# When is a compact operator differentiable?

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.

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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

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.
What if $T$ is not known to be linear? only continuous and compact? –  user17090 Dec 5 '11 at 17:26