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.

When is it possible to prove that a compact operator $T: V \to V$ where $V$ is a Banach space is also differentiable? Fréchet differentiable?

PS: There is a further information which might help. My operator $T$ associates to each vector field $j$ a vector field $b$ solution of a certain boundary value problem. I will write the whole set of equations if asked to.

share|improve this question
7  
I'm confused by this question. $T$ is a linear operator, correct? If $V$ is any Banach space, then $T$ is differentiable as soon as it is continuous. This is very easy to prove straight from the definition. When $V$ is finite dimensional then $T$ is always continuous and in fact compact, so this assumption is redundant. –  Nate Eldredge Dec 4 '11 at 18:09
    
@Nate: Thank you very much. This is all what I need! –  user17090 Dec 5 '11 at 6:48
    
You added "Fréchet differentiable" to your question: Both Gâteaux and Fréchet differentiability of continuous linear maps are easy consequences of the definitions; compact operators are always continuous. Also, when differentiability on Banach spaces comes unqualified it almost always means Fréchet differentiability, at least in the literature I know. –  t.b. Dec 5 '11 at 7:22
    
Thank you for the information. –  user17090 Dec 5 '11 at 7:35

1 Answer 1

up vote 0 down vote accepted

Reposting comment as answer, since it seems to be what the OP was looking for:

If $V$ is any Banach space, then $T$ is differentiable as soon as it is continuous. This is very easy to prove straight from the definition.

share|improve this answer
    
What if $T$ is not known to be linear? only continuous and compact? –  user17090 Dec 5 '11 at 17:26
3  
Which definition of "compact" do you use in this case? –  Mark Dec 5 '11 at 17:27
    
Anyway I refer to my related (but different!) question on MO mathoverflow.net/questions/82688/…, where the operator I am talking about is given explicitly. –  user17090 Dec 5 '11 at 17:34
    
In Wikipedia, they define a compact operator to be a linear operator which maps a bounded set to a relatively compact set. SO is "linear" a requirement of compactness? –  user17090 Dec 5 '11 at 17:50
1  
@Ali: that's usually the case, yes. –  Mark Dec 5 '11 at 18:09

Your Answer

 
discard

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