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

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There is no Schwartz function being one always they goes to zero at $\infinity$! – checkmath Apr 30 '12 at 1:21
Ok, so maybe my method of proof is not correct! But what about the result itself? – Kelly Apr 30 '12 at 2:44
The statement is untrue. – checkmath Apr 30 '12 at 3:02
There is no (Schwartz) function $g$ such that $\hat{g} = 1$. The Fourier transform sends functions from the Schwartz space to the Schwartz space. – t.b. Apr 30 '12 at 20:01

I think this statement is not true. Consider sequences $$ a_n=n\qquad b_n=\frac{a_n+a_{n+1}}{2} $$ and define functions $$ f_n(x)= \begin{cases} 0 & x<b_{n-1}\\ \left(\frac{x-b_{n-1}}{a_n-b_{n-1}}\right)^{n^2+1} & b_{n-1}\leq x<a_n\\ \left(1-\frac{x-a_{n}}{b_n-a_n}\right)^{n^2+1} & a_n\leq x<b_n\\ 0 & x \geq b_n \end{cases} $$ Now we define function $f(x)=\sum\limits_{n=-\infty}^\infty f_n(x)$. Its graph consist of uniformly disturbed peaks centered at integer points. Peaks become more sharp as $n$ tends to infinity.

enter image description here

We choose peaks sharp enough in order to $f\in L^2(\mathbb{R})$. One can show that $f$ is continuous and $$ \Vert f\Vert_{L^2}=\left(\sum\limits_{n=-\infty}^\infty\frac{1}{1+2n^2}\right)^{1/2}<+\infty $$ Since for all $x\in\mathbb{R}$ we have $0\leq f(x)\leq 1$ and for all $n\in\mathbb{Z}$ we have $f(n)=1$, then for all $x\in\mathbb{R}$ $$ f_o(x)=1 $$ Obviously, $f_o\notin L^2(\mathbb{R})$.

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The result is false consider a function that is continuous and in $L^2$ but $F(p)=1 ,\forall p\in \mathbb{Z}$, ( if you want construct small triangles). This kind of maximal operator is not in L^2, because $f_0\geq 1$.

For a function Consider $g(x)=\sqrt{ \max(1-|x|,0)}$ then define $h_n(x)=g(n^2 x)$.

Then consider $f(x)=\sum_{1}^{\infty} h_n(x)$.

I think that a nice question is if $f_0$ is L^2 then f=0.

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Is your function continuous? – Kelly Apr 30 '12 at 1:26
I think Kelly is right here, the continuity pretty much rules out the singular situations for $f_o(x)$. – Shuhao Cao Apr 30 '12 at 1:29
Would you mind giving a precise statement of the "nice question"? – Nicholas Stull Apr 30 '12 at 2:06
Sorry my statement is also false! For example $h_n$ satisfies your statement for all n! – checkmath Apr 30 '12 at 2:14
I thought about something rather interesting concerning this question. Let $$f(x) = \left\{\begin{array}{lcl} e^{-1/(1-x^2)}\right. & \ & -1 < x < 1\\ 0 & \ & otherwise\end{array}\right.$$ This function is $C^\infty$ (in particular, continuous), $L^2$, nonzero, and $f_0\in L^2$. That being said, I'm not sure how common these sorts of examples would be. – Nicholas Stull Apr 30 '12 at 14:24

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