Assume $A$ is open convex and $f$ convex continuous. Then $f$ is Gâteaux differentiable at $a \in A$ if and only if $\partial f (a)$ is a singleton This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments. See: SE blog: Answer own Question and MSE meta: Answer own Question.

Let $A$ be a subset of a normed space $X$ and $f:A \to \mathbb R$.

*

*Let $a \in \operatorname{int} A$. For $v \in X$, the right directional derivative $f_{+}^{\prime}(a)[v]$, the left directional derivative $f_{-}^{\prime}(a)[v]$, and the (bilateral) directional derivative $f^{\prime}(a)[v]$ are defined by:
$$
\begin{aligned}
f_{+}^{\prime}(a)[v] &= \lim _{t \to 0^+} \frac{f(a+t v)-f(a)}{t} \\
f_{-}^{\prime}(a)[v] &= \lim _{t \to 0^-} \frac{f(a+t v)-f(a)}{t} \\
f^{\prime}(a)[v] &= \lim _{t \to 0} \frac{f(a+t v)-f(a)}{t}.
\end{aligned}
$$
We say that $f$ is Gâteaux differentiable at $a$ if $f^{\prime}(a) \in X^{*}$.


*The subdifferential of $f$ at $a \in A$ is the set
$$
\partial f(a)=\left\{x^* \in X^* \mid f(x) - f(a) \ge \langle x^*, x-a \rangle \text { for each } x \in A\right\}.
$$
The elements of $\partial f(a)$ are called subgradients of $f$ at $a$.

Theorem: Assume $A$ is open convex and $f$ convex continuous. Then $f$ is Gâteaux differentiable at $a \in A$ if and only if $\partial f (a)$ is a singleton. In this case, $\partial f (a) = \{f'(a)\}$.

 A: We need the following lemmas

Lemma 1: Assume $A$ is open convex and $f$ convex. For $a\in A$ and $x^* \in X^*$, the following assertions are equivalent:

*

*(i) $x^* \in \partial f(a)$;

*(ii) $x^*(v) \leq f_{+}^{\prime}(a)[v]$ for each $v \in X$;

*(iii) $f_-^{\prime}(a)[v] \leq x^*(v) \leq f_{+}^{\prime}(a)[v]$ for each $v \in X$.



Lemma 2: Assume $A$ is open convex and $f$ convex continuous at $a\in A$. If $f^{\prime}(a) [v]$ exists finite for all $v \in X$, then $f$ is Gâteaux differentiable at $a$.


Lemma 3: Assume $A$ is open convex and $f$ convex continuous. For $a\in A$ and $v\in X$,
$$
f'_+(a)[v] = \max_{x^* \in \partial f (a)} \langle x^*, v \rangle.
$$

By this result, $\partial f (a) \neq \emptyset$. Let $x^* \in \partial f (a)$. By this result, $f'_+(a)[v]$ exists finite for all $v\in X$.
Let $f$ be Gâteaux differentiable at $a$. By Lemma 1, $f_-^{\prime}(a)[v] \leq x^*(v) \leq f_{+}^{\prime}(a)[v]$ and thus $f^{\prime}(a)[v] \leq x^*(v) \leq f^{\prime}(a)[v]$ for each $v \in X$. This implies $f'(a) = x^*$.
Let $\partial f (a)$ be a singleton. By Lemma 2, it suffices to show that $f'_+(a)[v] = f'_-(a)[v]$. This is indeed true from the fact that $f'_-(a)[v] =-f'_+(a)[-v]$ and Lemma 3.
