= Solution
The <clamped second-order Sobolev space> is $H_0^2(U)=\overline{C_c^\infty(U)}^{H^2}$. On a smooth bounded domain the <Sobolev trace theorem> characterizes it by zero value and zero <normal derivative> on the boundary. In particular, its whole first-order boundary jet is zero, since tangential derivatives of the zero trace also vanish. As above, assume $f\in L^2(U)$ and classical regularity up to the boundary.
For a <smooth> <weak solution>, compactly supported <test functions> and two <integrations by parts> give
$$
\int_U(\Delta^2u-f)v=0\quad(v\in C_c^\infty(U)),
$$
so $\Delta^2u=f$ pointwise. Membership in $H_0^2(U)$ supplies $u=\partial_nu=0$ on $\partial U$. Thus it is a classical solution of the <clamped biharmonic problem>.
Conversely, a $C^4(\overline U)$ classical solution with these traces belongs to $H_0^2(U)$. For every compactly supported <test function>, two <integrations by parts> give $\int_U\Delta u\Delta v=\int_Ufv$. Both sides are continuous for the $H^2$ <norm>, so the defining density of $C_c^\infty(U)$ in $H_0^2(U)$ extends this equality to every required test. \b[The two notions agree under the stated smoothness.]
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