3
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ One should note that $\|u^2\|_{L^2}^2 = \|u\|_{L^4}^4$. $\endgroup$
    – daw
    Commented Sep 27, 2022 at 16:24

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .