# Please confirm this (functionnal analysis)

1. Apr 3, 2008

### quasar987

I have a theorem here that I find a little surprising and I would like confirmation that I am interpreting it correctly.

The theorem says that for E an infinite dimensional Banach space (over K=R or C) and T:E-->E a compact operator, 0 is in the spectrum of T. That is to say, $$0\in\sigma(T)=\{\lambda\in K:\lambda I-T \mbox{ is not invertible}\}$$.

In particular, this means that as soon as E if infinite dimensional and T:E-->E is compact, then T is not invertible. There are no invertible compact operators on infinite dimensional Banach spaces!

2. Apr 3, 2008

### morphism

That's true. In the infinite dimensional case, the compact operators form a proper ideal of the bounded operators.

In fact more is true about the spectrum of a compact operator. Namely, if it isn't finite, then it consists of a (countable) sequence that converges to zero, and all these nonzero elements are eigenvalues.

3. Apr 5, 2008

### Aleph-0

Yes, but the claim is false : There ARE invertible compact operators in infinite dimensional Banach spaces, but the inverse has to be unbounded.
In other words, the spectrum of any compact operator in an infinite dimensional space always accumulate at 0 but it does not contain 0 in general.

4. Apr 5, 2008

### morphism

The word "invertible" refers to invertibility in the Banach algebra under consideration, which in this case is the space of bounded operators on E. It's in this sense that the spectrum of a compact operator must always contain 0. Of course if your definition of spectrum allows for unbounded inverses, then yes, there are invertible compact operators. But this is a very non-standard definition.

5. Apr 6, 2008

### mathwonk

if you know the basic result that in an infinite dimensional banach space the closed unit ball is never compact, then this result should not be surprizing at all. i.e., a compact operator maps bounded sets like the unit ball, into sets of compact closure, which are much smaller than the unit ball.

in particular if the map is injective, it must map an independent sequence of unit vectors into shorter and shorter images, forcing its set theoretic inverse to enlarge a sequence of unit vectors by more and more, hence to be unbounded.

say isnt aleph wrong here? i believe any bounded linear map which is bijective on a banach space actually has a bounded inverse. so actually I believe there are two senses in which his statement that invertible compact operators T exist in infinite dimensional banach spaces deviates from the usual meaning of "invertible", namely there are never any bijective ones, and even when there exist injective ones, their inverse are unbounded even on the image of T.

of course he might define a map as invertible if it is injective, but again that is not a very useful definition of invertible, since it only has a one sided inverse, not a genuine 2 sided inverse, and even that one sided inverse is unbounded on its domain.

of course it may be that necessity forces operator theorists to deal with such maps, and they may call them "invertible", but I have not run across this.

Last edited: Apr 6, 2008
6. Apr 6, 2008

### mathwonk

i checked with my operator theorist friends and apparently the term invertible is not used in this generality.

7. Apr 6, 2008

### Aleph-0

Ok I agree. The spectrum indeed contains zero.
About terminology, Kato in its standard book (Pertubation theory of linear operators) uses the term "invertible" for injective and always precises whether or not the inverse operator is bounded.