In the answer of general conditions for reverse poincare inequality by @user2070206, the author used the fact that for a function $f \in H^1_0(\Omega)$, where $\Omega$ is abounded Lipschitz domain, it holds that $$ \| f \|_{L^2(\Omega)} = \| f \|_{H^{-1}(\Omega)}. $$ I couldn't come up with a proof, neither did I find any reference that proofs this statement. Is it true? If so, how can it be proven?
What I have tried so far is that I can show that $\| f\|_{H^{-1}} \le \|f\|_{L^2}$. The other inequality isn't that easy anymore. I have $$ \|f\|_{H^{-1}} \le \sup_{v \in H^1_0, \|v\|_{H^1} = 1} (f, v)_{L^2}. $$ I tried to find a $v$, such that $\|v\|_{H^1}$ = 1 and $(f, v)_{L^2} = \|f\|_{L^2}$, but I couldn't find any.
The inequality $\|f\|_{L^2} \le c \|f\|_{H^{-1}}$ is wrong. I.e., there is no $c>0$ such that the inequality is true for all $f\in L^2$.
To see this, take $f$ and define $w\in H^1_0(\Omega)$ to be the weak solution of $-\Delta w = f$ subject to homogeneous Dirichlet boundary conditions.
Then $$ \|f\|_{H^{-1}} = \sup_{v \in H^1_0(\Omega)\|v\|_{H^1}\le 1} (f,v)_{L^2} = \sup_{v \in H^1_0(\Omega)\|v\|_{H^1}\le 1} (\nabla w, \nabla v)_{L^2} \le \|\nabla w \|_{L^2}. $$ Now take eigenfunctions of the Laplace, i.e., solutions to $-\Delta e = \lambda e$ and $e\ne0$. Testing the eigenvalue equation with $e$ yields $\|\nabla e\|_{L^2}^2 = \lambda \|e\|_{L^2}^2$. Then using the above estimate of the $H^{-1}$ norm for $e$ (setting ($f=\lambda e$ and $w=e$) gives $$ \|e\|_{H^{-1}} \le \lambda^{-1} \|\nabla e\|_{L^2} = \lambda^{-1/2} \|e\|_{L^2}. $$ Since the sequence of eigenvalues of the Laplacian tends to infinity, this proves the claim.