 Solution

The Lax-Milgram theorem states that on a real Hilbert space, a bounded bilinear form $a$ satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ for some $c>0$ represents every bounded linear functional uniquely: there is a unique $u$ with $a(u,v)=\ell (v)$ for all $v$.

The clamped trace conditions define a closed subspace $H_{\partial }^2(\Omega )$ of the Sobolev space $H^2(\Omega )$. On it take $a(u,v)={\int }_{\Omega }\Delta u\Delta v$ and $\ell (v)={\int }_{\Omega }fv$. Both are bounded by the Cauchy-Schwarz inequality. The Dirichlet elliptic regularity estimate on a smooth bounded domain gives $\Vert v{\Vert }_{H^2}\le C\Vert \Delta v{\Vert }_{L^2}$ for $v\in H^2\cap H_0^1$, so $a(v,v)\ge C^{-2}\Vert v{\Vert }_{H^2}^2$. The Lax-Milgram theorem now gives exactly one $u$ satisfyingTesting with compactly supported smooth $v$ and integrating by parts twice gives ${\Delta }^{2}u=f$ in distributions; the two boundary conditions hold as traces by the space's definition. This is the unique weak clamped biharmonic solution.