= Solution
The <Lax-Milgram theorem> gives, for a bounded <coercive bilinear form> on a real <Hilbert space> and every <bounded linear functional>, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The <Fredholm alternative> for a <compact operator> $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker(I+\mu K)^*$ in the Hilbert-space case.
<Uniform ellipticity> gives $\sum a_{ij}\xi_i\xi_j\geq\theta|\xi|^2$ for some $\theta>0$. Bounded coefficients make $B_\lambda$ bounded on $H^1$, and for $\lambda<-\gamma$,
$$
B_\lambda[u,u]\geq\theta\|Du\|_2^2+(-\gamma-\lambda)\|u\|_2^2\geq\min\{\theta,-\gamma-\lambda\}\|u\|_{H^1}^2.
$$
The forcing $v\mapsto\int fv$ is bounded on $H^1$. The <Lax-Milgram theorem> therefore gives \b[a unique <weak solution> for every $\lambda<-\gamma$].
When $c=0$, choose the shift $\lambda_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int(Kf)v=\int fv$. Thus $K=(L+I)^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, or
$$
\boxed{(I-K)u=Kf.}
$$
Conversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves <compactness> of $K$ as an $L^2$ operator.
By the supplied kernel assumption, $\ker(I-K)$ consists of constants. Also $K1=1$. Since $K$ is a <self-adjoint operator>, the <Fredholm alternative> condition is $Kf\perp1$, and
$$
\langle Kf,1\rangle=\langle f,K1\rangle=\int_\Omega f.
$$
Therefore \b[a <weak solution> exists exactly when the mean of $f$ is zero], and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.
For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the <weak formulation> explicitly defined in the paper, so does not require that stronger boundary identification.
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