Let $f(x)$ be a continuous function for all $x\in \mathbb R$, such that $f\in L^{2}(\mathbb R)$ (i.e., $\int_{-\infty}^{\infty}|f(x)|^{2}dx<\infty$), and define $f_{o}(x):=\sup_{|x-y|\leq 1}|f(y)|$
How to prove that $f_{o}\in L^{2}(\mathbb R)$, and $\|f_{o}\|_{L^{2}}\leq A\|f\|_{L^{2}}$, for some constant $A>0$?
- My progress is the follwoing, so correct me if I'm wrong, and advise me if I'm missing something:
We can construct a function $g\in S(\mathbb R)$ (Schwartz class) with $\hat{g}=1$, so $\hat{f}=\hat{f}\hat{g}$, hence $f=f*g$ (convolution), then
$f_{o}(x)\leq (|f|*g_{o})(x)$ which implies that $\|f_{o}\|_{L^{2}}\leq \|(|f|*g_{o})\|_{L^{2}}\leq \|f\|_{L^{2}} \|g_{o}\|_{L^{1}}$.