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

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.

• One should note that $\|u^2\|_{L^2}^2 = \|u\|_{L^4}^4$.
– daw
Commented Sep 27, 2022 at 16:24

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.
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.