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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.

bheinzek
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  • Not that important, but you probably should avoid saying "the first answer" as answers tied on votes can be shown in any order (which is the case there currently), and you never know if later the answer would be deleted, or if another answer would take the lead in votes, etc... It's just a technicality though. – Bruno B Jun 22 '23 at 11:14
  • Thank you, I will edit that later – bheinzek Jun 22 '23 at 11:29
  • The answer you linked to is wrong. There is another answer in that question with a counterexample – daw Jun 22 '23 at 12:56
  • This specific answer, answers the question in the context of PDEs – bheinzek Jun 22 '23 at 12:58
  • @daw Any ideas on where user2070206 went wrong? Just in case it's the equality that OP is asking about. I don't know negative Sobolev spaces all that well. – Bruno B Jun 22 '23 at 13:25
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    The inequality $|f|{L^2} \le c |f|{H^{-1}}$ is wrong. – daw Jun 22 '23 at 14:13

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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.

daw
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