# Weak$^*$ convergence of subgradients of a convex continuous function on a normed space

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. If $$f$$ is Gâteaux differentiable at $$a \in A$$, then $$|x_n - a| \to 0, x_n \in A, x_n^* \in \partial f(x_n) \implies x_n^* \stackrel{w^*}{\to} f'(a).$$

By this result, $$\partial f (a) = \{f'(a)\}$$ is a singleton. To simplify notation, let $$a^* = f'(a)$$. Assume the contrary that $$x_n^*$$ does not converge to $$a^*$$ in $$w^*$$-topology, i.e., there is $$v \in X$$ such that $$\langle x^*_n, v \rangle \not\to \langle a^*, v \rangle$$. There is $$\varepsilon > 0$$ (and by extracting a subsequence) such that $$| \langle x^*_n - a^*, v \rangle | \ge \varepsilon \quad \forall n.$$

By this result, $$f$$ is locally Lipschitz on $$A$$, i.e., there are $$r, L>0$$ such that $$B(a, r) \subset A$$ and $$f$$ is $$L$$-Lipschitz on $$B(a, r)$$. Then by this result, $$\bigcup_{x \in B(a, r)} \partial f (a) \subset L B_{X^*}.$$

By Banach-Alaoglu theorem (and by extracting a subsequence), there is $$x^* \in X^*$$ such that $$x_n^* \stackrel{w^*}{\to} x^*$$. We have for all $$x\in A$$, \begin{align} f(x) - f(x_n) &\ge \langle x_n^*, x-x_n \rangle \\ &= \langle x_n^*, x-a \rangle + \langle x_n^*, a-x_n \rangle \\ &\ge \langle x_n^*, x-a \rangle + -\|x_n^*\||a-x_n|. \end{align}

Notice that $$(x_n^*) \subset LB_{X^*}$$ is bounded. By taking the limit $$n \to \infty$$, we have $$f(x)-f(a) \ge \langle x_n^*, x-a \rangle \quad \forall x\in X.$$

So $$x^* \in \partial f (a)$$. It follows that $$x^* = a^*$$ and thus $$\langle x^*_n - a^*, v \rangle \to 0$$ which is a contradiction.

Update: @daw has pointed a mistake in above proof. Below is my fix.

By Banach-Alaoglu theorem, there are $$x^* \in X^*$$ a and a subnet $$(x^*_{\varphi(d)})_{d\in D}$$ of $$(x^*_n)$$ such that $$x^*_{\varphi (d)} \stackrel{w^*}{\to} x^*$$. Here $$D$$ is a directed set and $$\varphi:D \to \mathbb N$$ a monotone cofinal map. We have for all $$x\in A$$, \begin{align} f(x) - f(x_{\varphi (d)}) &\ge \langle x_{\varphi (d)}^*, x-x_{\varphi (d)} \rangle \\ &= \langle x_{\varphi (d)}^*, x-a \rangle + \langle x_{\varphi (d)}^*, a-x_{\varphi (d)} \rangle \\ &\ge \langle x_{\varphi (d)}^*, x-a \rangle - \| x_{\varphi (d)}^* \||a-x_{\varphi (d)}|. \end{align}

Now we need the following simple result for net convergence.

Lemma: Let $$(x_d)_{d\in D}, (y_d)_{d\in D}$$ be nets in $$\mathbb R$$. Assume $$x_d \to 0$$ and there is $$r$$ such that $$|y_d| \le r$$ for all $$d\in D$$. Then $$x_dy_d \to 0$$.

Proof: Let $$(-\varepsilon, \varepsilon)$$ be a neighborhood of $$0$$. Because $$x_d \to 0$$, there is $$d' \in D$$ such that $$x_d \in (-\varepsilon/r, \varepsilon/r)$$ for all $$d\in D$$ such that $$d' \le d$$. It follows that $$x_dy_d \in (-\varepsilon, \varepsilon)$$ for all $$d \in D$$ such that $$d' \le d$$. Hence $$x_dy_d \to 0$$.

Notice that $$(x_{\varphi (d)}^*) \subset LB_{X^*}$$ is bounded. By taking the limit and applying the Lemma, we have $$f(x)-f(a) \ge \langle x^*, x-a \rangle \quad \forall x\in X.$$

So $$x^* \in \partial f (a)$$. It follows that $$x^* = a^*$$ and thus $$\langle x^*_{\varphi(d)} - a^*, v \rangle \to 0$$ which is a contradiction.

• I think going to weak-star converging sub-sequences requires $X$ to be separable.
– daw
Commented Jun 23, 2022 at 12:31
• @daw I have updated my proof with a fix using net convergence. Could you have a check on my attempt? Commented Jun 23, 2022 at 15:12