Is the set of uniformly bounded non-decreasing functions a compact set with the 2 metric?

Question

Fix $M>0$. Let $\Phi = \{f|f:[a, b] \to [-M, M] \, \text{is an non-decreasing function} \}$. Define a metric $d: \Phi \times \Phi \to [0, \infty)$ by $(,)=(\int_{a}^b |(x)−(x)|^2 dx)^{1/2}$. Is the topology induced by compact?

Answer 1

The answer is yes.

Let $(f_n)_{n\in \mathbb{N}}$ be some sequence in $\Phi$ and let $(q_k)_{k\in \mathbb{N}}$ be some enumeration of $[a,b]\cap \mathbb{Q}$. By compactness of $[-M,M]$, we can extract a subsequence $(f_{n,1})_{n\in \mathbb{N}}$ such that $f_{n,1}(q_1)$ converges.

Inductively, let $(f_{n,k+1})_{n\in\mathbb{N}}$ be a subsequence of $(f_{n,k})_{n\in\mathbb{N}}$ such that $f_{n,k+1}(q_{k+1})$ converges. Combining all of these, we get that for every $k$, the tail $(f_{n,n})_{n\geq k}$ is a subsequence of $(f_{n,k})_{n\in\mathbb{N}}$ and therefore, $f_{n,n}(q_k)$ converges for every $k$ to some non-decreasing limit function $f:[a,b]\cap \mathbb{Q}\to \mathbb{R}$

Setting $f(x)=\sup_{q\in [a,x]\cap \mathbb{Q}} f(q)$ extends $f$ to a non-decreasing function on all of $(a,b]$ which, in particular, can only be discontinuous at countably many points. We claim that for every point $x$ of continuity of $f$, we must have $f_{n,n}(x)\to f(x)$.

Indeed, let $\varepsilon>0$ and note that by continuity at $x$, there exists some $\delta$ such that $|f(y)-f(x)|\leq \varepsilon/2$ for all $y\in [x-\delta, x+\delta]$ pick furthermore $n$ large enough that $|f_{n,n}(x)-f(q_-)|<\varepsilon/2$ and $|f_{n,n}(q_+)-f(x)|<\varepsilon/2$ for some choice of $q_-\in [x-\delta,x)\cap \mathbb{Q}$, $q_+\in (x,x+\delta]\cap \mathbb{Q}$ and $n$ large enough. Then, we get, again for $n$ large enough, since $f_{n,n}$ is non-decreasing, that $$ f(x)-\varepsilon \leq f(q_-)-\varepsilon \leq f_{n,n}(q_-)/2\leq f_{n,n}(x) \leq f_{n,n}(q_+)\leq f(q_+)+\varepsilon/2\leq f(x)+\varepsilon, $$ so that $|f(x)-f_{n,n}(x)|\leq \varepsilon$. We conclude that $f_{n,n}\to f$ almost everywhere.

Thus, we can apply dominated convergence since $|f_{n,n}|^2\leq M^2\in L^1([a,b])$ for all $n$ and get that

$$ \int_a^b |f_{n,n}|^2\textrm{d}x\to \int_a^b |f|^2\textrm{d}x, $$ which proves that $f_{n,n}$ converges to $f$ under $d$. Thus, $(\Phi,d)$ is compact.