# Equivalence of compact operators

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 $$(b) \Rightarrow (a)$$.

• What exactly is $B_X$? Feb 10, 2022 at 13:52
• $B_X = B(0,1) = \{x \in X: ||x|| < 1\}$ Feb 10, 2022 at 13:59

## 1 Answer

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?

• haha yes!! It was easy... Thank you mate :) Feb 10, 2022 at 14:07