0
  1. Can we say a compact operator is weak-to-norm continuous?
  2. What do we say about converse of question 1?
Ali Farzaneh
  • 329

2 Answers2

5

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).

Community
  • 1
David Mitra
  • 74,748
1

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

azarel
  • 13,120
  • 1
    Well, no one said we have a reflexive space here. – t.b. Aug 03 '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 03 '12 at 22:31
  • Sure, sorry that was a rather silly nitpick. It's a good hint for various converses... – t.b. Aug 03 '12 at 22:38