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(Apology in advance for the broadness of this question) I recently came across a relatively simple application where I needed to "balance" the "spreaded-out-ness" of a function with the "peaked-ness" of a function, and my approach was to average over a few norms of the function, and for the purposes of my application seems to be working well enough, though I'm interested in any possible existing theory of what I've done. I apologize in advance for not being very technical; my explanation is basically entirely intuitive.

Specifically, suppose a bounded $f\colon[0,1]\rightarrow\mathbb{R}_{\geq0}$ is given, and then sup-norm-normalize $f\leftarrow f/\|f\|_\infty$ so that $0\leq f\leq1$. Then as I understand, fractional powers $0<p<1$ will "lift the function away from zero", and so fractional "$p-$norms" will give a better idea of how much the function is spread out, whereas standard $p-$norms, $p\geq1$ will more accentuate the peaks as these powers "push the low spots closer to zero".

My painfully-simple solution was to just average the $\frac{1}{2}-$ and $2-$norms to "balance" the "spreaded-out-ness" and "peaked-ness" of the function in question, but I am very curious if there is any existing research on averaging norms to detect this type of structure of a function.

A little more precisely, I'm looking for literature related to quantities like $$\frac{1}{n}\sum_{k=1}^{n}\|f\|_{p_k}, \ \ 0<p_k<\infty$$ or even $$\frac{1}{p_2-p_1}\int_{p_1}^{p_2}\|f\|_pdp, \ \ 0<p_1<p_2<\infty.$$

Any ideas or direction will be greatly appreciated.

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  • $\begingroup$ I'm a little confused by the normalization --- it seems like you want the $p$-norms for $p \geq 1$ to ignore all but the highest parts of $f$, which works better the larger $p$ is. But by normalizing so that $\|f\|_\infty = 1$, you ensure that $\|f\|_p \to 1$ as $p \to \infty$, so very large values of $p$ don't tell you anything (unless you look closely at the discrepancy $1 - \|f\|_p$). Have I misunderstood? $\endgroup$
    – Nik Weaver
    Commented Mar 16, 2016 at 22:25
  • $\begingroup$ @NikWeaver In short, I normalized so I could still capture the spread-out-ness, since values $f(x)\geq1$ would be shrunk by the fractional norm, but I completely agree that the sup-norm (read large $p$...) encodes virtually nothing about the function in this particular application, which is in part why I chose the $2-$norm since it's everyone's "go to". If it helps the thought process/ideas where to go, we can just assume $f$ only takes values in $[0,1]$ pre-normalization and treat it as is. $\endgroup$
    – charlestoncrabb
    Commented Mar 17, 2016 at 0:22

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