= Solution
Use the usual <inner product> $\langle u,v\rangle=\int_0^1\overline u v\,dx$. Moving four <derivatives> by <integration by parts> leaves $L^*=D^4+1$ and gives the fourth-order boundary identity
$$
\langle u,Lv\rangle-\langle Lu,v\rangle
=\left[\overline u v'''-\overline {u'}v''+\overline {u''}v'-\overline {u'''}v\right]_0^1.
$$
Thus the differential expression is formally <self-adjoint>. If both <functions> satisfy the <clamped boundary conditions>, every displayed endpoint product vanishes, proving symmetry on the specified domain.
To check full <self-adjointness>, also identify the adjoint domain. Allow $v$ to run through the clamped domain. At either endpoint, $v''$ and $v'''$ can be chosen independently, for instance using a <polynomial> near that endpoint multiplied by a <smooth cutoff function>. The remaining boundary form is $[\overline u v'''-\overline {u'}v'']_0^1$. It vanishes for every such $v$ precisely when $u=u'=0$ at both endpoints. These are exactly the original <boundary conditions>. On the regular finite <interval>, the adjoint domain consequently equals $\{u\in H^4(0,1):u(0)=u'(0)=u(1)=u'(1)=0\}$, rather than a larger domain. Here $H^4$ is the <Sobolev space> of <functions> whose <weak derivatives> through order four are <square-integrable>.
\b[The operator together with these <boundary conditions> is fully <self-adjoint>.] As a further consistency check, <integration by parts> gives $\langle y,Ly\rangle=\int_0^1(|y''|^2+|y|^2)\,dx>0$ for every nonzero admissible $y$. The reusable calculation is the <Lagrange identity for a fourth-order differential operator>.
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