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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 (i). Could you please have a check on it?

Let $T:=I+K$ and $G := N(T)$. By Fredholm alternative, $\dim G < \infty$ and $R(T)$ is closed. Let $L$ be the topological complement of $G$. Then $L$ is closed, $G \cap L = \{0\}$ and $E=G \oplus L$. Let $\pi_G, \pi_L$ be the projections onto $G, L$ respectively. This implies

  • $I=\pi_G+\pi_L$,
  • $N(\pi_G) =L$ and $N(\pi_L) =G$.

First, we verify that $\widetilde T : L \to R(T), u \mapsto Tu$ is bijective. Assume $u,v \in L$ such that $Tu=Tv$. Then $u-v \in G \cap L$ 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 $u' :=\pi_L u \in L$ and $T u' =T (I-\pi_G)u=Tu-T\pi_G u=Tu=v$. Then $\widetilde T$ is surjective.

Because $R(T)$ is closed, it is itself a Banach space. By open mapping theorem, $\widetilde T$ has a bounded inverse $\widetilde T^{-1} : R(T) \to L$, i.e., $\widetilde T \widetilde T^{-1} =I_{R(T)}$ and $\widetilde T^{-1} \widetilde T = I_{L}$. By Hahn-Banach theorem, $\widetilde T^{-1}$ has a continuous linear extension $\widehat{\widetilde T^{-1}}:E \to E$.

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

  • Let $u \in G$. Then $(I-\pi_G)u=0$ and $Tu=0$. Then (i) holds.
  • Let $u \in L$. Then $(I-\pi_G)u=u$ and $$ \widehat{\widetilde T^{-1}} Tu= \widetilde T^{-1} \widetilde Tu=I_Lu=u. $$

The claim then follows.

Akira
  • 17,367

1 Answers1

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You was wrong in claiming

By Hahn-Banach theorem, $\widetilde T^{-1}$ has a continuous linear extension $\widehat{\widetilde T^{-1}}:E \to E$.

The vector-valued Hahn–Banach theorems do not hold for general Banach spaces. We fix this issue as follows. Let $\pi_{R(T)}$ be the projection onto $R(T)$. Then we define $$ M := \widetilde T^{-1} \pi_{R(T)}. $$

Analyst
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