# Convergence of Integral Implies Uniform convergence of Equicontinuous Family

Let $\{f_n\}$ be an equicontinuous family of functions on $[0,1]$ such that each $f_n$ is pointwise bounded and $\int_{[a,b]} f_n(x)dx \rightarrow 0$ as $n\rightarrow \infty$, for every $0\leq a \leq b \leq 1$.

Show $f_n$ converges to $0$ uniformly.

For this question I know that there exists a uniformly convergent subsequence $f_{n_k}$ by Arzela-Ascoli Theorem. For this uniformly convergent sequence I know $$\lim_{n\rightarrow \infty} \int_a^bf_{n_k}(x)dx = \int_a^b \lim_{n\rightarrow \infty}f_{n_k}(x)dx$$ Since the left side is zero If we assume $\lim_{n\rightarrow \infty}f_{n_k}(x)dx \neq \ 0$ for some $x\in [0,1]$ Then uniform continuity of the limit implies that it is $\neq 0$ on some interval which implies the integral is $\neq 0$ Which is a contradiction . Thus $f_{n_k}$ must converge uniformly to $0$. I don't see how to get to $f_n$ converges uniformly to $0$ though.

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tricki.org/article/… –  t.b. Jul 12 '11 at 18:37
@Theo: Is that really helpful? The technique you use to prove this is not in that list if I understood it correctly. –  Jonas Teuwen Jul 12 '11 at 20:28
@Jonas: I'm using the first bullet point: Every subsequence has a subsubsequence which converges to zero. I'm providing the missing argument. –  t.b. Jul 12 '11 at 20:32

You essentially proved: Every subsequence of $(f_n)$ has a (sub-)subsequence which converges uniformly to zero (why?).

The first bullet point in the link I provided in a comment asserts that then the sequence $(f_n)$ itself must converge to zero.

Here's why: Suppose $f_{n}$ does not converge uniformly to zero. Then there is an $\varepsilon \gt 0$ and a subsequence $f_{n_j}$ with sup-norm $\|f_{n_j}\|_{\infty} \geq \varepsilon$ for all $j$. This subsequence has again a convergent subsequence by Arzelà-Ascoli. Your argument shows that its limit must be zero, hence the subsequence must have uniform norm $\lt \varepsilon$ eventually, contradiction.

Here's the abstract thing:

Let $x_n$ be a sequence in a metric space $(X,d)$. Suppose that there is $x \in X$ such that every subsequence $(x_{n_j})$ has a subsubsequence $(x_{n_{j_k}})$ converging to $x$. Then the sequence itself converges to $x$.

Edit: As leo pointed out in a comment below the converse is also true: a convergent sequence obviously has the property that every subsequence has a convergent subsubsequence.

The proof is trivial but unavoidably uses ugly notation: Suppose $x_n$ does not converge to $x$. Then there is $\varepsilon \gt 0$ and a subsequence $(x_{n_j})$ such that $d(x,x_{n_j}) \geq \varepsilon$ for all $j$. By assumption there is a subsubsequence $x_{n_{j_k}}$ converging to $x$. But this means that $d(x,x_{n_{j_k}}) \lt \varepsilon$ for $k$ large enough. Impossible!

The way this is usually applied in "concrete situations" is to show

1. If a subsequence converges then it must converge to a specific $x$. This involves an analysis of the specific situation—this is usually the harder part and that's what you did.
2. Appeal to compactness to find a convergent subsubsequence of every subsequence—that's the trivial part I contributed.
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The result is true in general. –  leo Jul 13 '11 at 1:55
@leo: see edit. Is that what you were trying to tell me? –  t.b. Jul 13 '11 at 9:02
Maybe it's that you're working in the context of metric spaces instead of topological spaces? Or maybe, more generally, that this is an instance of the interplay between the "infinitely many" and "cofinitely many" quantifiers over the naturals? –  user83827 Jul 13 '11 at 12:22
@ccc: well, maybe... Good points of course, but I'd consider indulging myself in these issues as going rather far away from the original question. I guess only leo can enlighten me. –  t.b. Jul 13 '11 at 12:30
Yes @Theo, I should say, "It is true in general in the context of metric spaces". I no deal with the other topics that mention above yet. At least for this question. And, of course, using your notation, if $x_n\to x$, every subsequence have a subsubsequence that converges to $x$. –  leo Jul 14 '11 at 0:30