Interpolations of $ H^{-1}(\Omega) $ and $ H_0^1(\Omega) $. Let $ \Omega $ be a bounded Lipschitz domain in $ \mathbb{R}^d $. If $ u\in H^{-1}(\Omega)$ and $ u\in H_0^1(\Omega) $ with $ \|u\|_{H^{-1}(\Omega)}\leq A $ and $ \|u\|_{H_0^1(\Omega)}\leq B $. Can I obtain the inequality $ \|u\|_{L^2(\Omega)}\leq \sqrt{AB} $ ? I have known that if $ \|u\|_{\dot{H}^{-1}(\mathbb{R}^d)}\leq A $ and $ \|u\|_{\dot{H}^{1}(\mathbb{R}^d)}\leq B $, it can be inferred from
\begin{align}
\left(\int_{\mathbb{R}^d}|\widehat{u}(\xi)|^2d\xi\right)^{1/2}\leq \left(\int_{\mathbb{R}^d}|\xi|^2|\widehat{u}(\xi)|^2d\xi\right)^{1/4}\left(\int_{\mathbb{R}^d}|\xi|^{-2}|\widehat{u}(\xi)|^2d\xi\right)^{1/4} \, ,
\end{align}
that $ \|u\|_{L^2(\mathbb{R}^d)}\leq \sqrt{AB} $. I want to know whether it can be generalized to the case of bounded domains. Can you give me some references or hints?
 A: This is true.   One way to realise this is that one can write the $H^{-1}(\Omega)$ norm simply as
\begin{align}
\lVert u \rVert_{H^{-1}(\Omega)}^2 = & \langle (-\Delta)^{-\frac12} u, (-\Delta)^{-\frac12} u \rangle \, ,
\end{align}
where $(-\Delta)^{-1}:H^{-1}(\Omega) \to H^1_0(\Omega) $ is the inverse of the negative Dirichlet Lapalcian on $\Omega$. Thus, using the fact that $(-\Delta)^\alpha$ is self-adjoint for all $\alpha\in \mathbb R$
\begin{align}
\lVert u \rVert_{L^2(\Omega)}^2 =& \langle u, (-\Delta)^{-\frac{1}{2}}(-\Delta)^{\frac12} u \rangle \\
=& \langle (-\Delta)^{-\frac{1}{2}} u, (-\Delta)^{\frac12} u \rangle \\
\leq & \langle (-\Delta)^{-\frac{1}{2}} u, (-\Delta)^{-\frac12} u \rangle^{\frac12} \langle (-\Delta)^{\frac{1}{2}} u, (-\Delta)^{\frac12} u \rangle^{\frac12} \\
=& -\lVert u \rVert_{H^{-1}(\Omega)} \langle u, \Delta u \rangle^{\frac12} \\
=& \lVert u \rVert_{H^{-1}(\Omega)}\lVert u \rVert_{H^{1}_0(\Omega)} \, .
\end{align}
Thus,
\begin{align}
\lVert u \rVert_{L^2(\Omega)} \leq \lVert u \rVert_{H^{-1}(\Omega)}^{\frac12}\lVert u \rVert_{H^{1}_0(\Omega)}^{\frac12} \leq \sqrt{AB}\, .
\end{align}
Edit: Just some background you may find helpful. For a given $u \in H^{-1}(\Omega)$, $(-\Delta)^{-1}u$ is the unique Dirichlet solution $v \in H^1_0(\Omega)$ of the elliptic PDE
$$
-\Delta v= u \, .  
$$
It is a positive self-adjoint operator and so by standard functional calculus arguments one can define its square root. Now to convince yourself of the definition of $H^{-1}(\Omega)$ I have provided, note that $(-\Delta)^{-1}u$ is exactly the Reisz representative of $u$ in $H^1_0(\Omega)$. That is to say  for any $f \in H^1_0(\Omega)$ and with $v$ as defined above
\begin{align}
& \langle v,f \rangle_{H^1_0(\Omega)}\\
 =& \int_{\Omega }\nabla v \cdot \nabla f \, \mathrm{d}x \\
=& - \int_{\Omega }\Delta v \, f \, \mathrm{d}x \\
=&\langle u, f\rangle \, .
\end{align}
Thus (since the Riesz map is unique), we have that $(-\Delta)^{-1}:H^{-1}(\Omega)\to H^1_0(\Omega)$ is an isometric isomorphism. Therefore,
\begin{align}
\lVert u \rVert_{H^{-1}(\Omega)}^2 =& \lVert (-\Delta)^{-1}u \rVert_{H^{1}_0(\Omega)}^2 \\
=& \langle (-\Delta)^{-1}u,u \rangle \\
=&\langle (-\Delta)^{-\frac12} u, (-\Delta)^{-\frac12} u \rangle\, .
\end{align}
I have of course glossed over some small details but I am happy to explain them if you find something confusing.
