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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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Around each integer $n$ consider the interval $I_n=(n-\frac{1}{2n^2},n+\frac{1}{2n^2})$ and build the triangles given by the endpoints of $I_n$ and the point $(n,1)$ in the plane. We have then a piecewise function which can be extended by zero outside the $I_n$. This gives us the graph of a continuous, square integrable function $f$ for which $f^{\#}\equiv 1$.

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    Yes, my fears are confirmed by your example! I had faith that oscillations would be a problem. Your peaks in the triangles exhibit what I felt very clearly. +1!2012-05-05
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    I found that $\|f\|_{W(L^{\infty},L^{2})}=\|f^{\#}\|_{L^{2}}$, I'm confusing??!!2012-05-05
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    @Kinda What is that first norm?2012-05-05
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    It is the Amalgam space norm.2012-05-06
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    So, do you mena that if the function doesn't oscillate or have peaks like above then the result will be true?2012-05-07
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    What if the function $f$ is differentiable on $\mathbb R$, in this case I think the fears above will not exist! Right!2012-05-08