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

Question

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?

Answer 1

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.

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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.

Answer 2

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 $Tx_n\rightarrow T x$.

Answer 3

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.