Proposition 4 (Fredholm Alternative)

Let E be a Banach space and let $h$ be a compact endomorphism of E. Let $\lambda$ be an element of K $-\{0\}$. Let F be the kernel of $1_E-\lambda h$ and G the kernel of $1_{E'}-\lambda^th$.

a) The spaces F and G have the same finite dimension $n\geqslant 0$;

b) For $y\in E$, there exists $x\in E$ such that $x-\lambda h(x) =y$ if and only if $\langle y, \ell \rangle = 0$ for all $\ell \in G$;

c) For $\ell \in E'$, there exists $f\in E'$ such that $f-\lambda^th(f) =\ell$ if and only if $\langle x, f\rangle = 0$ for all $x\in F$.

In particular, one and only one of the following conditions is valid:

(i) The space F is nonzero ;

(ii) For every $y\in E$, there exists $x\in E$ such that $x-\lambda h(x) =y$.

Replacing $h$ by $\lambda h$, we reduce to the case where $\lambda = 1$.

The endomorphism $^th$ is compact by corollary 1 of III, p. 9. The endomorphisms $w= 1_E-h$ of E and $w'= 1_{E'}-^th$ of $E'$ are therefore Riesz endomorphisms (cor. 2 of prop. 2 of III, p. 75). In particular, their kernels F and G have finite dimension. Since $w'=^tw$ and the index of $w$ is zero, the dimension of F is equal to that of G by formula (3) of III, p. 43.

For an element $y$ of E, the equation $x-h(x) =y$ has a solution $x\in E$ if and only if $x$ belongs to the image of $w$. Since the dual of the cokernel of $w$ identifies with the kernel G of Ker($^tw$) (EVT, IV, p. 27, prop. 2), this is the case if and only if $\langle y, \ell \rangle = 0$ for all $\ell \in G$.

For an element $\ell$ of $E'$, the equation $f-^th(f) =\ell$ has a solution $f\in E'$ if and only if $f$ belongs to the image of $w'$. Since the image of $^tw$ is the orthogonal of the kernel F of $w$ (EVT, IV, p. 27, prop. 2) this is the case if and only if $\langle x, f\rangle = 0$ for all $x\in F$.