If $\mathcal{F}$ is pointwise bounded, then $\mathcal{F}$ is locally equi-Lipschitz and locally equi-bounded

Question

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Let $X$ be a Banach space and $\mathcal{F}$ a family of real-valued functions on $X$.

• We say that $\mathcal{F}$ is pointwise bounded if, for each $x \in X$, the set $\{f(x) \mid f \in \mathcal{F}\}$ is bounded.

• We say that $\mathcal F$ is pointwise equi-continuous if, for each $x\in X$, for each $\varepsilon>0$, there exists $\delta>0$ such that for all $f\in \mathcal F$, we get $\|y-x\| < \delta \implies |f(y)-f(x) < \varepsilon|$.

• We say that $\mathcal F$ is locally equi-bounded if, for each $x\in X$, there exist a neighborhood $U$ of $x$ and $m \in \mathbb R$ such that for all $f\in \mathcal F$, for all $y\in U$ we get $|f(y)| \le m$.

• We say that $\mathcal F$ is locally equi-Lipschitz if, for each $x\in X$, there exist a neighborhood $U$ of $x$ and a constant $L>0$ such that $f$ is $L$-Lipschitz continuous on $U$ for all $f\in \mathcal F$.

• We say that $\mathcal F$ is uniformly equi-continuous if, for each $\varepsilon>0$, there exists $\delta>0$ such that for all $f\in \mathcal F$, we get $\|y-x\| < \delta \implies |f(y)-f(x)| < \varepsilon$.

Theorem: Let $C$ be an open convex subset of $X$. Let $\mathcal{F}$ be a family of lower semi-continuous convex functions on $C$. If $\mathcal{F}$ is pointwise bounded, then $\mathcal{F}$ is locally equi-Lipschitz and locally equi-bounded.

Answer 1

We need the following lemmas:

Lemma 1: Let $C$ be an open convex subset of a Banach space $X$ and $f: C \to \mathbb{R}$ convex. If $f$ is l.s.c., then $f$ is continuous on $C$.

and

Lemma 2: Let $X$ be a n.v.s. Recall that $B(x, r)$ (resp. $\overline B(x, r)$) denotes the open (resp. closed) ball of radius $r$ and center $x$. Fix $a \in X, r>0, \varepsilon \in (0, r)$, and $m, M \in \mathbb R$. Let $f: \overline B(a, r) \to \mathbb R$ be convex.

1. If $f(x) \le m$ for all $x \in \overline B(a, r)$, then $|f(x)| \le |m| + 2|f(a)|$ for all $x \in B(a, r)$.
2. If $|f(x)| \le M$ for all $x \in \overline B(a, r)$, then $f$ is $\frac{2M}{\varepsilon}$-Lipschitz on $\overline B(a, r - \varepsilon)$.

We define $g:X \to \mathbb R$ by $g(x) := \sup_{f\in \mathcal F} f(x)$. Clearly, $g$ is also lower semi-continuous because, for all $\alpha \in \mathbb R$, we have $$ \begin{align} \{x\in X \mid g(x) \le \alpha\} &= \{x\in X \mid f(x) \le \alpha \text{ for all } f\in \mathcal F\} \\ &= \bigcap_{f \in \mathcal F} \underbrace{\{x\in X \mid f(x) \le \alpha\}}_{\text{closed in } X}. \end{align} $$

By Lemma 1, $g$ is continuous on $C$. It follows that $g$ is locally bounded on $C$. This in turn implies $\mathcal F$ is locally equi-bounded from above on $C$. The claim then follows from Lemma 2.