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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 $\|f_{o}\|_{L^{2}}\leq \|(|f|*g_{o})\|_{L^{2}}\leq \|f\|_{L^{2}} \|g_{o}\|_{L^{1}}$.

  • 0
    There is no Schwartz function being one always they goes to zero at $\infinity$!2012-04-30
  • 0
    Ok, so maybe my method of proof is not correct! But what about the result itself?2012-04-30
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    The statement is untrue.2012-04-30
  • 1
    There is no (Schwartz) function $g$ such that $\hat{g} = 1$. The Fourier transform sends functions from the Schwartz space to the Schwartz space.2012-04-30

2 Answers 2