TITLE: What distinguishes weak and strong convergence of bounded linear operator in Banach spaces?
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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 $M(S^n)$) $^{[1]}$ 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 $\|S^ne_j\|\to 0$.

[1] 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.