Given Banach spaces $E$ and $F$.
Consider a bounded operator: $$T:E\to F:\quad\|T\|<\infty$$
Certainly one has: $$T\text{ compact}\implies\mathcal{R}T\text{ separable}$$ What about the converse?
2026-03-31 08:41:23.1774946483
Compact Operators: Separable Range
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It's true that compact implies separable range. You can find a proof of this in Abramovich/Aliprantis' Problems in Operator Theory.
The other direction does not hold. To see this consider the identity operator on $\ell^2$. Since $\ell^2$ is separable the range of the identity is (since it equals the whole space). But the unit ball is compact if and only if the space is finite dimensional hence the identity operator is not compact.