Some sort of interpolation equation

Question

I have recently stumbled across a interpolation inequality for $u \in H^3(\Omega)$ when $\Omega$ is a lipschitz domain \begin{equation} \vert \vert u \vert \vert_{H^1(\Omega)} \leq C \vert \vert u \vert \vert_{L^2(\Omega)}^{2/3} \vert \vert u \vert \vert_{H^3(\Omega)}^{1/3} \end{equation} and was curious if I can show it. I know that for $u \in H^2$ we have \begin{equation} \vert \vert \nabla u \vert \vert_{H^1(\Omega)}^2 \leq C \vert \vert u \vert \vert_{L^2(\Omega)} \vert \vert u \vert \vert_{H^2(\Omega)} \end{equation} and so I started estimating the square of $\vert \vert u \vert \vert_{H^1(\Omega)}$ \begin{align} \vert \vert u \vert \vert_{H^1(\Omega)}^2 &= \vert \vert u \vert \vert_{L^2(\Omega)}^2 + \vert \vert \nabla u \vert \vert_{L^2(\Omega)}^2 \\ &= \vert \vert u \vert \vert_{L^2(\Omega)}^2 + \left(-\int_{\Omega}{\mbox{div}(\nabla u) u } + \int_{\partial \Omega}{(\nabla u \cdot \nu) u} \right) \\ &= \vert \vert u \vert \vert_{L^2(\Omega)}^2 + \left(-\int_{\Omega}{\Delta u u } + \int_{\partial \Omega}{(\nabla u \cdot \nu) u} \right) \\ &\leq \vert \vert u \vert \vert_{L^2(\Omega)}^2 + \left(\vert \vert \Delta u \vert \vert_{L^2(\Omega)} \vert \vert u \vert \vert_{L^2(\Omega)} + \int_{\partial \Omega}{(\nabla u \cdot \nu) u} \right) \\ &= (\vert \vert u \vert \vert_{L^2(\Omega)}^{2/3} \vert \vert u \vert \vert_{L^2(\Omega)}^{1/3})^2 + \left(\vert \vert \Delta u \vert \vert_{L^2(\Omega)} \vert \vert u \vert \vert_{L^2(\Omega)} + \int_{\partial \Omega}{(\nabla u \cdot \nu) u} \right) \\ &\leq (\vert \vert u \vert \vert_{L^2(\Omega)}^{2/3} \vert \vert u \vert \vert_{H^3(\Omega)}^{1/3})^2 + \left(\vert \vert \Delta u \vert \vert_{L^2(\Omega)} \vert \vert u \vert \vert_{L^2(\Omega)} + \int_{\partial \Omega}{(\nabla u \cdot \nu) u} \right) \end{align} where $\nu$ is a unit normal field on $\partial \Omega$.My next idea would be to use the extension operator $F : H^2(\Omega) \to H^2(\mathbb{R}^d)$ with $|| Fu ||_{H^2(\mathbb{R}^d)} \leq C ||u ||_{H^2(\Omega)}$ and $|| Fu ||_{L^2(\mathbb{R}^d)} \leq C ||u ||_{L^2(\Omega)}$ in the estimation above such that \begin{align} \vert \vert Fu \vert \vert_{H^1(\mathbb{R}^d)}^2 &\leq (\vert \vert Fu \vert \vert_{L^2(\mathbb{R}^{d})}^{2/3} \vert \vert Fu \vert \vert_{H^2(\mathbb{R}^{d})}^{1/3})^2 \\ &+ \vert \vert \Delta F u \vert \vert_{L^2(\mathbb{R}^d)} \vert \vert Fu \vert \vert_{L^2(\mathbb{R}^{d})} \\ &\leq C^2(\vert \vert u \vert \vert_{L^2(\Omega)}^{2/3} \vert \vert u \vert \vert_{H^3(\Omega)}^{1/3})^2 \\ &+ C\vert \vert \Delta F u \vert \vert_{L^2(\mathbb{R}^d)} \vert \vert u \vert \vert_{L^2(\Omega)} \end{align} where the boundary integral vanishes.

Edit: We can w.l.o.g asume $||u||_{L^2(\Omega)} \leq 1$ since we can choose $\epsilon > 0$ and look at $\frac{u}{||u||_{L^2(\Omega)} + \epsilon}$ or $\frac{u}{||u||_{L^2(\Omega)}}$ and use the estimates of $||\Delta Fu ||_{L^2(\mathbb{R}^d)}$ by iterativ partial integration and Cauchy-Schwarz and use the fact that $||u||_{L^2(\Omega)} \leq 1$ to get a similar estimate to $C^2(\vert \vert u \vert \vert_{L^2(\Omega)}^{2/3} \vert \vert u \vert \vert_{H^3(\Omega)}^{1/3})^2$. The divsion by $||u||_{L^2(\Omega)} + \epsilon$ or $||u||_{L^2(\Omega)}$ will not change the inequality since the factors will cancel each other out. The case $||u||_{L^2(\Omega)} = 0$ is trivial. I will post the answer if someone is interested.

Answer 1

We can apply the interpolation inequality $$ \|u\|_{H^1}^2 \le c \|u\|_{L^2} \|u\|_{H^2} $$ to the partial derivatives of $u$ (i.e., replace $u$ by $\partial_{x_i}u$) to get $$ \|u\|_{H^2}^2 \le c \|u\|_{H^1} \|u\|_{H^3}, $$ using this in the first inequality gives $$ \|u\|_{H^1}^2 \le c \|u\|_{L^2} \|u\|_{H^2} \le c^{3/2} \|u\|_{L^2}\|u\|_{H^1}^{1/2} \|u\|_{H^3}^{1/2}, $$ dividing by $\|u\|_{H^1}^{1/2}$ and raising everything to power $2/3$ yields the claimed estimate.