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This question has been asked here showing $\lVert \Delta u \rVert_{L^2(U)} + \lVert u \rVert_{L^2(U)}$ is equivalent to $\|u\|_{H^2(U)}$ norm for $H^2(U)$ space assuming the domain $U$ is sufficient smooth. However I came across some problem when following the hint below the post, I do as follows:

Consider $u \in H^2(U)$ it satisfies the equation :

$$\Delta u = f \\ u = g\ \text{ on }\ \partial U $$

By the elliptic reguarity we have $$\|u\|_{H^2} \le C(\|u\|_{L^2 } + \|f\|_{L^2} + \|g\|_{H^2(U)})$$

The problem here differs from the problem for equivalent norm in $H = H^1_0 \cap H^2$ is that this equation above does not has homogenuous boundary condition, therefore the regularity result has $\|g\|_{H^2(U)}$ in it correct.I have no idea how to bound this term.

yi li
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    apply the estimate for the problem with homogeneous boundary conditions to $u-g$. – daw May 21 '22 at 11:20
  • also the statement is FALSE without assumptions on $\Omega$, see any textbook on PDEs, e.g., Evans – daw May 21 '22 at 11:21
  • Can you elaborate more , $|u|{H^2} - |g|{H^2(U)}\le |u-g|{H^2(U)} \le |f- \Delta g|{L^2(U)}$ still involves norm of $g$, I don't have enough knowledge to deal with norm of $g$ @daw – yi li May 21 '22 at 12:29

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