0
$\begingroup$

I'm reading about compact operators and I'm trying to prove the following statement:

Let $X,Y$ be Banach spaces and $T:X \to Y$ a linear operator. Then the following are equivalent:

(a) T is compact.

(b) The set $T(B_X)$ is a relatively compact subset of $Y$.

The definition of compact operator that I know is: $T:X \to Y$ is a compact operator if it sends bounded sets of $X$ to relatively compact sets of $Y$. With this it's very easy to show $(a) \Rightarrow (b)$, but I don't know how to proceed on $(b) \Rightarrow (a)$.

$\endgroup$
2
  • $\begingroup$ What exactly is $B_X$? $\endgroup$
    – Meowdog
    Feb 10, 2022 at 13:52
  • $\begingroup$ $B_X = B(0,1) = \{x \in X: ||x|| < 1\}$ $\endgroup$
    – D.R. Cstr
    Feb 10, 2022 at 13:59

1 Answer 1

Reset to default
1
$\begingroup$

Hint. If $B \subseteq X$ is bounded, then by boundedness there is $R > 0$ such that $$ B \subseteq \{x \in X: \|x\| < R\} = RB_X $$ Now $T(B) \subseteq T(RB_X) = RT(B_X)$ by linearity. As multiplication by $R$ is a homeomorphism, $RT(B_X)$ is relatively compact by (b). Can you conclude?

$\endgroup$
1
  • $\begingroup$ haha yes!! It was easy... Thank you mate :) $\endgroup$
    – D.R. Cstr
    Feb 10, 2022 at 14:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .