1
$\begingroup$

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.

  • 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.2011-12-04
  • 0
    @Nate: Thank you very much. This is all what I need!2011-12-05
  • 0
    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.2011-12-05
  • 0
    Thank you for the information.2011-12-05

1 Answers 1

1

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.

  • 0
    What if $T$ is not known to be linear? only continuous and compact?2011-12-05
  • 3
    Which definition of "compact" do you use in this case?2011-12-05
  • 0
    Anyway I refer to my related (but different!) question on MO http://mathoverflow.net/questions/82688/how-to-prove-that-1-is-not-an-eigenvalue-of-tx, where the operator I am talking about is given explicitly.2011-12-05
  • 0
    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?2011-12-05
  • 1
    @Ali: that's usually the case, yes.2011-12-05