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Let $X, Y$ be Banach spaces and $T \in B(X, Y )$ be linear operator. Show that if $x_n \rightharpoonup x$ in $X$, then $Tx_n \rightharpoonup Tx$ in $Y$ .

My attempt

since $T \in B(X, Y )$, $\|T\| < \infty$ , given that $x_n \rightharpoonup x$, choose $N$ such that $\|\langle x_n,z\rangle-\langle x,z\rangle\| \le \frac{\epsilon}{\|T\|} , \ \forall n>N , \forall z\in X. \ \ \ $ Then;

\begin{align} \|\langle Tx_n,z\rangle-\langle Tx,z\rangle\| & = \|\langle T(x_n-x),z\rangle\|\\ & \le \|T\|\|\langle (x_n-x),z\rangle\| \\ & = \|T\|\|\langle x_n,z\rangle-\langle x,z\rangle\| \\ & \le \|T\|\frac{\epsilon}{\|T\|} = \epsilon , \ \ \forall z \in X \end{align}

domath
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1 Answers1

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There are several problems with what you wrote:

  • You seem to be treating $X$ as a Hilbert space, and assuming that $Y=X$. In a general Banach space there is no inner product nor Riesz Representation Theorem.

  • You cannot choose $N$ uniformly on $z$, as you wrote.

  • You write norms when applied to numbers, which is not necessarily wrong (the absolute value in $\mathbb C$ is actually a norm) but it's weird.

  • Your first estimate makes no sense, even in a Hilbert space.

How to do the exercise:

In a Banach space, that $x_n\to x$ weakly means that (the numbers) $f(x_n)\to f(x)$ for all $f\in X^*$. The exercise is then trivial, since given $T\in B(X,Y)$ and $g\in Y^*$, then $g\circ T\in X^*$. Then $g(Tx_n)\to g(Tx_n)$ by hypothesis.

Martin Argerami
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  • so if I assume the Hilbert spaces and change as the following would it make sense? (I did not get your 4th point.) $\forall z\in X$ choose $N$ such that $|\langle x_n,z\rangle-\langle x,z\rangle| \le \frac{\epsilon}{|T|} , \ \forall n>N , . \ \ \ $ Then; $\forall z \in X$

    \begin{align} |\langle Tx_n,z\rangle-\langle Tx,z\rangle| & = |\langle T(x_n-x),z\rangle|\ & \le |T||\langle (x_n-x),z\rangle| \ & = |T||\langle x_n,z\rangle-\langle x,z\rangle| \ & \le |T|\frac{\epsilon}{|T|} = \epsilon \end{align}

     – domath Mar 31 '20 at 17:44
  • No. My 4th point says that the inequality $|\langle Tx,z\rangle|\leq |T|,|\langle x,z\rangle|$ is wrong. – Martin Argerami Mar 31 '20 at 18:01
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    For instance try $$T=\begin{bmatrix} 0&1\0&0\end{bmatrix},\ \ x=\begin{bmatrix} 0\1\end{bmatrix},\ \ z=\begin{bmatrix} 1\0\end{bmatrix}. $$ – Martin Argerami Mar 31 '20 at 18:02
  • I do not understand you in this statement "Then $g(Tx_n)\to g(Tx_n)$ by hypothesis. " the hypothesis was on $x_{n}$ and $x$ not $Tx_{n}$ and $Tx$, could you please explain this? –  Apr 15 '20 at 03:36
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    Call $g\circ T$ as $h$, and write $h(x_n)\to h(x)$. – Martin Argerami Apr 15 '20 at 03:55