I was reading "Hilbert Space Operators in Quantum Physics" by Jiˇrí Blank • Pavel Exner • Miloslav Havlícek ˇ
A linear everywhere defined operator $C$ on $H$ is said to be compact if it maps any bounded subset of $H$ to a precompact set. Since $H$ is a metric space, this is equivalent to the requirement that any bounded sequence $(x_n)⊂H$ contains a sub-sequence $(x_{n_k} )$ such that $(Cx_{n_k} )$ converges.
I am not able to deduce the following statement.
any bounded sequence $(x_n)⊂H$ contains a sub-sequence $(x_{n_k} )$ such that $(Cx_{n_k} )$ converges.
My attempt:-
$(Cx_{n_k} )$ is a sequence in locally compact space. ($\because$ C is a compact operator.) I am not able to proceed further. Could you help me? I also know that A set $M$ in a complete metric space $X$ is precompact iff it is completely bounded. In particular, if X is a finite–dimensional normed space, then $M$ is precompact iff it is bounded.
The comment by Eric Wofsey is correct: If $C$ is compact, then $C(A)$ being precompact means that $\overline{C(A)}$ is compact.
Let $C: H \to H$ be a linear operator on $H$, and suppose $C$ is a compact operator. Now, let $A = \{x_n\} \subset H$ be a bounded sequence. Since $A$ is bounded, the image sequence $C(A) = \{Cx_n\}$ is precompact. Hence, $\overline{C(A)}$ is compact. At this point, if you know compactness in a metric space is equivalent to every sequence having a convergent subsequence, we are done because $\{Cx_n\}$ is a sequence in the compact space $\overline{C(A)}$.