= Solution
The <Lax-Milgram theorem> states that on a real <Hilbert space>, a bounded bilinear form $a$ satisfying $a(v,v)\geq c\|v\|^2$ 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 $\|v\|_{H^2}\leq C\|\Delta v\|_{L^2}$ for $v\in H^2\cap H_0^1$, so $a(v,v)\geq C^{-2}\|v\|_{H^2}^2$. The <Lax-Milgram theorem> now gives exactly one $u$ satisfying
$$
\int_\Omega\Delta u\,\Delta v=\int_\Omega fv\qquad(v\in H_\partial^2(\Omega)).
$$
Testing with compactly supported smooth $v$ and integrating by parts twice gives $\Delta^2u=f$ in distributions; the two boundary conditions hold as traces by the space's definition. \b[This is the unique weak clamped biharmonic solution.]
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