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So this problem comes from the functional analysis course I took last semester. Although simple, I thought was pretty cool. In the actual exercise they had broken it down into three parts, which I did not show in the problem post. Here I will show the steps as claims.

Claim 1: The operator $latex {\Delta^2: A \rightarrow L^2(\Omega)}$ is a bijection

Proof: To show injectivity we show that this operator has trivial kernel. One inclusion is obvious, we just need to show that $latex {\Delta^2 u = 0~\implies u=0}$. To see this we recall from elliptic regularity theory for the Laplacian that if $latex {\Delta \phi = f}$ for some $latex {f\in H^k(\Omega)}$ then

$latex \displaystyle \norm{\phi}_{H^{k+2}}\leq C\norm{f}_{H^k}.$

Now back to our question which can be written as $latex {\Delta(\Delta u)=0}$, and so applying the estimate we can see that $latex {\norm{\Delta u}_{H^{2}}\leq 0}$ and so $latex {\Delta u = 0}$. Applying the estimate one more time shows that $latex {u=0}$. For surjectivity we begin taking some $latex {f\in L^2(\Omega)}$ and let $latex {u\in H^1_0(\Omega)}$ be the unique solution to $latex {\Delta u = f}$, and by the regularity theory $latex {u \in H^2(\Omega)}$. Then let $latex {v \in H^1_0(\Omega)}$ be the unique solution $latex {\Delta v = u}$. It follows that $latex {v\in A}$ and $latex {\Delta(\Delta v)=f}$. (Super cool right?!) $latex \Box$

Claim 2: If $latex {\Delta^2 u =f }$ then

$latex \displaystyle \forall \phi \in A: \int_{\Omega}u\Delta^2 \phi = \int_{\Omega}\phi f.$

Proof: Let $latex {\phi\in A}$, multiplying $latex {\Delta^2 u =f }$ by $latex {\phi}$ and integrating we get

$latex \displaystyle \int_{\Omega} \Delta^2 u \phi = \int_{\Omega} f\phi.$

Since $latex {\phi \in H^1_0(\Omega)}$ we can integrate by parts once with vanishing boundary terms on the left hand side (just replace $latex {\phi}$ with a sequence of compactly supported smooth functions, integrate by parts using the definition of weak derivative passing limits by dominated convergence) to get

$latex \displaystyle \int_{\Omega} \Delta(\Delta u) \phi =-\int_{\Omega} \nabla \Delta u \nabla \phi .$

Now, $latex {\Delta u \in H^1_0}$ (since it is the solution to $latex {\Delta v =f}$) and so we can integrate by parts once more, again with vanishing boundary terms and we get

$latex \displaystyle \int_{\Omega} f\phi=\int_{\Omega} \Delta u \Delta\phi. \ \ \ \ \ (1)$

Now if we try to integrate by parts again we will run into a dead end, try it! Instead we play the same game with $latex {\int_{\Omega} u \Delta^2 \phi }$ to get

$latex \displaystyle \int_{\Omega} u \Delta^2 \phi = \int_{\Omega} \Delta u \Delta\phi . \ \ \ \ \ (2)$

Equating (1) and (2) gives the claim. $latex \Box$

Now we are ready to finish the problem, which is to prove some regularity result on weak solutions to $latex {\Delta^2 u =f }$ for $latex {f\in L^2}$; that is $latex {u\in A}$. By surjectivity of $latex {\Delta^2}$ there exists some $latex {\tilde{u}\in A}$ such that $latex {\Delta^2 \tilde{u} =f}$. We just need to conclude now that $latex {\tilde{u}=u}$. By claim 2 we have that

$latex \displaystyle \forall \phi \in A: \int_{\Omega}(u-\tilde{u})\Delta^2 \phi = 0,$

and since $latex {\Delta^2}$ is bijective we have that this is equivalent to

$latex \displaystyle \forall g \in L^2: \int_{\Omega}(u-\tilde{u})g = 0.$

Setting $latex {g = u-\tilde{u}}$ we get that $latex {u-\tilde{u}=0}$.