Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
The statement is untrue. –  checkmath Apr 30 '12 at 3:02
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. –  t.b. Apr 30 '12 at 20:01

2 Answers 2

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, $f_o\notin L^2(\mathbb{R})$.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.