Brezis' exercise 6.25(ii): the existence of $\widetilde M, \widetilde P \in \mathcal L(E)$ such that $(I+K) \circ \widetilde M = I-\widetilde P$

Question

Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. Let $I:E \to E$ be the identity map. For $T \in \mathcal L(E)$, we denote by $N(T)$ its kernel and by $R(T)$ its range.

I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $E$ be a real Banach space and $K \in \mathcal{K}(E)$. Prove that there exist $M, \widetilde{M} \in \mathcal{L}(E)$ and finite-rank projections $P, \widetilde{P} \in \mathcal{L}(E)$ such that

• (i) $M \circ(I+K)=I-P$,
• (ii) $(I+K) \circ \widetilde{M}=I-\widetilde{P}$.

Note: A projection operator $P$ is a bounded linear operator such that $P \circ P=P$.

There are possibly subtle mistakes that I could not recognize in below attempt of (ii). Could you please have a check on it?

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Let $T:=I+K$. By Fredholm alternative, $R(T)$ is closed and $\operatorname{codim} R(T) = \dim N(T) < \infty$. Let $L$ be the topological complement of $R(T)$. Let $G$ be the topological complement of $N(T)$. Then

• $L$ is closed, $L \cap R(T) = \{0\}$ and $E=L \oplus R(T)$,
• $G$ is closed, $G \cap N(T) = \{0\}$ and $E=G \oplus N(T)$.

Let $\pi_G, \pi_L, \pi_{R(T)}, \pi_{N(T)}$ be the projections onto $G, L, R(T), N(T)$ respectively. This implies

• $I=\pi_L + \pi_{R(T)} = \pi_G + \pi_{N(T)}$,
• $N(\pi_L) =R(T)$ and $N(\pi_{R(T)}) =L$,
• $N(\pi_G) =N(T)$ and $N(\pi_{N(T)}) =G$.

First, we verify that $\widetilde T : G \to R(T), u \mapsto Tu$ is bijective. Assume $u,v \in G$ such that $Tu=Tv$. Then $u-v \in N(T) \cap G$ and thus $u-v=0$. Then $\widetilde T$ is injective. Let $v \in R(T)$. There is $u\in E$ such that $v=Tu$. Then $u' :=\pi_G u \in G$ and $T u' =T (I-\pi_{N(T)})u=Tu-T\pi_{N(T)} u=Tu=v$. Then $\widetilde T$ is surjective.

Because $G, R(T)$ are closed, they are Banach spaces. By open mapping theorem, $\widetilde T$ has a bounded inverse $\widetilde T^{-1} : R(T) \to G$, i.e., $\widetilde T \widetilde T^{-1} =I_{R(T)}$ and $\widetilde T^{-1} \widetilde T = I_{G}$.

We define $\widetilde M := \widetilde T^{-1} \pi_{R(T)}$ and $\widetilde P := \pi_L$. Because $E=L \oplus R(T)$, it suffices to verify that (i) holds for $u \in L \cup R(T)$.

• Let $u \in L$. Then $(I-\pi_L)u=0$ and $\pi_{R(T)}u=0$. Then (i) holds.
• Let $u \in R(T)$. Then $(I-\pi_L)u=u$ and $$ T\widetilde T^{-1} \pi_{R(T)}u = \widetilde T \widetilde T^{-1} u=I_{R(T)}u=u. $$

The claim then follows.