# Characterize subdifferential of a convex function by directional derivative

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. 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 $$v \in X$$.

As a corollary, we obtain that $$\partial f(a)$$ is fully determined by the values of $$f$$ in any neighborhood of $$a$$.

Below, we use $$x^*(v)$$ and $$\langle x^*, v \rangle$$ interchangeably. Notice that $$f'_-(a)[v] = -f'_+(a)[-v]$$, so (ii) is equivalent to (iii).
Let's prove (i) implies (ii). Let $$x^* \in \partial f(a)$$, i.e., $$f(x)-f(a) \ge \langle x^*, x-a \rangle \quad \forall x\in A.$$ Then $$f(a+tv)-f(a) \ge t\langle x^*, v \rangle$$ and thus $$\frac{f(a+t v)-f(a)}{t} \ge \langle x^*, v \rangle \quad \forall t>0.$$ The claim then follows by taking the limit $$t \to 0^+$$.
Let's prove (ii) implies (i). Notice that $$f$$ is convex, so the map $$\varphi:(0, +\infty) \to \mathbb R, t \mapsto \frac{f(a+t v)-f(a)}{t}$$ is increasing. Assume $$\langle x^*, v \rangle \le f_{+}^{\prime}(a)[v]$$ for each $$v \in X$$. Then $$\langle x^*, v \rangle \le \varphi (t)$$ for all $$t>0$$. We pick $$t=1$$ and $$v=x-a$$. Then $$\langle x^*, x-a \rangle \le \frac{f(a+1(x-a))-f(a)}{1}.$$ This completes the proof.