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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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    Hm. $A$fter 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.2019-01-08

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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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    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