Let $p > 2$ and $\Omega\subset R^n$ an open bounded subset of $R^n$. Moreover let $u\in W_0^{1, p}(\Omega)$. I want to prove that an inequality of this type holds \begin{align*} \int_{\Omega} \vert\nabla u\vert^{p - 2} \Vert\nabla u\Vert_{L^2} dx\leq \left(\int_{\Omega} \vert\nabla u\vert^{p}dx\right)^{\frac{1}{p}}, \end{align*} but I don't know how to proceed. Could anyone help?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Let $f$ be the function $\nabla u$. We have to show an inequality involving only $f$, so we assume $f\in L^p(\Omega)$. So $f^p$ is integrable, and we do not know more about $f$. The condition $p>2$, the generality of $\Omega$ and $f$ are together too weak to be able to control the L.H.S. by using the R.H.S. - for instance consider $\Omega=\Bbb R$, and a piecewise linear function $f\ge 0$ taking the value $0$ outside $[-\epsilon, M+\epsilon]$, and the value $C$ on $[0,M]$, and being linear on $[-\epsilon,0]$, and on $[M,\epsilon]$. Then passing with $\epsilon $ to $0$ (and assuming an inequality (of the type) as the given one) we would expect an equality of the shape: $$ M\; C^{p-2}\cdot\sqrt M\;C\qquad \le\qquad M^{1/p}\; C\ . $$ But this is too much to have in general.
-
$\begingroup$ Thank you! But I don't understand why you take $\Omega = R$ if i ask $\Omega$ to be an open bounded set. $\endgroup$ Feb 8, 2020 at 7:45
-
1$\begingroup$ I wanted only to have some counterexample in a simple case. For a general (open) $\Omega\subseteq \Bbb R^N$ take some spherical region of diameter $M+2\epsilon$ centered in $a$, say, inside of it, and consider a similar function $F$ which is $C$ on the ball centered in $a$ and diameter $M$, and zero outside a ball centered in $a$ with diameter $M+2\epsilon$. Then, if an equality of the type in the OP would exist, we would get an equality of the same shape as above, with $M$ replaced by the volume of the ball ($=(M/2)^N$ times one ball volume). Such an inequality fails for $C\to \infty$. $\endgroup$ Feb 10, 2020 at 12:25