If $u\in H^2(\mathbb{R}^2)$, does $u$ satisfy Lipschitz continuity $$\lvert u(x)-u(y)\rvert \leq C\lvert x-y\rvert\,$$ where $x,y\in \mathbb{R}^2,$ for some constant $C$?
I believe this statement is correct, since $H^2(\mathbb{R}^2)$ has a Banach algebra property, and thus, $u(x)$ is bounded and continuous. However I am not sure how to show Lipschitz continuity directly, from properties of $H^2$ space, where $$H^2(\mathbb{R}^2)=\bigl \langle u(x):\mathbb{R}^2\rightarrow \mathbb{R}^2\lvert u,\nabla u, \Delta u\in L^2(\mathbb{R}^2)\bigr \rangle,$$ and the $H^2(\mathbb{R}^2)$ norm defined from the Fourier Transform, $$\lvert| u\rvert|=\int_{\mathbb{R}^2}(1+|\xi|^2)^2|\hat{u}|^2d\xi<\infty.$$