Equality in an integral estimate

Question

Let $u,v\in W^{1,p}(\Omega)$ with $1<p<\infty$ and $\Omega$ be a bounded smooth domain in $\mathbb{R}^N$. Suppose, we have $$\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla v\,dx=\left(\int_{\Omega}|\nabla u|^p\,dx\right)^\frac{p-1}{p}\left(\int_{\Omega}|\nabla v|^p\,dx\right)^\frac{1}{p} $$ Suppose the equality $$ \int_{\Omega}|\nabla u|^p\,dx=\int_{\Omega}|\nabla v|^p\,dx $$ holds. Then can we say that $u=v$ in $\Omega$?

I tried in the following way: Since the first given equality, which is the equality in Holder's inequality holds, and by the second given equality holds, we have $|\nabla u|=|\nabla v|$ in $\Omega$. Then I am unable to proceed.

Can someone please help to give an argument?

Answer 1

You missed one step. Actually, you're first using the Cauchy-Schwarz inequality pointwise: $$ |\nabla u|^{p-2} \langle \nabla u, \nabla v \rangle \le |\nabla u|^{p-2} \cdot |\nabla u| \cdot |\nabla v|, $$ and then Holder's inequality: $$ \int_{\Omega} |\nabla u|^{p-2} \langle \nabla u, \nabla v \rangle \le \int_{\Omega} |\nabla u|^{p-1} \cdot |\nabla v| \le \left(\int_{\Omega}|\nabla u|^p \right)^\frac{p-1}{p}\left(\int_{\Omega}|\nabla v|^p \right)^\frac{1}{p}. $$ If there's an equality, you can use equality conditions for both:

• Equality in Cauchy-Schwarz (for each point separately) tells you that for a.e. $x \in \Omega$ the vectors $\nabla u(x)$ and $\nabla v(x)$ are proportional (with a nonnegative ratio).
• Equality in Holder tells you that the functions $|\nabla u|$ and $|\nabla v|$ are proportional. Combined with the previous observation, this implies that the proportionality ratio of $\nabla u(x), \nabla v(x)$ is independent of $x$, so in fact the functions $\nabla u$ and $\nabla v$ are proportional.
• If additionally $\int_{\Omega}|\nabla u|^p=\int_{\Omega}|\nabla v|^p$, then this proportionality ratio has to equal $1$.

In consequence, $\nabla u = \nabla v$ a.e. In other words, $u$ and $v$ locally differ by a constant. Without some normalization (e.g., $u,v$ have zero trace on the boundary), nothing more can be said.

Answer 2

These assumptions imply $\nabla u=\nabla v$. The claim is trivially true if one of $\nabla u$ or $\nabla v$ is identically zero.

Using the assumptions and Hoelder inequality, we get $$ 0 \le\int_\Omega |\nabla u|^{p-2} \nabla (v-u) \cdot \nabla (v-u) \le \|\nabla u\|_{L^p}^p -2 \|\nabla u\|_{L^p}^{p-1}\|\nabla v\|_{L^p}^1 +\|\nabla u\|_{L^p}^{p-2}\|\nabla v\|_{L^p}^2\\ = \|\nabla u\|_{L^p}^{p-2}(\|\nabla u\|_{L^p}-\|\nabla v\|_{L^p})^2=0. $$ This implies $\nabla u = \nabla v$ for almost all $x$ for which $\nabla u(x)\ne0$.

Since we have equality in Hoelders inequality, the functions $|\nabla u|^p$ and $|\nabla v|^p$ are linearly independent. This implies $\nabla u(x)=0$ if and only if $\nabla v(x)=0$.

And $\nabla u = \nabla v$ follows. If $u,w\in W^{1,p}_0$ this implies $u=v$.

In general, $u=v$ does not follow, as the assumptions are fulfilled trivially if we set $v:=u+1$.