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Let $\{f_n\}$ be absolutely continuous functions on $[0,1]$, $f_n \geq 0$ $\forall n \in \mathbb{N}$. Suppose that $\lim_{n \to \infty} \int^1_0 f_n dx = 0$. Moreover, suppose that $\forall$ $\epsilon > 0$ $\exists$ $\delta > 0$ such that $\int_{E} |f_n'(x)|dx$ < $\epsilon$ $\forall n \in \mathbb{N}$, if $m(E) < \delta$. Prove that $f_n$ converges to $0$ uniformly on $[0,1]$.

Can someone help show this? Thanking you in advance.

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the $\mathbb L^1$ convergence gives you pt-wise almost convergence of a subsequence. Once you get it close to zero on a reasonable dense set, use the derivative condition to tie it down everywhere. Since uniform convergence is metric, the 'subsequence' business is easily dealt with. I think you replace the almost sure with convergence in prob, it migh even be simpler, as it state the fctn is small in every small interval. – mike Apr 21 '12 at 12:49
up vote 3 down vote accepted

Suppose that $f_n$ does not converge uniformly to $0$. Then, there is an $\epsilon>0$ so that for all $N$ there is an $n>N$ and an $x_n$ so that $$ f_n(x_n)>2\epsilon\tag{1} $$ Find $\delta>0$ so that $|E|<\delta$ implies $$ \int_E|f_n^\prime(x)|\,\mathrm{d}x<\epsilon\tag{2} $$ Suppose that $n$ is such that $(1)$ is true. Let $$ E_n=\left\{x\in[0,1]:|x-x_n|<\dfrac{\delta}{3}\right\}\tag{3} $$ Note that $\dfrac\delta3\le|E_n|\le\dfrac{2\delta}{3}<\delta$ (since $x_n$ could be near the boundary of $[0,1]$).

Then $(2)$ says that for $x\in E_n$ we have $$ |f_n(x)-f_n(x_n)|\le\int_{E_n}|f_n^\prime(t)|\,\mathrm{d}t<\epsilon\tag{4} $$ The triangle inequality applied to $(1)$ and $(4)$ says that for $x\in E_n$, $f_n(x)>\epsilon$. Thus, $$ \int_{E_n}f_n(x)\,\mathrm{d}x>\epsilon\frac{\delta}{3}\tag{5} $$ Since there are arbitrarily large $n$ so that $(1)$ is true, $(5)$ contradicts the assumption that $$ \lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=0\tag{6} $$

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How did you get $$ |f(x) - f(x_n)| \leq \int_{E_n} |f^\prime_n (x)| dx$$ in equation (4)? – Rudy the Reindeer Apr 25 '12 at 19:08
because it should be $$ |f_n(x)-f_n(x_n)|\le\int_{E_n}|f_n^\prime(x)|\,\mathrm{d}x $$ as it does now. :-) – robjohn Apr 25 '12 at 19:17
And how do you get that? – Rudy the Reindeer Apr 25 '12 at 19:20
$$f_n(x)-f_n(x_n)=\int_{x_n}^xf'(x)\,\mathrm{d}x$$ and the $x$ inside the integral is a dummy variable, but for clarity, it should probably be something else. – robjohn Apr 25 '12 at 19:26
Nice! + 1. This deserves more upvotes. – Rudy the Reindeer Apr 25 '12 at 20:05

Following Mike's comment, pass to a subsequence to get almost everywhere convergence to 0. Thus there is a set $A$ of measure 1 with $f_n(x) \to 0$ for all $x \in A$. In particular $A$ is dense. Now the assumption about the $f'$ guarantees that the family $\{f_n\}$ is equicontinuous (since $|f(x)-f(y)| \le \int_x^y |f'(t)|\,dt$). One can now follow the proof of the Arzela-Ascoli theorem to obtain the uniform convergence.

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