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The eigenvalues of $L$ are $\pi /\sqrt{2}$, $-\pi /\sqrt{2}$ and zero, none equal to one. On $E=\mathrm{span}\{1,\cos 2x\}$ the map $L-I$ is an invertible two-dimensional operator; on $E^{\perp }$ it is $-I$. Hence it is invertible on all of $L^2(0,\pi )$. Thus $(L-I)y=\sin x$ has exactly one solution. This is the finite-rank Fredholm alternative; no range obstruction survives the nonresonant identity term.

Rewrite the equation as $(L-I)y=\cos x$. On $E=\mathrm{span}\{\cos 2x,\sin 2x\}$, the operator $L-I$ has eigenvalues $\pi /2-1$ and $\pi /4-1$, both nonzero. On $E^{\perp }$ it equals $-I$. The orthogonal decomposition $L^2(0,\pi )=E\oplus E^{\perp }$ therefore makes $L-I$ bijective with bounded inverse. In particular there is exactly one solution for this forcing. This is also the finite-rank Fredholm alternative: $1$ is not an eigenvalue of $L$, so adding the identity term removes the image obstruction present in $Ly=f$.

Put $E=\mathrm{im}S$, a finite-dimensional subspace, and decompose $H=E\oplus E^{\perp }$. The restrictionis injective, because a nonzero vector in its kernel would be a $\lambda$-eigenvector of $S$. By finite-dimensionality it is therefore surjective.

Given $y=y_E+y_{\perp }$, first put $x_{\perp }=-y_{\perp }/\lambda$. Then $S{x}_{\perp }\in E$, so choose $x_E\in E$ such thatFor $x=x_E+x_{\perp }$ we obtain $(S-\lambda I)x=y$. Hence $S-\lambda I$ is surjective. This is the finite-rank Fredholm alternative.