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I was wondering about something related to compact operators. If we have a compact operator $T:X \mapsto Y$ and a bounded sequence $(x_n)n$, then we know that there is a convergent subsequence $(Tx_{n_k})$.

My question is: When can we deduce from this that there is a $w \in X$ such that $\lim_{k \rightarrow \infty}(Tx_{n_k})=Tw$? Is there a theorem that deals with this case?

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2 Answers 2

up vote 2 down vote accepted

Sufficient criteria for the existence of a $w\in X$ with $Tx_{n_k}\to Tw$ are

  1. $\mathcal{R}(T)$ is finite-dimensional (well, closed, but if a compact operator has closed range, the range is finite-dimensional).
  2. $X$ is a reflexive Banach space. Then $x_{n_k}$ has a weakly convergent subsequence, say to $w$, and since a continuous operator is also weakly continuous, $Tx_{n_{k_m}} \xrightarrow{w} Tw$. Since $Tx_{n_k} \to y$ in the norm topology, we also have $Tx_{n_k} \xrightarrow{w} y$, and hence $y = Tw$.

There may be further criteria, but these are the ones I know off-hand.

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I believe you need that $X$ is reflexive.

Let's replace the subsequence $x_{n_k}$ by $x_n$ for ease of notation. If $X$ is reflexive, then the sequence $x_n$ converges weakly to some $x \in X$. (every bounded sequence in a reflexive Banach space has a weakly convergent subsequence)

Then, because $x_n \to x$ weakly, $Kx_n \to Kx$ strongly because $K$ is compact, and we have it.

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