= Solution
{c}
Take $l>0$ and use the <inner product> $\langle u,v\rangle=\int_0^l\overline u v\,dx$ on the <Hilbert space> $L^2(0,l)$. The <differential operator> has domain
$$
D(\mathcal L)=\{u\in H^2(0,l):u(0)=u(l)=0\},
$$
where $H^2$ is a <Sobolev space>. Twice applying <integration by parts> gives the <Lagrange identity for a Sturm-Liouville operator>
$$
\langle u,\mathcal Lv\rangle-\langle\mathcal Lu,v\rangle
=[\overline u v'-\overline{u'}v]_0^l.
$$
The <Dirichlet boundary conditions> make this expression zero for $u,v\in D(\mathcal L)$, and the <formal adjoint> is $\mathcal L^*=d^2/dx^2+1$. To obtain the domain of the <adjoint operator>, initially impose the <boundary conditions> only on $u$. The remaining expression is $-[\overline{u'}v]_0^l$. Since $u'(0)$ and $u'(l)$ vary independently, it vanishes for all such $u$ exactly when $v(0)=v(l)=0$. The adjoint domain is therefore the original domain: this is full <self-adjointness>, including the <boundary conditions>.
Consequently the <eigenvalues> are real, <eigenfunctions> for distinct <eigenvalues> are <orthogonal>, and regular <Sturm-Liouville theory> supplies a complete <orthonormal eigenbasis>. Here these objects can be exhibited explicitly:
$$
\boxed{\phi_n(x)=\sqrt{\frac2l}\sin\frac{n\pi x}{l},\qquad
\lambda_n=1-\left(\frac{n\pi}{l}\right)^2,\qquad n=1,2,\ldots.}
$$
Indeed $\mathcal L\phi_n=\lambda_n\phi_n$, and the <Fourier sine basis> is complete in $L^2(0,l)$. There are no additional <eigenvalues>: for a nonzero <eigenfunction> $u$, <integration by parts> gives
$$
\lambda=1-\frac{\int_0^l|u'|^2\,dx}{\int_0^l|u|^2\,dx}<1,
$$
so $u''+(1-\lambda)u=0$ and the endpoint values force $\sqrt{1-\lambda}\,l=n\pi$. All these <eigenvalues> are simple. The <eigenvectors> are precisely the nonzero multiples of these <eigenfunctions>, regarded as <vectors> in the <Hilbert space>.
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