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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. For $a\in A$ and $v\in X$, $$ f'_+(a)[v] = \max_{x^* \in \partial f (a)} \langle x^*, v \rangle. $$

Analyst
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  • Since $f$ is defined on $A$ and $A$ is open, isn't the continuity of $f$ directly implied by its convexity? Since a convex fonction is locally Lipschitz on the interior of its domain – P. Camilleri Dec 20 '23 at 10:44
  • @P.Camilleri What you said is only true if $X$ is finite-dimensional. – Analyst Dec 22 '23 at 08:51

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By this result, $\partial f (a) \neq \emptyset$. By this result, $f'_+(a)[v]$ exists finite for all $v\in X$. By this result, $\langle x^*, v \rangle \leq f'_+ (a)[v]$ for all $v \in X$. Let's construct an optimal $x^*$. We need the following lemma.

Lemma: Let $X$ be a t.v.s., $f: X \to \mathbb R \cup \{+\infty\}$ convex, $A$ an affine subspace of $X$, and $g : A \to \mathbb R$ affine such that $g \le f$ on $A$. Assume that $\operatorname{int} (\operatorname{dom} f) \cap A \neq \emptyset$. Then there is an affine extension $\hat g : X \to \mathbb R$ of $g$ such that $\hat g \le f$. If, moreover, $f$ is continuous at some point in $\operatorname{dom} f$, then $\hat g$ is continuous.

We extend $f$ to the whole $X$ by setting $f(x) := +\infty$ for all $x \notin A$. We define $$ g: a +\mathbb R v \to \mathbb R, a+tv \mapsto f(a)+tf'_+(a)[v]. $$

Then $g(a) = f(a)$. By Lemma, there is a continuous affine extension $\hat g:X \to \mathbb R$ such that $\hat g \le f$. There are $x^* \in X^*$ and $\alpha \in \mathbb R$ such that $\hat g = x^*+\alpha$.

Because $\hat g (a) = f(a)$, $\hat g(x) = \langle x^*, x-a\rangle +f(a)$. Because $\hat g \le f$, then $f(x)-f(a) \ge \langle x^*, x-a\rangle$ for all $x\in X$.

We have $\hat g(a+v) = \langle x^*, v\rangle +f(a)$ and $g(a+v) = f(a)+f'_+(a)[v]$. Then $\langle x^*, v\rangle = f'_+(a)[v]$. This completes the proof.

Analyst
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