Equicontinuity and $L_1$-convergence implies uniform convergence on compact sets

Question

I am trying to figure out, whether the following statement holds: Let $(f_k)_{k\in \mathbb{N}} =: \mathcal{F} \subseteq C(K)$ a sequence of continous functions, where $K\subseteq \mathbb{R}^n$ compact. Assume $\mathcal{F}$ is equicontinuous and the sequence $f_k$ converges to $f$ with respect to the $L_1$ norm. Does this imply that $f_k \rightarrow f$ uniformly on $K$?

I feel like I might be able to apply some version of Arzelà–Ascoli, but I am not entriely sure about the details.

Answer 1

We have the following result. The main idea of the proof comes from this answer.

Let $E\subseteq \mathbb{R}^{d}$ be a non-empty Borel set. Assume there is an increasing sequence $(K_{n})_{n\in\mathbb{N}}$ of non-empty compact subsets of $\mathbb{R}^{d}$ such that $E = \bigcup_{n\in\mathbb{N}} K_{n}$ and that every compact subset of $E$ is contained in a member of the sequence $(K_{n})_{n\in\mathbb{N}}$. (This condition is automatically satisfied if $E$ is a non-empty open or closed subset of $\mathbb{R}^{d}$.) Let $(f_{n})_{n\in\mathbb{N}}$ be an equicontinuous sequence in $C(E)\cap L_{p}(E)$ and $f\in L_{p}(E)$. Suppose the sequence $(f_{n})_{n\in\mathbb{N}}$ converges to $f$ in $L_{p}(E)$. Then the following hold.

1. There exists a $g\in C(E)$ such that $(f_{n})_{n\in\mathbb{N}}$ converges to $g$ uniformly on compact subsets of $E$ and $f = g$ almost everywhere.

2. If $f\in C(E)$ and $|U\cap E| > 0$ for all open $U\subseteq \mathbb{R}^{d}$ with $U\cap E \neq \emptyset$ (where $|A|$ is the Lebesgue measure of a measurable subset $A$), then $(f_{n})_{n\in\mathbb{N}}$ converges to $f$ uniformly on compact subsets of $E$.

We first show that every subsequence of $(f_{n})_{n\in\mathbb{N}}$ has a subsequence converging uniformly to some $g\in C(E)$ that is equal to $f$ almost everywhere. Let $(f_{n_{j}})_{j\in\mathbb{N}}$ be a subsequence of $(f_{n})_{n\in\mathbb{N}}$. By the Arzelà-Ascoli theorem there is a subsequence $(f_{n_{j_{k}}})_{k\in\mathbb{N}}$ of $(f_{n_{j}})_{j\in\mathbb{N}}$ which converges uniformly on compact subsets of $E$ to some $g\in C(E)$. In particular, the sequence $(f_{n_{j_{k}}})_{k\in\mathbb{N}}$ converges almost everywhere to $g$, and so does every subsequence of $(f_{n_{j_{k}}})_{k\in\mathbb{N}}$. On the other hand, since $(f_{n_{j_{k}}})_{k\in\mathbb{N}}$ converges to $f$ in $L_{p}(E)$, there is a subsequence of $(f_{n_{j_{k}}})_{k\in\mathbb{N}}$ converging to $f$ almost everywhere. Hence we have $g = f$ almost everywhere.

Since at most one member of $C(E)$ can be equal almost everywhere to $f$, we deduce from above that there is some $g\in C(E)$ which is equal to $f$ almost everywhere such that every subsequence of $(f_{n})_{n\in\mathbb{N}}$ has a subsequence converging to $g$ uniformly on compact subsets of $E$. By this result applied to the sequence $(f_{n})_{n\in\mathbb{N}}$ in $C(E)$, we conclude that $(f_{n})_{n\in\mathbb{N}}$ converges to $g$ uniformly on compact subsets of $E$. This shows the first part.

Now suppose that $f\in C(E)$ and that $|U\cap E| > 0$ for all open $U\subseteq\mathbb{R}^{d}$ with $U\cap E \neq \emptyset$. By the first part there is some $g\in C(E)$ such that $(f_{n})_{n\in\mathbb{N}}$ converges to $g$ uniformly on compact subsets of $E$ and $f = g$ almost everywhere. The set $A = \{x\in E : f(x) \neq g(x)\}$ is a relatively open subset of $E$ with $|A| = 0$, so by assumption we have $A = \emptyset$. Hence $f = g$ pointwise and the second part follows.