When does $v \in H_0^1$ and $|v|_{L^2}=0$ imply that $\|v\|_{H_0^1}=0$?

Question

Let us assume that the boundary of the domain in the definition of the Sobolev spaces $L^2$ and $H_0^1$ is sufficiently smooth.

Let $|\cdot |$ denote the norm in $L^2$. Then for a function $v$ in $H_0^1$, the norm is given via $\|v\|^2=|v|^2+|\nabla v|^2$.

In general, one cannot bound the $H_0^1$-norm by the $L^2$-norm, as the gradient of a function, cannot be bounded by the function values.

What if for $v\in H_0^1$, one has $|v|=0$. Does this imply that $|\nabla v|=0$?

I have tried to come to terms with this in 1D. Consider an interval $(a,b)$ and $u\in L^2(a,b)$, with a weak derivative $u'\in L^2(a,b)$ and $u(a) = u(b) = 0$. Then, $u$ is absolutely continuous almost everywhere, and one has $u(x) = \int_a^xu'(s)ds$. Then, $0=|u|^2=\int_a^b(\int_a^xu'(s)ds)^2dx$ which somehow should give that $\int_a^bu'(s)^2ds$ is zero as well...

Answer 1

HINT: The $H^1_0$ together with $\|v \|_{L^2}=0$ will make $v=0$ almost everywhere. Try proving this fact using contradiction.

Answer 2

Got it, thanks to Shuhao Cao.

Let $\Omega$ be the domain under consideration. From $|v| = 0$, we conclude that $v=0$ a.e. . If this was not the case, i.e., if there was a $\Omega_0\subset \Omega$ of nonzero measure where $v\neq 0$, then, $|v| > 0$ and we would get a contradiction.

Thus, $v=0$ almost everywhere on $\Omega$. Then, for every coordinate direction $x_i$, we get that $\int \partial_{x_i} v \phi = -\int v \partial_{x_i}\phi =0$ for all $\phi \in \mathcal C_0^\infty (\Omega)$ and - by the fundamental lemma of variational calculus - that $\partial_{x_i}v = 0$ a.e. . Accordingly, $|\nabla v| = 0$. qed