I'm self-studying using Applied Analysis by John Hunter and Bruno Nachtergaele. In chapter 5 on Banach space, the authors defined strong convergence and weak convergence as followed:
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A sequence ($T_{n}$) in $\mathcal{B}(X,Y)$ converges strongly if:
$\lim_{n\to\infty} T_{n}x = Tx$ for every $x \in X$ -
We say that $T_{n}$ converges weakly to $T$ in $\mathcal{B}(X,Y)$ if the pointwise values $T_{n}x$ converge weakly to $Tx$ in Y.
Then they say they will not consider the weak convergence of operators in this book…, but I'm confused between the two. They look quite identical to me. So, what is/are the difference(s) and why the difference(s) is/are important to keep in mind?
Best Answer
Strong operator topology is easier to use because norm estimates are easier to manipulate, but sometimes it is too strong to prove in applications. Some natural sequences of operators in Fourier analysis and quantum theory, for instance, would only converge weakly (see examples below). Wikipedia has a somewhat terse article on strong and weak topologies. Some additional subtleties are discussed under What is the dual space in the strong operator topology? The classical "encyclopedia" on operators in Banach spaces, including dual spaces and operator topologies, is Linear Operators by Dunford and Schwartz.
Formally, the difference is explicit, we have that for all $x$:
Strong: $||T_nx-Tx||\xrightarrow[n\to\infty]{}0$ (not to be confused with uniform convergence $||T_n-T||\xrightarrow[n\to\infty]{}0$)
Weak: $\langle\varphi,T_nx-Tx\rangle\xrightarrow[n\to\infty]{}0$ for any $\varphi$ in the dual space.
Of course, the former implies the latter since $|\langle\varphi,T_nx-Tx\rangle|\leq||\varphi||\,||T_nx-Tx||$, but the converse is false. Let the Banach spaces be $X=Y=l_2$ (square summable sequences), and $e_n$ be the sequence with zeros everywhere except at $n$-th position, where it has $1$ (these form the standard basis in $l_2$). Define $T_n$ to be the operator that acts on a sequence $x\in l_2$ as $T_nx=x_1e_n$. Then $$\langle\varphi,T_nx\rangle=x_1\langle\varphi,e_n\rangle=x_1\varphi_n\xrightarrow[n\to\infty]{}0$$ for any $\varphi\in l_2$ (identifying $l_2$ with its dual space) because $\sum_n|\varphi_n|^2<\infty$. Therefore, $T_n$ converge to $0$ weakly.
But $T_n$ does not converge strongly. Since strong convergence implies weak convergence and limits are unique if it did it would have converged to $0$ also. But $||T_nx||=|x_1|\,||e_n||=|x_1|$ for any $x\in l_2$, which does not converge to $0$ whenever $x_1\neq0$. In fact, $e_n$ are well known to converge to $0$ weakly in $l_2$, but not by norm, this was the idea behind the construction of $T_n$.
Another (more natural) example of an operator sequence converging weakly but not strongly is given by the sequence of powers $S^n$, where $S$ is the right shift operator $S(x_1,x_2,\dots)=(0,x_1,x_2,\dots)$. It is the simplest example of a creation operator, they occur in quantum field theory.