I've found the following statement on a book:
It is easy to see that given any $p < 2$, if a function $h$ on a probability space has very small support, then its $L_p$ norm is much smaller than its $L_2$ norm.
What I believe "very small support" means is: Given a $h:(\Omega, \Sigma, \mu) \to (\mathbb{R},B(\mathbb{R})$ measurable function, where $\mu(\Omega) = 1$. Define $$S = supp(h) = \{\omega \in \Omega : h(\omega) \neq 0\}.$$ Then "very small support" would mean that $$\mu(S) \mbox{ is close to 0}.$$
I'd like to know that, if we have a given $\varepsilon>0$ such that $\mu(S)<\varepsilon$, than that would imply what about $\|h\|_p$ and $\|h\|_2$? It means that there's a $\delta>0$ small such that
$$\left(\int_\Omega |h(\omega)|^pd\mu(\omega)\right)^{\frac{1}{p}}\leq \delta \left(\int_\Omega |h(\omega)|^2d\mu(\omega)\right)^{\frac{1}{2}}?$$ (that is $\|h\|_p \leq \delta \|h\|_2$)
And if so, why is that true? The most I can do is
$$\|h\|_p\leq \left[\sup_{\omega \in S}\{|h(\omega)|^p\}\mu(S)\right]^{\frac{1}{p}}\leq \sup_{\omega \in S}\{|h(\omega)|\}\mu(S)^{\frac{1}{p}}\leq \sup_{\omega \in S}\{|h(\omega)|\}\mu(S)^{\frac{1}{2}}$$
then I can't go back to the $L_2$ norm. It seems it should be simple to prove but I couldn't do it.
This topic appeared in the chapter 5 about "Hypercontractivity" on the book "Noise sensitivity of Boolean functions and percolation [Garban, Steif]".
Thanks in advance.
We want to have an estimate of the form $$ \def\norm#1{\left\|#1\right\|} \norm{f}_p \le C\norm{f}_2 $$ (Generalized) Hölder comes to mind. We have $$ \frac 12 + \frac 1r = \frac 1p \iff \frac 1r = \frac 1p - \frac 12 = \frac{2-p}{2p} \iff r = \frac{2p}{2-p} $$ Therefore, as $f = f\chi_S$ $$ \norm{f}_p \le \norm f_2 \norm{\chi_S}_{r} $$ Now $$ \norm{\chi_S}_r = \mu(S)^{1/r}.$$ Altogether: $$ \norm f_p \le \mu(S)^{(2-p)/2p} \norm f_2 $$