# How to prove this estimate in $W_0^{1, p}(\Omega)$?

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?

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.
• Thank you! But I don't understand why you take $\Omega = R$ if i ask $\Omega$ to be an open bounded set. Feb 8, 2020 at 7:45
• 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 $C\to \infty$. Feb 10, 2020 at 12:25