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Let $C$ be an open convex subset of a Banach space $X$ and $\mathcal F$ a collection of real-valued continuous functions on $C$. We say that $\mathcal{F}$ is pointwise bounded if, for each $x \in C$, the set $\{f(x) \mid f \in \mathcal{F}\}$ is bounded.

Theorem: Assume $X$ is separable. Let $(f_n)$ be a pointwise bounded sequence of l.s.c. convex functions on $C$. Then there exists a subsequence $(f_{\varphi(n)})$ of $(f_n)$ that converges pointwise and uniformly on compact subsets to a continuous convex function on $C$.

Akira
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1 Answers1

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We need the following lemmas.

Lemma 1: Let $\mathcal{F}$ be a family of l.s.c. convex functions on $C$. If $\mathcal{F}$ is pointwise bounded, then $\mathcal{F}$ is locally equi-Lipschitz and locally equi-bounded.

Lemma 2: Let $(f_{n})$ be a sequence of l.s.c. convex functions on $C$ that converges pointwise on $C$ to a (convex) function $f: C \to \mathbb{R}$. Then $f$ is continuous and the convergence is uniform on compact sets.

Let $D$ be a countable dense subset of $C$. By diagonal argument, we can extract a subsequence $(f_{\varphi(n)})$ such that $(f_{\varphi(n)}(x))$ is convergent for each $x \in D$.

Fix $a\in C$ and $\varepsilon>0$. By Lemma 1, $(f_n)$ is locally equi-Lipschitz, so there is $r,L>0$ such that $f_n$ is $L$-Lipschitz on $B(a, r) \subset C$ for all $n \in \mathbb N$. There is $b \in D$ such that $\|b-a\| < \min\{r, \varepsilon /(2L)\}$. There is $N\in \mathbb N$ such that $$ |f_{\varphi(n)} (b) - f_{\varphi(m)} (b)| < \varepsilon \quad \forall m,n>N. $$

It follows that for all $m,n>N$, we get $$ \begin{align} |f_{\varphi(n)} (a) - f_{\varphi(m)} (a)| &\le |f_{\varphi(n)} (b) - f_{\varphi(n)} (a)| +|f_{\varphi(m)} (b) - f_{\varphi(m)} (a)| +|f_{\varphi(n)} (b) - f_{\varphi(m)} (b)| \\ &\le L\|b-a\| +L\|b-a\| + \varepsilon \\ &\le 2L \varepsilon /(2L)+ \varepsilon = 2\varepsilon. \\ \end{align} $$

It follows that $(f_{\varphi(n)}(a))$ is Cauchy and thus convergent. The rest then follows from Lemma 2.

Akira
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