# Some sort of interpolation equation

I have recently stumbled across a interpolation inequality for $$u \in H^3(\Omega)$$ when $$\Omega$$ is a lipschitz domain $$$$\vert \vert u \vert \vert_{H^1(\Omega)} \leq C \vert \vert u \vert \vert_{L^2(\Omega)}^{2/3} \vert \vert u \vert \vert_{H^3(\Omega)}^{1/3}$$$$ and was curious if I can show it. I know that for $$u \in H^2$$ we have $$$$\vert \vert \nabla u \vert \vert_{H^1(\Omega)}^2 \leq C \vert \vert u \vert \vert_{L^2(\Omega)} \vert \vert u \vert \vert_{H^2(\Omega)}$$$$ and so I started estimating the square of $$\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.

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.