Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.