Why is $a$ not coercive when it is defined on $H^1(\Omega)$?

Question

Given the Poisson's equation with homogeneous Neumann boundary conditions and the associated bilinear form $$a(u,v) = \int_{\Omega}\nabla u \cdot \nabla v \, dx$$ on $H^1(\Omega)$, why is $a$ not coercive? Whereever I look it says we need $H_0^1(\Omega)$ for coercivity, but I am not sure why $H^1(\Omega)$ wouldn't work. Thanks for the help.

Answer 1

This is because $1 \in H^1(\Omega)$ and $a(1,1) = 0$. Here, $1$ is the constant function with value $1$ everywhere.