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So I'm stuck at the following result, about compact operators on Hilbert spaces (which I think it's called Fredholm's theorem) from Stein. It's exercise 29 from Chapter 4.

Let $T$ be a compact operator on a Hilbert space $\mathcal{H}$, and assume $\lambda \neq 0$.

a) Show that the range of $\lambda I - T$ is closed.

b) Show that this is not true for $\lambda = 0$.

c) Show that the range of $\lambda I - T$ is all of $\mathcal{H}$ if and only if the nullspace of $\overline{\lambda} I - T^{*}$ is trivial.

Sorry if it's already here on the forum! Thanks

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This are standard fact which you can find in most functional analysis books – Norbert Apr 18 '12 at 9:27
Conway's Functional Analysis is a good reference for this. – Nicholas Stull Apr 18 '12 at 12:30
Thanks a lot; got it. What about $\lambda = 0$. Do you know any simple counterexample? – Anna Apr 19 '12 at 6:26
@Anna you can answer your own question. – Davide Giraudo Apr 19 '12 at 20:38 – leo Apr 19 '12 at 20:54

Consider $\mathcal H:=\ell^2(\mathbb C)$ the Hilbert space of complex sequences $\{z_k\}$ such that $\sum_{k=1}^{+\infty}|z_k|^2<\infty$, endowed with the canonical inner product. We denote $e_n$ the sequence which is $1$ at $n$, $0$ for the other integers and define
$$Tz:=\sum_{n=1}^{+\infty}\frac{\langle e_n,z\rangle}{n^2} e_n.$$ Then $T$ is linear and compact (we can in fact choose a sequence which is convergent to $0$, see this question for instance).

Its range $R(T)$ contains all the sequences such that their terms vanish for $n$ large enough (indeed, $T\left(\sum_{k=1}^nk^2\alpha_ke_k\right)=\sum_{k=1}^n\alpha_ke_k$), so the closure of $R(T)$ is $\mathcal H$. But $R(T)$ is not $\mathcal H$, since for example the sequence $\sum_{n\geq 1}\frac 1{n^2}e_n$ is in $\mathcal H$ but never reached by an element of $\mathcal H$.

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