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.
Best Answer
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 $$ \lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=0\tag{6} $$