I am trying to show the following result.

Let $D\subset\Bbb R^3$ be a bounded open set with smooth boundary. For any $f\in H^{-1}(D)$, let $\phi$ be the unique weak solution to the following Dirichlet problem $$-\Delta\phi=f\\\phi|_{\partial D}=0$$ Then, $\|\nabla\phi\|_{L^2}=\|f\|_{H^{-1}}$

It is easy to show that $\|\nabla\phi\|_{L^2}\geq\|f\|_{H^{-1}}$, so I just need $\|f\|_{H^{-1}}\geq\|\nabla\phi\|_{L^2}$. I tried to use the characterization of $H^{-1}(D)$. Namely, $$\|f\|_{H^{-1}}^2=\inf\sum_{i=0}^3\|f^i\|_2^2$$ where the infimum ranges over all $f^0,f^1,\cdots,f^3\in L^2$ such that $$\langle f,v\rangle=\int_Df^0v+\sum_{i=1}^3\int_Df_i\partial_iv$$ for all $v\in H_0^1(D)$.

Hence, I just need to prove $\sum_{i=0}^3\|f^i\|_2\geq\|\nabla\phi\|_2$ for all such $f^0,f^1,\cdots,f^3\in L^2$. Now, I have $$\int|\nabla\phi|^2=\int_Df^0\phi+\sum_{i=1}^3\int_Df_i\partial_i\phi$$ So,

\begin{align*} \int|\nabla\phi|^2 &\leq\int_D|f^0||\phi|+\sum_{i=1}^3\int_D|f_i||\partial_i\phi|\\ &\leq\epsilon(\|\phi\|_2^2+\|\nabla\phi\|_2^2)+(4\epsilon)^{-1}\sum_{i=0}^3|f_i\|_2^2\\ &\leq\epsilon C\|\nabla\phi\|_2^2+(4\epsilon)^{-1}\sum_{i=0}^3\|f_i\|_2^2 \end{align*}

So, $$(1-\epsilon C)\|\nabla\phi\|_2^2\leq(4\epsilon)^{-1}\sum_{i=0}^3\|f_i\|_2^2$$ I want to choose some $\epsilon$ such that $1-\epsilon C=(4\epsilon)^{-1}$. However, since $C>1$, this is impossible.

I am completely running out of ideas. Any hints or help? Thanks!


It depends on your definition of $\|f\|_{H^{-1}}$.

Let us equip $H_0^1$ with the scalar product $$(u,v)_{H_0^1} = \int \nabla u \nabla v \, \mathrm dx.$$ Then, the weak formulation of Poisson's equation is $$(\phi,v)_{H_0^1} = f(v) \quad\forall v \in H_0^1.$$ Hence, the solution $\phi \in H_0^1$ is just the Riesz representative of $f \in H^{-1} = (H_0^1)'$. This yields $$\|\nabla \phi\|_{L^2} = \|\phi\|_{H_0^1} = \|f\|_{H^{-1}}$$ if you define the norm in $H^{-1}$ by $$\|f\|_{H^{-1}} := \sup_{u \ne 0} \frac{f(u)}{ \|\nabla u\|_{L^2} }.$$

Note that $\|\nabla\phi\|_{L^2} = \|f\|_{H^{-1}}'$ does not hold if you define $$\|f\|_{H^{-1}}' := \sup_{u \ne 0} \frac{f(u)}{ \sqrt{\|\nabla u\|^2_{L^2} + \|u\|_{L^2}^2} }.$$

  • $\begingroup$ Could you please give me an example of your last sentence? Namely, the equality doesn't hold if the $H^{-1}$ is defined in the "usual way"? Thank you so much! $\endgroup$ – YYF May 2 '16 at 11:45
  • $\begingroup$ What do you mean by example? You can take any $f \ne 0$, since this always implies $\|f\|_{H^{-1}} > \|f\|_{H^{-1}}'$. $\endgroup$ – gerw May 2 '16 at 12:39
  • $\begingroup$ You're right! Thanks!! $\endgroup$ – YYF May 2 '16 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.