# Local maximum function

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. –  Patrick Da Silva May 5 '12 at 19:30
My fears are mostly caused by the fact that the supremum can behave very strangely if $f$ can oscillate. –  Patrick Da Silva May 5 '12 at 19:36
But $f$ is locally bounded on $\mathbb R$, does this imply anything! –  Kinda May 5 '12 at 19:50
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). –  Patrick Da Silva May 5 '12 at 19:54
Yes you'r right! In fact I expect something like $\|f^{\#}\|_{L^{2}}\leq M \|f\|_{L^{2}}$, for some $M$. –  Kinda May 5 '12 at 19:59
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 $f^{\#}\equiv 1$.
I found that $\|f\|_{W(L^{\infty},L^{2})}=\|f^{\#}\|_{L^{2}}$, I'm confusing??!! –  Kinda May 5 '12 at 21:01