Suppose $f \in L^{\infty}(\mathbb{R})$, $f_{h}(x) = f(x+h)$, and
$$
\lim_{h \rightarrow 0}||f_{h} – f||_{\infty} = 0
$$
Prove that there exists a uniformly continuous function $g$ on $\mathbb{R}$ such that $f = g$ a.e
Firstly, I used the theorem such that if $f$ is Borel measurable and integrable, then there exists continuous function $g$ having compact support. However, I can't show $f$ is integrable, and Borel measurable.
And I wonder about a notation $||f_{h} – f||_{\infty}$ means $||f(x+h)-f(x)||_{\infty}$.
Can anybody help?
Best Answer
Given $h>0$ let $$ f^h(x)=\frac{1}{2\,h}\int_{x-h}^{x+h}f(y)\,dy=\frac{1}{2\,h}\int_{-h}^{h}f(x-y)\,dy. $$ Lebesgue's differentiation theorem implies that $\lim_{h\to0}f_h(x)=f(x)$ almost everywhere. Moreover the family $f^h$ is uniformly bounded, since $|f^h(x)|\le\|f\|_\infty$ for all $h>0$.
Let's prove now that $f^h$ is equicontinuous. Given $x,x'\in\mathbb{R}$ $$ |f^h(x)-f^h(x')|\le\frac{1}{2\,h}\int_{-h}^{h}|f(x-y)-f(x'-y))|\,dy. $$ Let $\epsilon>0$ be given. By hypothesis there exists $\delta>0$ such that $$ |x-x'|<\delta\implies|f(x-y)-f(x'-y))|\le\epsilon\quad\forall y\in\mathbb{R}. $$ Thus, if $|x-x'|<\delta$, then $|f^h(x)-f^h(x')|\le\epsilon$.
The Ascoli-Arzela theorem implies that there is a sequence $h_n>0$ such that $f^{h_n}$ converges uniformly.