What does it tell you about two functions if their $L^p$ norms are the same for all $p\in[1,\infty]$? Certainly they could be related by composition with a diffeomorphism with Jacobian of norm 1, or even one could be a "pulled apart version of the other one" in the sense of $x^2\chi_{[0,2]}$ vs $x^2 \chi_{[0,1]} + (x-1)^2 \chi_{[2,3]}$. To try to ignore the second type of issue, I'll restrict to smooth functions, and ask the following precise question:

Given $f,g\in C^\infty(\mathbb{R})$ such that $\Vert f \Vert_{L^p(\mathbb{R})} = \Vert g \Vert_{L^p(\mathbb{R})} <\infty$ for all $p\in [1,\infty]$ is it necessarily true that $f(x) = g(\pm x + C)$ for some constant $C$ and for a choice of either $+$ or $-$ for all $x \in \mathbb{R}$?

As Qiaochu Yuan shows in his answer, smoothness doesn't solve the issue of "pulling apart" at all. Thus, I am interested in the following:

What is the "smallest" (in whatever sense) but still nonempty subset of $S \subset [1,\infty]$ such that there is $f,g\in C^\infty(\mathbb{R})$ such that $S$ is the set of $p$ such that $\Vert f \Vert_{L^p} \neq \Vert g \Vert_{L^p}$?

  • $\begingroup$ It's not clear to me whether the complement of S can even contain an interval. Does anyone have such an example? $\endgroup$ – Qiaochu Yuan Nov 11 '10 at 0:14

The complement of $S$ in $(1,\infty)$ can't contain an interval or even a sequence converging to a point in $(1,\infty)$. Let $f$ and $g$ be two real functions on ${\mathbb{R}}$ all whose moments exist. Assume $\int|f|^p=\int|g|^p$ for all $p\in S^c$. put $h(z)=\int|f|^z-\int|g|^z$. where $z\in \mathbb{C}$. $h$ is an analytic functions in $\{Real z>1\}$ which we assume is not constantly zero (since $S$ is not empty) so can't vanish on a converging sequence.


No. You can take $f, g$ to be sums of smooth bump functions with disjoint support. As long as $f, g$ each have bumps of the same size and shape you can put them in arbitrary locations (and orientations). Maybe you should require something like $f, g$ non-negative everywhere.

  • $\begingroup$ Ok, well I see that I've failed to remove the "pulling apart" issue.. $\endgroup$ – Otis Chodosh Nov 10 '10 at 21:00
  • $\begingroup$ I've edited the question to see if there is any insight on the second question. $\endgroup$ – Otis Chodosh Nov 10 '10 at 21:03

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