$L^4/L^2$ norm gives the area of the support

Question

At a conference I heard a speaker saying that physicists use the ratio of the $L^4$ over the $L^2$ norm to estimate the size of the support of a function. I could not show this on my own or find any reference. Help or references are most welcome!

Answer 1

An intuition from Terry Tao which I find useful is contained in Example 1 here: https://terrytao.wordpress.com/2009/01/09/245b-notes-3-lp-spaces/

Example 1 remarks that:

if $f(x)=A1_E(x)$, where $A$ is a number, $E$ is some set and $1_E$ is the function that equals $1$ on $E$ and $0$ elsewhere,

then $\lVert f\rVert_p= \lvert A\rvert \lvert E\rvert^\frac{1}{p}$.

Here $\lvert A\rvert $ is interpreted as the "height" and $\lvert E\rvert$ as the "width" of the function $f$. This remark provides the intuition that the $L^p$ norms measure functions according to the rule $$ \text{height}\cdot\text{(width)}^\frac1p. $$

Plugging this intuition into the ratio $\lVert f\rVert_2/\lVert f\rVert_4$, we obtain $$ \frac{\lVert f \rVert_2}{\lVert f\rVert_4}\sim \text{(width)}^{\frac14}, $$ which agrees with the physicist's interpretation you gave.

Answer 2

By replacing the function $f$ by $g=\frac{\lvert f\rvert}{\lVert f\rVert_2}$ we may assume $g\geq 0$ with $\lVert g\rVert_2=1$.

We will estimate the measure of $$ A=\{x: g(x)\geq\theta\}=\{x: g(x)^2\geq\theta^2\} $$ where $\theta\in(0,1)$.

First we have the Markov's inequality bound $$ \mu(A)\leq\frac1{\theta^2}\lVert g^2\rVert_1=\frac1{\theta^2} $$ giving us an upper bound on $\mu(A)$.

On the other hand, we can write $$ g^2=g^2 1_{g<\theta}+g^21_{g\geq\theta} $$ so integrating $$ 1=\int g^2 1_{g<\theta}\,\mathrm{d}\mu + \int g^21_{g\geq\theta}\,\mathrm{d}\mu $$ the first term is at most $\theta^2$ (assuming $\mu$ is a probability measure. We don't have "useful" bound if $\mu$ is not finite) and the second term is, by Cauchy-Schwarz $$ \int g^21_{g\geq\theta}\,\mathrm{d}\mu \leq \sqrt{\int g^4\,\mathrm{d}\mu\cdot\int 1_{g\geq\theta}\,\mathrm{d}\mu}=\sqrt{\lVert g\rVert_4^4\mu(A)} $$ So that gives $$ \mu(A)\geq\frac{(1-\theta^2)^2}{\lVert g\rVert_4^4} $$

In summary, we have estimated, for $0<\theta<1$, $$ \left(\frac{\lVert f\rVert_2}{\lVert f\rVert_4}\right)^4(1-\theta^2)^2 \leq \mu(\{\lvert f\rvert\geq\theta\lVert f\rVert_2\})\leq\frac1{\theta^2}\left(\frac{\lVert f\rVert_2}{\lVert f\rVert_4}\right)^2. $$