ iff there exist $(x_n) \subset A$ and $(\lambda_n) \subset \mathbb{R}^{+}$ satisfying \begin{cases} x_n \longrightarrow x \\ \lambda_n(x_n-x) \longrightarrow y \end{cases} I have dealt with the easy inclusion $T_{L \cap K}(x) \subset T_{K}(x)\cap T_{L}(x)$.
For the other inclusion, let $z \in T_{K}(x)\cap T_{L}(x)$ then there exist $(x_n) \subset K, (y_n) \subset L, (\lambda_n) \subset \mathbb{R}^{+}$ and $(\mu_n) \subset \mathbb{R}^{+}$ satisfying \begin{cases} x_n \longrightarrow x \\ \lambda_n(x_n-x) \longrightarrow z \end{cases} and \begin{cases} y_n \longrightarrow x \\ \mu_n(y_n-x) \longrightarrow z \end{cases} We have to find a sequence $(z_n) \subset L \cap K$ and a sequence $(\gamma_n) \subset \mathbb{R}^{+}$ so that \begin{cases} z_n \longrightarrow x \\ \gamma_n(z_n-x) \longrightarrow z \end{cases} My idea is as follows:
If the set $A=\left\{n \in \mathbb{N}: x_n=y_n \right\}$ is infinite, we can easily construct a subsequence of $(x_n)$ satisfying the required thing.
In the case that $A$ is finite, I know we have to use the assumptions that $K$ and $L$ are two closed convex set and that $0 \in \operatorname{Int}(K-L)$ but I'm stuck here.
Anybody can help me? Any help would be appreciated!
