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Could any one help me in proving this question please:

Let $X$ be a reflexive Banach space and $T: X \rightarrow X$ a linear operator. Prove that $T$ is bounded iff $$x_{n} \rightharpoonup x \quad \Rightarrow \quad Tx_{n} \rightharpoonup Tx.$$

I found this question here:

if $x_n \rightharpoonup x$ in $X$, then $Tx_n \rightharpoonup Tx$ in $Y$ , for $T \in B(X, Y )$

But I do not understand if this is an answer for the question or no, I got confused from the information there, could anyone clarify what is written for me please?

Intuition
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3 Answers3

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The reverse implication is a simple consequence of the closed-graph theorem: Let $(x_n)$ be a sequence such that $x_n\to x$ and $Tx_n\to y$. We need to show $Tx=y$. Since every strongly convergent sequence is weakly convergent, it follows $x_n\rightharpoonup x$, and by the assumption $Tx_n \rightharpoonup Tx$. Since weak limits are unique, $Tx=y$, the graph of $T$ is closed, and $T$ is continuous.

Martin Argerami
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daw
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The answer you mention shows your implication $\implies$.

Reflexivity of $X$ is not needed (but I have left below an argument that uses it). The key observation is that any weakly convergent sequence (not net!) is bounded. This follows from the Uniform Boundedness Principle: for each $f\in Y^*$, the sequence (of numbers) $\{f(Tx_n)\}$ is bounded; that is, $$\sup\{|f(Tx_n)|:\ n\}<\infty$$ for each $f\in Y^*$. Using $\{\widehat{Tx_n}\}\subset Y^{**}$ as the family $F$ in the UBP, we get that $$ \sup\{\|Tx_n\|:\ n\}=\sup\{|f(Tx_n)|:\ n\in\mathbb N,\ \|f\|=1\}<\infty $$ Now, if $T$ were unbounded, there would exist a sequence $x_n$ such that $x_n\to0$ and $\|Tx_x\|>n$. This would give a sequence with $x_n\rightharpoonup0$ such that $Tx_n$ does not converge weakly.

Using reflexivity of $X$:

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 subsequence 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.

Martin Argerami
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  • I am not sure if reflexivity of $X$ is necessary here. Do you have a counterexample?

    How about the following argument? If $T$ is not bounded, then there exists ${x_n}$ such that $x_n\to 0$ while $|Tx_n|\geq n$. Now, since $x_n\rightharpoonup 0$, we know from the assumption that $Tx_n \rightharpoonup 0$. But this means that $|Tx_n|$ is bounded, so we have a contradiction?

     – user58955 Apr 14 '20 at 01:13
  • I think the statement should hold in general for $T:X\to Y$ for Banach spaces $X$ and $Y$ without any reflexivity assumption. – user58955 Apr 14 '20 at 01:27
  • You are right. I have edited the answer. – Martin Argerami Apr 14 '20 at 02:41
  • The reverse implication is a simple consequence of the closed-graph theorem. – daw Apr 14 '20 at 07:06
  • In the reverse direction proof: I do not understand this sentence: "Because of linearity, it is enough to show that $x_n\to0$ implies $Tx_n\to0$." Could you clarify it please? –  Apr 15 '20 at 08:28
  • $x_n\to x\ \iff\ (x_n-x)\to0$. And $Tx_n\to Tx\ \iff\ T(x_n-x)\to0$. – Martin Argerami Apr 15 '20 at 08:38
  • And then you rename $(x_{n} - x)$ by another name or what ? as you build up a new sequence. –  Apr 15 '20 at 08:44
  • Yes, I rename it as $x_n$. – Martin Argerami Apr 15 '20 at 09:01
  • Do you have a proof for this "Because $X$ is reflexive, closed balls are weakly compact"? –  Apr 15 '20 at 09:11
  • I have a proof to this: " If $X$ is a reflexive Banach space then its closed unit ball is weakly sequentially compact." So I need $X$ to be complete not just normed space. –  Apr 15 '20 at 09:19
  • Do you have a proof for that $X$ is complete? –  Apr 15 '20 at 09:20
  • Yes; duals are always complete. – Martin Argerami Apr 15 '20 at 09:24
  • Where is the dual here sorry I can not see? –  Apr 15 '20 at 09:29
  • Do you know what "reflexive" means? – Martin Argerami Apr 15 '20 at 09:30
  • yess "A normed space $X$ is reflexive if the natural embedding $J: X \rightarrow X^{**}$ is surjective." –  Apr 15 '20 at 09:32
  • Do you mean that it is isometric isomorphic to the double dual? –  Apr 15 '20 at 09:33
  • Do you have a proof for that the duals are complete? –  Apr 15 '20 at 09:34
  • Yes, the double dual is a dual. – Martin Argerami Apr 15 '20 at 09:36
  • what about my last comment? –  Apr 15 '20 at 09:38
  • LMGTFY: https://www.google.com/search?q=dual+space+is+complete&oq=dual+space+is&aqs=chrome.1.69i57j0l3.9750j0j7&client=tablet-android-samsung-nf-rev1&sourceid=chrome-mobile&ie=UTF-8 – Martin Argerami Apr 15 '20 at 09:45
  • Perfect! +1 … it helped me a lot thank you! –  Apr 15 '20 at 10:08
  • Also how can I change my closed unit ball to your general balls? as I only took the statement for unit balls? –  Apr 15 '20 at 12:51
  • Do you mean in the line before last "to every subseqence ${Tx_{n}}$? –  Apr 15 '20 at 13:19
  • why the weak limit is the same as the strong limit for the same sequence? –  Apr 15 '20 at 16:12
  • "Now, if were unbounded, there would exist a..." @MartinArgerami. How do I get this ? – mudok Apr 01 '23 at 13:36
  • $T$ unbounded means that for every $c>0$ there exists $x$ with $|Tx|>c|x|$. If you divide by $|x|$ you see that this is equivalent to the existence of $x$ with $|x|=1$ and $|Tx|>c$. So you can find $y_n$ with $|y_n|=1$ and $|Ty_n|>n^2$. Now take $x_n=y_n/n$. – Martin Argerami Apr 01 '23 at 17:58
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Start by writing down $||Tx_n-Tx||=||T(x_n-x)||$ by linearity, try using the definition of boundedness of the operator $T$ to obtain an inequality involving $||x_n-x||$. Now you want to use that $x_n\rightarrow x$, think about what value of $\epsilon$ you want to use in the definition of convergence to show that $Tx_n\rightarrow T x$.

Basel J.
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  • -1: This makes no sense. It's not related to weak convergence, does not deal with the reverse implication (which is the non-trivial one), nor is related to the fact that $X$ is reflexive. – Martin Argerami Apr 13 '20 at 02:53
  • The person referenced an answer in which they were trying to prove the forwards direction, so I was trying to clarify how one would approach that. But thanks for clarifying the backwards direction. – Basel J. Apr 13 '20 at 03:07