Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
This are standard fact which you can find in most functional analysis books –  no identity 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
    
meta.math.stackexchange.com/q/3286/8271 –  leo Apr 19 '12 at 20:54

1 Answer 1

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 $\mathcal H$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.