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

Question

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

Answer 1

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.

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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.