Estimation of gradients in Poisson's equation

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 $$\|f\|_{H^{-1}}' := \sup_{u \ne 0} \frac{f(u)}{ \sqrt{\|\nabla u\|^2_{L^2} + \|u\|_{L^2}^2} }.$$
• 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! – YYF May 2 '16 at 11:45
• What do you mean by example? You can take any $f \ne 0$, since this always implies $\|f\|_{H^{-1}} > \|f\|_{H^{-1}}'$. – gerw May 2 '16 at 12:39