# Local Maximum Function - Again!

So, I'm posting this question again because there is new given information for the problem, and I think people just answering new posted questions:

Let $f(x)$ be a continuous and locally bounded function on $\mathbb R$, then the local maximum function is defined by $$f^{\#}(x)=\sup_{y\in[x-1,x+1]}|f(y)|$$ Edit: We can replace the "$1$" in the "sup" by any nember $d\leq 1$.

I'm trying to 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.e., something like $\|f^{\#}\|_{L^{2}(\mathbb R)}\leq C\, \|f\|_{L^{2}(\mathbb R)}$.

The fears in the prevous answers were that $f$ could have (sharp) peaks, which makes $f^{\#}\notin L^{2}(\mathbb R)$.

Now, given the above information about $f$, and assuming that $f$ is continuously differentiable, with bounded derivaive on $\mathbb R$, and that $f$ is bounded on $\mathbb R$, does this change the argument to have the result which I'm looking for!

Link to old post: Local maximum function

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My intuition tells me that if you restrict attention to $f$'s having $\sup|f|, \sup|f'|\leq M$ for a single constant $M$ something could be said, but I'm unable to prove this. –  Jose27 May 8 '12 at 20:41
I have the same feeling, so if both $f$ and $f'$ are bounded we can choose $M=\max\{M_{1},M_{2}\}$, where $|f|\leq M_{1}$ and $|f'|\leq M_{2}$. –  Kinda May 8 '12 at 21:50

## 1 Answer

And basically the same example works all right...

Call $s$ the elementary spike defined by $s(x)=(1-|x|)^+$ and consider $f(x)=\sum\limits_{i=1}^ks(n(x-i))$ for some parameters $k$ and $n$ to be determined.

Then $\|f\|_2$ behaves like $k/n$ and $f^\#=1$ on the interval $[1,k]$ hence $\|f^\#\|_2$ is at least of the order of $\sqrt{k}$. If $k\ll n^2$, this ruins the possibility that $\|f^\#\|_2\leqslant C\|f\|_2$ uniformly over such functions $f$, for any finite $C$.

To get $C^\infty$ counterexamples, keep the idea and replace $s$ by a $C^\infty$ function with support $[-1,1]$.

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