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Let $E,F$ be Banach spaces. Let $\mathcal L(E, F)$ be the space of bounded linear operators from $E$ to $F$, and $\mathcal K(E, F)$ its subspace consisting of compact operators. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. We have Theorem 6.6 in Brezis' Functional Analysis, i.e.,

Fredholm alternative Let $I:E \to E$ be the identity map and $T \in \mathcal K(E, E)$. 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) $\dim N(I-T)=\dim N(I-T^*)$.

With exactly the same proof as the one of Fredholm alternative, I have showed that

S1 Let $E_0$ be a closed subspace of $E$. Let $I:E_0 \to E$ be the identity map and $T \in \mathcal K(E_0, E)$. Then

(a) $\dim N(I-T) < \infty$,

(b) $R(I-T)$ is closed.

To be honest, I'm not sure whether my proof of S1 is correct or not. There are possibly subtle mistakes that I could not recognize.

S2 Let $E_0$ be a closed subspace of $E$. Let $I:E_0 \to E$ be the identity map and $T \in \mathcal K(E_0, E)$. 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) $\dim N(I-T)=\dim N((I-T)^*)$.

My questions are as follows:

Among conclusions (a), (b), (c), (d) of S2, which ones remain true? Can we prove those valid ones by Fredholm alternative?

Thank you so much for your elaboration!

Akira
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  • Did you tried some simple examples in the finite-dimensional situation? Then, $I - T$ can be identified with an arbitrary matrix in $\mathbb R^{n \times m}$ with $n > m$. – gerw Jun 21 '23 at 07:31
  • S2 b,c are false: if $T=0$ then $R(I-T)=E_0 \ne E$. – daw Jun 21 '23 at 10:00
  • @daw I can see that S2(c) is false. Could you explain why S2(b) is false? – Akira Jun 21 '23 at 10:07
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    I thought that $N((I-T^*))^\perp$ would be $E$ so different from $R(I-T)$, but this is wrong. – daw Jun 21 '23 at 12:08

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