1
$\begingroup$

Strong Operator Topology (definition-$1$): We called a sequence $\{T_n\}$ in $\beta(X,Y),$ for $X,~Y$ are Banach spaces and $\beta(X,Y)$ denotes the family of bounded linear operators from $X$ to $Y$, converges strongly to $T$ if for all $x$ in $X$, $T_nx$ converges to $Tx$. That is, we can say $$\lim_{n \to \infty}\|T_nx-Tx\|=0,~~\text{ for each }~x \in X.$$ Now the associated topology is known as Strong Operator Topology.

Strong Operator Topology (definition-$2$): Let us consider a family of maps $\{F_x:\beta(X,Y) \to Y \text{ such that } F_x(T)=Tx,~T \in \beta(X,Y)\}$ where $x\in X.$ Then Strong Operator Topology is defined as the weak topology generated by the family of maps $F_x$ for each $x \in X$.

I am not able understand the equivalence of definition $1$ and $2$. How can I prove both the definitions are same (equivalent)? I know the definition of weak topology in Banach spaces. But I am not able to understand the equivalence of aforesaid definitions. Can you please help me to understand these definitions. Thank you for your time.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

We want to show that a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 1 if and only if a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 2.

Suppose that $T_{\lambda}\to T$ with respect to $\textbf{def}$ 1, then we know that $T_{\lambda}x\to Tx$ for each $x\in X$ meaning that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T).$$ Putting this in context, for each $F_x\in \{F_x:\beta(X,Y)\to Y \text{ such that }F_x(T)=Tx, T\in\beta(X,Y)\}$ we have $F_x(T_{\lambda})\to F_x(T)$. So $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2.

On the other hand, if $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2, then we know that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T),$$ for each $x\in X$ and as a result $T_{\lambda}x\to Tx$ for each $x\in X$ and we have convergence with $\textbf{def}$ 1. It is simply unpacking the definitions.

$\endgroup$
3
  • $\begingroup$ Why is it enough to argue with convergence of sequences? With your argument the cocountable topology and the discrete topology were always the same (for both the only converging sequences are eventually constant), which is not true. $\endgroup$ Dec 29, 2021 at 20:25
  • 1
    $\begingroup$ Here $T_{\lambda}$ is a net, two topologies are the same if they have the same converging nets. $\endgroup$
    – esteban
    Dec 29, 2021 at 20:27
  • $\begingroup$ Aahhh, I am a moron. Now, I understand why you picked $\lambda$ as an index. That is of course the way to go (+1). $\endgroup$ Dec 29, 2021 at 20:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .