Let $E$ be a Banach space. Let $\mathcal K(E)$ be the space of all compact (bounded linear) operators from $E$ to $E$. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. Let $I:E \to E$ be the identity map.
Below is the Fredholm alternative in Brezis' Functional Analysis.
Theorem 6.6 Let $T \in \mathcal K(E)$. Then
(a) $\dim N(I-T) < \infty$,
(b) $R(I-T)$ is closed, and more precisely $R(I-T) = N(I-T^*)^\perp$,
(c) $N(I-T) = \{0\} \iff R(I-T) = E$,
(d) $\dim N(I-T)=\dim N(I-T^*)$.
The proof of (d) by the author is by contradiction. My goal is to give a more direct proof of (d).
Could you elaborate if the map $I-S$ in my below attempt is actually injective?
My attempt By (a), $\dim N(I-T) < \infty$. Then $N(I-T)$ has a (closed) complement subspace $G$ in $E$, i.e., $E= N(I-T) \oplus G$. Let $\pi_1 : E \to N(I-T)$ be the (continuous linear) projection map. Then $\pi_1$ is surjective with $N (\pi_1) = G$.
We have $T \in \mathcal K(E) \iff T^* \in \mathcal K(E^{*})$. By (a), $\dim N(I-T^*) < \infty$. By (b), $\operatorname{codim} R(I-T) < \infty$. Then $R(I-T)$ has a (closed) complement subspace $L$ in $E$, i.e., $E= R(I-T) \oplus L$. Let $\pi_2 : E \to L$ be the (continuous linear) projection map. Then $\pi_2$ is surjective with $N(\pi_2) = R(I-T)$.
Let $\Lambda:= \pi_2 \circ \pi_1:E \to L$. Then $\Lambda$ is linear continuous. We have $\dim L = \dim N(I-T^*) < \infty$. Then $\Lambda$ has finite rank and thus $\Lambda \in \mathcal K(E)$.
Let's prove that $\Lambda$ is surjective. Let $S := T-\Lambda$. Then $S \in \mathcal K(E)$. First, we'll show that $I-S$ is injective. Indeed, if $(I-S)u=0$ then $(I-T)u + \Lambda u =0$. Because $E= R(I-T) \oplus L$ and $R(\Lambda) \subset L$, we get $(I-T)u=\Lambda u=0$. Then $u \in N(I-T) \cap N(\Lambda)$. We want to prove $u=0$.
