Inequality with norm in Space $L^2(\Omega)$

Question

Let $\Omega \subset \mathbb{R}^N$ a bounded domain.

Let $v \in L^2(\Omega)$. It is possible to make an estimate of the type $$\|v^2\|_{L^2(\Omega)} \leq \|v\|^k_{L^2(\Omega)}$$, for some $k \in \mathbb{R}$.

Using Holder's inequality, I'm able to get something like $$\|v^2\|_{L^2(\Omega)} \leq \|v^3\|_{L^2(\Omega)}\|v\|_{L^2(\Omega)}.$$ But what I really want is to get that square out of the norm. Thanks.

Answer 1

Let $A\subset \Omega$ satisfy $0<|A|<1.$ For $u= |A|^{-1/4}1\hspace{-2.5pt}{\rm I}_A$ we have $$\|u^2\|_2=1,\quad \|u\|_2=|A|^{1/4}<1$$ Hence for any constant $k>0$ the inequality does not hold.

Answer 2

By Cauchy-Schwarz inequality you obtain $$ \| v \|_{L^2(\Omega)}^2 = \int_\Omega v^2 dx \leq \|1\|_{L^2(\Omega)}\|v^2\|_{L^2(\Omega)} = |\Omega|^{1/2} \|v^2\|_{L^2(\Omega)}, $$ hence the inequality you wish for (get that square out of the norm) holds in the other direction with a constant depending on the size of the domain.