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.
  1. Can we say a compact operator is weak-to-norm continuous?
  2. What do we say about converse of question 1?
share|improve this question
    
Is the underlying space a Banach space or a Hilbert space? –  J. Loreaux Aug 3 '12 at 22:28
2  
Please read: meta.math.stackexchange.com/questions/1803/… –  t.b. Aug 3 '12 at 22:29

2 Answers 2

Hint for (2): the closed unit ball is weakly compact in any reflexive Banach space (as pointed out by t. b.)

share|improve this answer
    
Well, no one said we have a reflexive space here. –  t.b. Aug 3 '12 at 22:24
    
@t.b. That's right. I was reading some paper on $C^*$-algebras so I had a Hilbert space on my mind. I'll leave the answer in case the OP was about reflexive spaces. –  azarel Aug 3 '12 at 22:31
    
Sure, sorry that was a rather silly nitpick. It's a good hint for various converses... –  t.b. Aug 3 '12 at 22:38

For part 2., more can be said: If $T:X\rightarrow Y$ is weak-to-norm continuous where $X$ and $Y$ are Banach spaces, then $T$ has finite rank. One can prove this using an argument similar to those in this post and this post (Use the fact that the inverse image under $T$ of the unit ball of $Y$ is weakly open to find $f_1,\ldots,f_n$ in $X^*$ so that $\Vert Tx\Vert\le 1$ whenever $|x_i^*x|<1$ for all $i$. Then show that the kernal of $T$ must contain the subspace $\cap_{i=1}^n \text{ker}( f_i)$ of $X$. This subspace has codimension at most $n$; thus $T$ has rank at most $n$).

In light of this fact, your first question becomes: is every compact operator of finite rank? The answer of course is "no", in general (however, recall that a compact operator between Banach spaces is weak-to-norm sequentially continuous).

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.