# What distinguishes weak and strong convergence of bounded linear operator in Banach spaces?

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:

• 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?

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.

• Thank you for the example and especially for the meta answer of "why" which is often not discussed. And I think the problem is that the authors haven't discussed dual space when they talk about the weak convergence, but I see the distinction now! May 16, 2017 at 15:39
• @userrandom There was a typo in my definition of $T_n$, I fixed it and added why there is no strong convergence. I also added another example of an operator sequence converging weakly but not strongly. May 16, 2017 at 19:40
• Just want to say that this is very clear explanation, I had a hard time with counterexample so thanks @Conifold Oct 21, 2018 at 23:56

This is not an alternative to Conifold's excellent answer but only a remark.

I like to think at weak convergence of operators as "convergence of the matrix elements". Indeed, if $X=Y$ is a Hilbert space with orthonormal basis $\{e_1, e_2, e_3 \ldots\}$, any linear operator $T$ satisfies $$Tx= \left(\sum_{i=1}^\infty \langle e_j | Te_i\rangle x_i\right)e_j$$ where $x_i=\langle x| e_i\rangle$, so $T$ can be (formally) associated with the infinite matrix $$M=\begin{bmatrix} \langle e_1 | Te_1\rangle & \langle e_1 | Te_2\rangle & \langle e_1 | Te_3\rangle & \ldots \\ \langle e_2 | Te_1\rangle & \langle e_2 | Te_2\rangle & \langle e_2 | Te_3\rangle & \ldots \\ \ldots & \ldots & \ldots &\ldots \\ \end{bmatrix}$$ Weak convergence of $T_n$ to $T$ is equivalent to the convergence of every entry of $M_n$ to the corresponding entry of $M$ together with the fact that $\| T_n\| \le C <\infty$. Link.

As an example, consider the shift operator $$S(x_1, x_2, x_3 \ldots)= ( 0, x_1, x_2, x_3, \ldots),\quad S\colon \ell^2\to \ell^2$$ introduced in Conifold's answer. The associated matrix with respect to the standard orthonormal basis $e_n=(0\ldots 0 ,1,0\ldots)$ (the $1$ is in the $n$th position) is: $$M(S)=\begin{bmatrix} 0 & 0 & 0& \ldots \\ 1&0&0&\ldots \\ 0 & 1 & 0 & \ldots \\ 0& 0&1 &\ldots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$ The $n$th power of $S$ is the operator that shifts $n$ times to the right: $S(x_1, x_2, x_3\ldots) = (0\ldots 0, x_1, x_2\ldots),$ and the associated matrix is $$M(S^n)=\begin{bmatrix} 0 & 0 & 0 & \ldots \\ 0 & 0 & 0 & \ldots \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots \\ 1 & 0 & 0 & \ldots \\ 0 & 1 & 0 & \ldots \\ 0 & 0 & 1 & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$ where the first non-vanishing row is the $n+1$th one. This representation makes it visually clear why the sequence $S^n$ converges weakly but does not converge strongly.

Weak convergence. Each matrix element is $0$ for sufficiently big $n$. Moreover, the operator norm of $S^n$ (equal to the sup of the $\ell^2$ norm of the columns of $M(S^n)$) $^{}$ is $1$ for all $n$, so in particular it is bounded. Therefore $S^n$ converges weakly to $0$.

Failure of strong convergence. For each $e_j$, the norm of $S^ne_j$ equals the $\ell^2$ norm of the $j$the column of $M(S^n)$, so it is $1$ for all $n$. Thus it is not true that $\|S^ne_j\|\to 0$.

 This is false. The sup of the $\ell^2$ norm of the columns is smaller than the actual operator norm and the inequality can be strict. Anyway, in the case at hand the operator norm is clearly $1$ for all $n$, since the matrix is always essentially the same up to the insertion of some vanishing rows, which clearly do not change the norm.