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?
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asked May 31, 2014 at 17:09
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1$\begingroup$ It's true that compact implies separable range. The other direction I suspect does not hold. Why not try to find a counterexample? $\endgroup$– Rudy the ReindeerMay 31, 2014 at 17:12
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1 Answer
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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.
answered May 31, 2014 at 17:19
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$\begingroup$ Can u explain me what compact operators makes special apart from having separable range? I would like to understand the idea behind this definition. $\endgroup$ May 31, 2014 at 17:27
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$\begingroup$ On a Hilbert space compact operators are limits of finite rank operators. For further information about where they turn up, see en.wikipedia.org/wiki/Compact_operator. $\endgroup$– UrgjeMay 31, 2014 at 18:16
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$\begingroup$ Yes sure but I would like to understand what makes them sooo special - I mean there must be some reasoning for considering them as an extra class of operators themselves (somethink like "bounded operators are precisely the continuous one"). $\endgroup$ May 31, 2014 at 19:25
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1$\begingroup$ @Freeze_S Compact operators have much nicer properties than arbitrary operators. You can think of them as behaving a bit like operators on finite dimensional vector spaces. $\endgroup$ May 31, 2014 at 19:40
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1$\begingroup$ @MattN.: I opened a new thread on this question: math.stackexchange.com/questions/816352/… $\endgroup$ May 31, 2014 at 20:15