= Solution
The <spectral theorem for compact self-adjoint operators> used here concerns a compact <self-adjoint operator>: its nonzero <eigenvalues> are real, have finite multiplicity and can accumulate only at zero; the orthogonal complement of its kernel is the closed span of its orthonormal eigenvectors. Self-adjointness is essential for this eigenbasis statement.
Let $Kf$ be the unique zero-parameter <weak solution>. Uniqueness gives linearity. Testing with $Kf$ and using coercivity gives $\|Kf\|_{H^1}\leq\nu^{-1}\|f\|_2$. The <Rellich-Kondrachov theorem> makes the embedding $H_0^1(\Omega)\hookrightarrow L^2(\Omega)$ compact on this bounded $C^1$ domain, so $K$ is compact on $L^2$. Symmetry gives
$$
\langle Kf,g\rangle=B[Kg,Kf]=\langle f,Kg\rangle.
$$
It is self-adjoint and positive. It is injective: if $Kf=0$, the weak equation pairs $f$ to zero with all $H_0^1$ tests, dense in $L^2$, so $f=0$. Moreover $\langle Kf,f\rangle=B[Kf,Kf]>0$ for $f\ne0$.
The parameter equation is equivalent to $u=K(f+\lambda u)$, that is
$$
\boxed{u-\lambda Ku=Kf.}
$$
A solution of this operator equation in $L^2$ lies in $H_0^1$, because its right-side representation is $K(f+\lambda u)$, and the equivalence follows in both directions. For $\lambda\leq0$, the form $B[u,v]-\lambda\langle u,v\rangle$ remains coercive; the preceding minimization argument gives a unique <weak solution>.
The <spectral theorem> now supplies a complete <orthonormal basis> $(\phi_j)$ of $L^2$ with $K\phi_j=\kappa_j\phi_j$, $\kappa_j>0$, and $\kappa_j\to0$. The elliptic <eigenvalues> are $\lambda_j=1/\kappa_j>0$, with finite multiplicity and tending to infinity, and $L\phi_j=\lambda_j\phi_j$ weakly. If $\lambda$ is not one of them, then
$$
\boxed{u=\sum_j\frac{\langle f,\phi_j\rangle}{\lambda_j-\lambda}\phi_j.}
$$
This converges even in $H_0^1$: its squared energy <norm> is $\sum_j\lambda_j|\langle f,\phi_j\rangle|^2/|\lambda_j-\lambda|^2$, finite because the multiplier is bounded over $j$. Partial sums solve the weak equation with the corresponding $L^2$ projections of $f$. Passing to the limit proves the equation for every test; the same coefficients force uniqueness. Thus the expansion represents \b[the unique <weak solution> away from the elliptic <eigenvalues>].
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