Proof that bilinear form $(u,v)_{L^2(\Omega)} \mapsto (\nabla u, \nabla v)_{L^2(\Omega)}$ is not coercive in $H^1(\Omega)$

Question

Let $\Omega \in \mathbb{R}^n$ be a bounded and regular domain. Take a look at the following bilinear form: \begin{align} (u,v)_{L^2(\Omega)} \mapsto (\nabla u, \nabla v)_{L^2(\Omega)} \end{align} In Alain Miranville's "The Cahn–Hilliard Equation: Recent Advances and Applications", the author claims that the above bilinear form is not coercive on $H^1(\Omega)$, i.e.

\begin{align} (\nabla u, \nabla u)_{L^2(\Omega)} = \lVert \nabla u \rVert^2_{L^2(\Omega)} \geq m \lVert u \rVert^2_{L^2(\Omega)} \end{align},

for a fixed $m>0$ is not true for all $v\in H^1(\Omega)$.

I feel like there must be an easy counterexample to verify the above statement (as it seems to be a very basic fact) but I fail.

Can somebody give me a hint?

Answer 1

Take $n=1$, $\Omega = (0,1)$ to be in a very basic 1D setting. What function $u(x)$ can you think of which has a very very small derivative (or even better $u' = 0$), but is not zero?