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I am trying to understand the proof of the reverse direction in the first answer in this link:

Prove that $T$ is bounded iff $x_{n} \rightharpoonup x \quad \Rightarrow \quad Tx_{n} \rightharpoonup Tx.$

Here is the proof I am speaking about:

Using reflexivity of $X$:

For the converse, you need to use that $X$ is reflexive. We want to show that $T$ is continuous: that is, if $x_n\to x$, then $Tx_n\to Tx$. Because of linearity, it is enough to show that $x_n\to0$ implies $Tx_n\to0$. So suppose that $x_n\to0$. Then of course $x_n\rightharpoonup0$, which by hypothesis implies that $Tx_n\rightharpoonup0$. Because every weakly convergence sequence is bounded, the sequence $\{Tx_n\}$ is bounded. Because $X$ is reflexive, closed balls are weakly compact. So there exists a convergent subsequence $\{Tx_{n_k}\}$. Say $Tx_{n_k}\to y$. Since strong convergence implies weak convergence, $y=0$. Now we can apply this last reasoning to every sequence of $\{Tx_n\}$: that is, any subsequence of $\{Tx_n\}$ has a subsequence that converges to $0$: so $Tx_n\to0$. Thus $T$ is continuous, so bounded.

My questions are:

1-Is there is a typo in this part of the second line from below"apply this last reasoning to every sequence of $\{Tx_n\}$," Should it be every subsequence instead?

2-Also I do not know how to prove this lemma:

Lemma Let $X$ be a topological space and $\mathbf{x}=(x_n)_{n\in \mathbb{N}}$ be a sequence of elements of $X$. If every subsequence of $\mathbf{x}$ contains a subsequence convergent to $x$ then $x_n \to x$.

Could anyone provide me with the proof please?

3- I do not understand the finalization of the proof in the last 2 lines, could anyone explain it for me?

1 Answers1

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To answer your questions:

1) You are correct, it should say subsequence.

2) You can prove this by contradiction. We prove it for the case of metric spaces (since you are in a normed vector space this is enough). Assume that $x_n \not\to x$. Then there exists $\epsilon > 0$ and an increasing sequence $(n_k)_{k \geq 1}$ such that $d(x, x_{n_k}) \geq \epsilon$ for all $k \geq 1$. This subsequence cannot have a further subsequence converging to $x$ leading us to a contradiction.

3) I'm not sure what you are asking. The proof shows that for any $x_n \to 0$ we have $Tx_n \to 0$. By linearity this shows that $x_n \to x$ implies $Tx_n \to Tx$. This is equivalent to continuity of $T$ which in turn is equivalent to boundedness of $T$.

pg_star
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  • For (3) I am speaking about the last 2 lines in the proof, why we should include them in the proof? –  Apr 15 '20 at 14:09
  • (2) I can not see what contradiction we will arrive to? –  Apr 15 '20 at 14:10
  • If there wasn't a contradiction there would be a subsequence $(x_{n_{k_m}}){m \geq 1}$ such that $d(x{n_{k_m}}, x) \to 0$ as $m \to \infty$. Can you see how this contradicts the definition of the sequence $(x_{n_k})_{k \geq 1}$? – pg_star Apr 15 '20 at 14:14
  • Also what is the $x$ of the lemma in our case? –  Apr 15 '20 at 14:15
  • For your proof you would take $y$ to be '$x$'. – pg_star Apr 15 '20 at 14:17
  • I mean the statement of the lemma has 2 x's, one is a sequence and the other is a limit. I am asking about the one representing the sequence? –  Apr 15 '20 at 14:18
  • Is it our $Tx_n$? –  Apr 15 '20 at 14:21
  • I think I see the contradiction now thanks for clarifying this part. –  Apr 15 '20 at 14:23