How to prove that $(f_n)$ is equi-Lipschitz and converges uniformly on compact sets?

Question

I'm reading Theorem 0.9 in this lecture note.

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Below is my attempt where I got stuck at the end. Could you elaborate on how to finish the proof?

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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 $\| \cdot\|_\infty$.

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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 $L>0$.

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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 $n \to \infty$.

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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 $\| \cdot\|_\infty$.

Answer 1

$\left( \right. g_{\varphi(n)} \to g$ 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 $x,y \in K$.

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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}

Answer 2

Theorem. In a Banach space, the closed convex hull of a compact set is compact.

Lemma. Let $C$ be an open convex set in a Banach space $X$. If $K \subset C$ is compact, then the closed convex hull of $K$, $\operatorname{Cl}( \operatorname{conv}K)$ is contained in $C$.

$Proof$. Suppose $K \subset C$ is compact. Since $\{C\}$ is an open cover of $K$, there is a number $\rho>0$ so that $B(p,\rho) \subset C$ for every $p \in K$ (The Lebesgue Number Lemma).

Fix $x \in \operatorname{conv} (K)$. So $x=\sum_{1}^{m} t_k p_k$ for some $p_1, p_2, \ldots, p_m \in K$ and $t_1, t_2, \ldots, t_m \in \mathbb{R}_{>0}$ with $\sum_{1}^{m} t_k=1$.

Since open balls are convex and the convex hull of the Minkowski sum equals the Minkowski sum of the convex hulls, we have \begin{aligned}B(x, \rho)=x+\rho B(0,1)&=\sum_1^m t_kp_k +\left(\sum_1^m t_k \rho\right) B(0,1) \\&=\sum_1^m t_k \left[ p_k + \rho B(0,1) \right] \\&=\sum_1^m t_k B(p_k, \rho). \end{aligned}

Suppose $y \in B(x, \rho)$. Then $y=\sum_{1}^{m} t_k y_k$ for some $y_k \in B(p_k, \rho) \subset C$ $(k=1,2, \ldots, m)$. Since $C$ is convex and $y_k \in C$ for $k=1,2, \ldots, m$, we thus have that $y=\sum_{1}^{m} t_k y_k \in C$.

Therefore $B(x,\rho) \subset C$ for every $x \in \operatorname{conv} K$. This means $d(\operatorname{conv} K, X \setminus C ) \geq \rho$, where $d(E,F)=\inf \{\| x-y\| : x \in E, \, y \in F \}$ $(E,F \subset X)$.

Let $\varepsilon \in (0 , \rho)$. Suppose $y \in \operatorname{Cl}(\operatorname{conv} K)$. Then there is $x \in \operatorname{conv} K$ so that $\|x-y\|<\varepsilon$. So if $z \in X \setminus C$, then we have \begin{aligned}\|y-z\|&=\|(y-x)-(z-x)\| \\&\geq |\|x-y\|-\|x-z\|| \\&=\|x-z\|-\|x-y\| \quad \quad \quad \quad (\|x-z\| \geq \lambda > \|x-y\|) \\&>\rho-\varepsilon. \end{aligned} Thus $d\left(\operatorname{Cl}(\operatorname{conv} K), X \setminus C \right)\geq \rho$ as $\{\| y-z\| : y \in \operatorname{Cl}(\operatorname{conv} K), \, z \in X \setminus C \}$ is bounded below by $\rho$. In other words, $\operatorname{Cl}(\operatorname{conv} K) \subset C$.

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Other theorem. If the functions in the family $\mathcal{F}=\{f_\alpha: X \rightarrow \mathbb{R}\, |\, \alpha \in A \}$ are locally equi-Lipschitz on an open convex set $C$ in a Banach space $X$, then the functions in $\mathcal{F}$ are equi-Lipschitz on compact subsets of $C$.

$Proof.$ Let $K \subset C$ be compact.

Since $K$ is a compact set in a Banach space, the closed convex hull of $K$, $K':=\operatorname{Cl}\left( \operatorname{conv} K\right)\subset C$ is compact [by the Theorem] (and hence bounded). Since $K' \subset C$ [by the Lemma], for each $x \in K' $ there is a neighborhood $U_x$ and a constant $L_x \geq 0$ so that every function in the family $\mathcal{F}$ is $L_x$-Lipschitz on $U_x$. Since $K'$ is compact and $\{U_x: x \in K'\}$ covers $K'$, there is a finite subcollection $\{U_{x_1}, U_{x_2}, \ldots, U_{x_j} \}$ that covers $K'$. Choose $L_0:=\max\{L_{x_1}, L_{x_2}, \ldots, L_{x_j} \}$, and let $[j]:=\{1, 2, \ldots, j\}$. If $x,y \in U_{x_k}$ for some $k \in [j]$, then we have $$|f_{\alpha}(x)-f_{\alpha}(y)|\leq L_0 \|x-y \| \text{ whenever } f_{\alpha} \in \mathcal{F}.$$ Since $K'$ is a compact set in a Banach space and $\{U_{x_1}, U_{x_2}, \ldots, U_{x_j} \}$ is an open cover of $K'$, there is a number $\lambda>0$ such that for every $x \in K'$, there is $k \in [j]$ so that $B(x,\lambda) \subset U_{x_k}$ (The Lebesgue Number Lemma).

So we may otherwise suppose $x,y \in K'$ are such that $B(x,\lambda) \subset U_{x_k}$ and $y \notin U_{x_k}$ for some $k \in [j]$. Since $K'$ is convex, we have \begin{aligned}|f_{\alpha}(x)-f_{\alpha}(y)| &\leq L_0 \cdot \operatorname{diam} K' \\& \leq \left(L_0 \operatorname{diam} K' \right)\cdot \frac{\|x-y\|}{\lambda} \quad \quad \quad [y \notin U_{x_k} \implies y \notin B(x,\lambda)]. \end{aligned} Thus if we take $L:=\left(L_0 \operatorname{diam} K' \right)/\lambda$, then we have $$|f_{\alpha}(x)-f_{\alpha}(y)| \leq L \|x-y\| \text{ whenever } x,y \in K \text{ and } f_{\alpha} \in \mathcal{F}.$$