I'm reading Theorem 0.9 in this lecture note.
Below is my attempt where I got stuck at the end. Could you elaborate on how to finish the proof?
Let $C$ be an open convex subset of a Banach space $X$.
- 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 $f\in \mathcal F$.
Theorem: Let $(f_{n})$ be a sequence of lower semi-continuous 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.
Lemma 1: 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.
Lemma 2: Let $(X, \| \cdot\|)$ be a normed vector space, $C$ its open convex subset, and $f:C \to \mathbb R$ convex. Then the following statements are equivalent.
- (i) $f$ is locally Lipschitz on $C$;
- (ii) $f$ is continuous on $C$;
- (iii) $f$ is continuous at some point of $C$;
- (iv) $f$ is locally bounded on $C$;
- (v) $f$ is upper bounded on a nonempty open subset of $C$.
Arzelà–Ascoli theorem: Let $Y$ be a compact Hausdorff space and $C(Y)$ the space of real-valued continuous functions on $Y$. Then a subset $F$ of $C(Y)$ is relatively compact in the topology induced by the supremum norm $\| \cdot\|_\infty$ if and only if it is pointwise equi-continuous and pointwise bounded.
-
Clearly, $f$ is convex. Convergent sequence is bounded, so $(f_n)$ is pointwise bounded. By Lemma 1, $(f_n)$ is locally equi-Lipschitz and locally equi-bounded. Fix $a\in C$. There is a neighborhood $U$ (in the subspace topology of $C$) of $a$ and $m \in \mathbb R$ such that
$$
|f_n(x)| \le m \quad \forall n \in \mathbb N, \forall x\in U.
$$
It follows that $|f(x)| \le m$ for all $x \in U$. Hence $f$ is locally bounded. Then by Lemma 2, $f$ is continuous. -
Fix a compact subset $K \subset C$. Let $g_n := f_n \restriction K$ and $g := f \restriction K$. Then $(g_n)$ is pointwise bounded and pointwise equi-continuous. By Arzelà–Ascoli theorem, there exist $h \in C(K)$ and a subsequence $\varphi \in \mathbb N^\mathbb N$ such that $g_{\varphi(n)} \to h$ in $\| \cdot \|_\infty$. This implies $g_{\varphi(n)} \to h$ pointwise and thus $h=g$. Hence $g_{\varphi(n)} \to g$ in $\| \cdot\|_\infty$.
However, I'm stuck at showing $g_{n} \to g$ in $\| \cdot\|_\infty$.
Update 1: A proof that $(f_n)$ is equi-Lipschitz on compact set can be derived from here. For the sake of completeness, I represent it below. We will show that all $f_n$'s and $f$ share the same Lipschitz constant on $K$.
For each $x\in C$, there is $r_x,L_x,m_x>0$ such that $f_n$ is $L_x$-Lipschitz and bounded by $m_x$ on $B_C(x, r_x)$ for all $n$.
Let $B_K(x, r_x/2) :=\{y\in K \mid \|y-x\|<r_x/2\}$. Then $\{B_K(x, r_x/2) \mid x\in K\}$ is an open cover of $K$. There is a finite subset $\{x_1, \ldots, x_n\} \subset K$ such that $\{B_K(x_1, r_{1}/2), \ldots, B_K(x_n, r_{n}/2)\}$ covers $K$.
Let $M := 2\max \{m_1, \ldots, m_n\}$. Fix $y,z\in K$. Assume $y \in B_K(x_i, r_{i}/2)$ and $z \in B_K(x_j, r_{j}/2)$.
-
If $y \in B_K(x_i, r_{i}) \cap B_K(x_j, r_{j})$, then $y,z \in B_K(x_j, r_{j})$, so $|f_n(y)-f_n(z)| \le L_j\|y-z\|$.
-
If $z\in B_K(x_i, r_{i}) \cap B_K(x_j, r_{j})$, then $y,z \in B_K(x_i, r_{i})$, so $|f_n(y)-f_n(z)| \le L_i\|y-z\|$.
-
If $y,z \notin B_K(x_i, r_{i}) \cap B_K(x_j, r_{j})$, then $\|y-z\| \ge \|y- x_j\| – \|x_j-z\| \ge r_j – r_j/2 = r_j/2$. So
$$
|f_n(y)-f_n(z)| \le M = \frac{M}{\|y-z\|}\|y-z\| \le \frac{M}{r_j/2} \|y-z\|.
$$
Let
$$
L := \max \left \{L_1, \ldots, L_n, \frac{M}{r_1/2}, \ldots, \frac{M}{r_n/2} \right\}.
$$
It follows that
$$
|f_n(y)-f_n(z)| \le L \|y-z\| \quad \forall y,z\in K, \forall n \in \mathbb N.
$$
Taking the limit $n \to \infty$, we got
$$
|f(y)-f(z)| \le L \|y-z\| \quad \forall y,z\in K.
$$
This means $g,g_1,g_2,\ldots$ share the same Lipschitz constant $L>0$.
Update 2: Below I prove that $g_n \to g$ uniformly.
Fix $\varepsilon>0$. Because $K$ is compact, it is totally bounded. There are $x_1, \ldots, x_m \in K$ such that $\{B_K(x_1, \varepsilon/(4L)), \ldots, B_K(x_m, \varepsilon/(4L))\}$ covers $K$. There is $N \in \mathbb N$ such that
$$
|f_n(x_i)-f(x_i)| < \frac{\varepsilon}{2} \quad \forall n>N, \forall i=1, \ldots,m.
$$
Let $c_x$ be one of the $x_i$'s that are closest to $x \in K$. Then $\|x-c_x\| < \varepsilon/(4L)$ for all $x\in K$. It follows that, for all $n>N$, we have
$$
\begin{align}
\sup_{x\in K} |f_n(x)-f(x)| &\le \sup_{x\in K} |f_n(x)-f_n(c_x)| + \sup_{x\in K} |f(x)-f(c_x)| + \sup_{x\in K} |f(c_x)-f_n(c_x)| \\
&\le L\sup_{x\in K} \|x-c_x\|+ L\sup_{x\in K} \|x-c_x\| + \sup_{x\in K} |f(c_x)-f_n(c_x)| \\
&\le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align}
$$
It follows that $\|g_n-g\|_\infty \to 0$ as $n \to \infty$.
Update 3: It seems I have finally got an approach via Arzelà–Ascoli theorem.
- Fix a compact subset $K \subset C$. Let $g_n := f_n \restriction K$ and $g := f \restriction K$. Clearly, $(g_n)$ converges pointwise to $g$. We need the following well-known result.
Urysohn subsequence principle: A sequence $(x_n)$ converges to $x$ if and only if every subsequence of $(x_n)$ has a further subsequence which converges to $x$.
- Let $\psi \in \mathbb N^\mathbb N$ be a subsequence. Then $(g_{\psi(n)})_n$ is pointwise bounded and pointwise equi-continuous. By Arzelà–Ascoli theorem, there exist $h \in C(K)$ and a subsequence $\varphi \in \mathbb N^\mathbb N$ of $\psi$ such that $g_{\varphi(n)} \to h$ in $\| \cdot \|_\infty$. This implies $g_{\varphi(n)} \to h$ pointwise and thus $h=g$. Hence $g_{\varphi(n)} \to g$ in $\| \cdot\|_\infty$. By Urysohn subsequence principle, $g_{n} \to g$ in $\| \cdot\|_\infty$.
Best Answer
$\left( \right. g_{\varphi(n)} \to g$ uniformly on $K \left. \right)$ $\land$ $\forall n \in \mathbb{N} \left( \right. g_{\varphi(n)}$ $L$-Lipschitz on $K \left. \right)$ $\implies$ $g$ is $L$-Lipschitz on $K$:
Let $\varepsilon>0$ be given. There is a positive integer $N$ so that $|g_{\varphi(n)}(x)-g(x)|<\varepsilon/2$ whenever $n \geq N$ and $x \in K$. Since $g_{\varphi(n)}$ is $L$-Lipschitz for each $n \in \mathbb{N}$, we thus have
\begin{aligned} |g(x)-g(y)|&=|g(x)-g_{\varphi(n)}(x)+g_{\varphi(n)}(x)-g_{\varphi(n)}(y)+g_{\varphi(n)}(y)-g(y)| \\& \leq |g(x)-g_{\varphi(n)}(x)|+|g_{\varphi(n)}(x)-g_{\varphi(n)}(y)|+|g_{\varphi(n)}(y)-g(y)| \\& < L\|x-y\|+\varepsilon \quad \text{whenever } n \geq N \text{ and }x,y\in K. \end{aligned}
Since $\varepsilon>0$ was arbitrary, we may conclude that $|g(x)-g(y)| \leq L\|x-y\|$ for all $x,y \in K$.
Let $\varepsilon>0$ be given, and let $\mathcal{G}=\{g_n: n \in \mathbb{N} \}$.
Let $L$ be a Lipschitz constant for the family of functions $\mathcal{G}':=\mathcal{G} \cup \{g\}$ on $K$. Choose $M:=\max\{1, L\}$.
Since $K$ is compact and $\{B\left(x,\frac{\varepsilon}{4M}\right) : x \in K\}$ covers $K$, there is a finite subcollection $\{B\left(x_1,\frac{\varepsilon}{4M}\right), B\left(x_2,\frac{\varepsilon}{4M}\right), \ldots, B\left(x_i,\frac{\varepsilon}{4M}\right) \}$ that covers $K$. For each of $k=1,2, \ldots, i$, there is a positive integer $N_k$ such that $|g_n(x_k) - g(x_k)|<\varepsilon/2$ whenever $n \geq N_k$. Choose $N:=\max\{N_1, N_2, \ldots, N_i\}$. Then we have $|g_n(x_k) - g(x_k)|<\varepsilon/2$ whenever $n \geq N$ for $k=1, 2, \ldots, i$. So if $n \geq N$ and $x \in K$, we thus have \begin{aligned}|g_n(x)-g(x)|&=|g_n(x)-g_n(x_k)+g_n(x_k)-g(x_k)+g(x_k)-g(x)| \\& \leq |g_n(x)-g_n(x_k)|+|g_n(x_k)-g(x_k)|+|g(x_k)-g(x)| \\& \leq 2M\|x_k-x \| + |g_n(x_k)-g(x_k)| \\& < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{aligned}