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Let $f(x)$ be a continuous and locally bounded function on $\mathbb R$, then the local maximum function is defined by $$ f^{\#}(x)=\sup_{y\in[x-1,x+1]}|f(y)| $$

Can we find a relation between the $L^{2}$ norm of $f^{\#}$ and the $L^{2}$ norm of $f$ ? (if we know that $\|f\|_{L^{2}(\mathbb R)}<\infty$)

(I'm not sure if this is related to the the Amalgam space $W(L^{\infty},L^{2})$ !!)

I don't know exactly the next step for $$\int_{-\infty}^{\infty}|f^{\#}(x)|^{2}dx=\int_{-\infty}^{\infty}\big|\sup_{y\in[x-1,x+1]}|f(y)| \big|^{2}dx$$

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    $\| f^{\#} \|_{L^2(\mathbb R)} > \| f \|_{L^2(\mathbb R)}$ if $f \neq f^{\#}$? I fear it is the only thing we can say in general, but then again I haven't given much thought about it.2012-05-05
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    My fears are mostly caused by the fact that the supremum can behave very strangely if $f$ can oscillate.2012-05-05
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    But $f$ is locally bounded on $\mathbb R$, does this imply anything!2012-05-05
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    Perhaps you could try bounding the norm of $\| f^{\#} \|$ by above, but the fact that $f$ is locally bounded, for me, only means that $f^{\#}$ is well-defined (at least for the moment, that is all I see).2012-05-05
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    Yes you'r right! In fact I expect something like $\|f^{\#}\|_{L^{2}}\leq M \|f\|_{L^{2}}$, for some $M$.2012-05-05
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    It is quite easy to show that if $\| f \|_{L^2(\mathbb R)}$ is finite, $f$ continuous and locally bounded implies that $|f(x)| \le M$ for some constant $M$, for all $x \in \mathbb R$. Also, since $f$ is $L^2$, $f(x) \to 0$ when $|x| \to \infty$. Would that be a starting point?2012-05-05
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    I don't see this, could you please help me proving it (that $|f(x)|\leq M$)?2012-05-05
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    Hm. After thinking about it I don't think this is true at all. It suffices to produce continuous peaks that give very low area in contribution. Never mind my comment. I had an argument in mind but when you asked me to write it explicitly I saw a failure in what I was trying to do.2012-05-05

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