Compact operator in reflexive space compact

In summary, the conversation is discussing an unproven statement in a functional analysis book about the image of a closed unit sphere through a compact linear operator on a linear variety of a Banach space being compact if the Banach space is reflexive. The person is asking if anyone knows a proof of this statement and provides some useful facts, including a proof that they found. They are also trying to understand more advanced concepts and welcome anyone to join the discussion.
  • #1
DavideGenoa
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Hi, friends! I find an interesting unproven statement in my functional analysis book saying the image of the closed unit sphere through a compact linear operator, defined on a linear variety of a Banach space ##E##, is compact if ##E## is reflexive.
Do anybody know a proof of the statement?
##\infty## thanks!
 
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  • #2
Some facts which may be useful:
- Given a Banach space, the unit ball of its dual space is weak*-compact.
- Given a reflexive Banach space, the the weak and weak* topologies on its dual space coincide.
- Any closed convex subset of a Banach space is weakly closed.
- Any continuous map with compact domain has compact range.
 
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  • #3
I have found a proof what I didn't know of the propositions you quote here. My text states a lot of things of a more advanced level than the elements of theory that it explains, but I'm curious and want to try to understand as much as I can...
While searching I have also found this page, which would resolve the problem if I understood why Because A is compact, ##Ax_{n_j}\to Ax## strongly. I only know that ##\forall f\in X^{\ast}## ##f(x_{n_j})\to f(x)## strongly...
I ##\infty##-ly thank you and anybody wishing to join the thread!
 
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1. What is a compact operator in a reflexive space?

A compact operator in a reflexive space is a linear transformation between two reflexive spaces that maps bounded sets to relatively compact sets. In other words, the images of bounded sets under a compact operator are contained within a compact subset of the target space.

2. How does a compact operator differ from a bounded operator?

While both compact and bounded operators map bounded sets to bounded sets, a compact operator also maps bounded sets to relatively compact sets. This means that the image of a bounded set under a compact operator is contained within a compact subset of the target space, whereas the image of a bounded set under a bounded operator is simply contained within the target space itself.

3. What is the importance of compact operators in functional analysis?

Compact operators play a vital role in functional analysis as they are used to study properties of linear transformations between reflexive spaces. They also have many applications in various areas of mathematics, including differential equations, integral equations, and spectral theory.

4. How are compact operators related to compact sets?

The concept of compact operators is closely related to that of compact sets. A compact set is a set that is closed and bounded, and a compact operator maps bounded sets to relatively compact sets. In fact, a compact operator can be defined as a continuous linear transformation that maps every bounded set to a relatively compact set.

5. Can a compact operator be unbounded?

No, a compact operator is always a bounded operator by definition. This means that its operator norm, which measures the maximum magnitude of its output for a given input, is finite. However, a bounded operator is not necessarily a compact operator, as it may not map bounded sets to relatively compact sets.

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