TITLE: Counter example: strong convergence does not imply convergence in operator norm
QUESTION [5 upvotes]: Consider a family of operators $T_n\in\mathcal{L}(X)$ where $X$ is a separable Hilbert space. Find examples in which $T_n$ converges strongly to $T\in \mathcal{L}(X)$, i.e.

$\|Tx-T_nx\|\to 0$ as $n\to \infty$ for all $x\in X$,

but not in operator norm topology, i.e.

$\|T-T_n\|_{op}\to 0$ as $n\to \infty$.

REPLY [2 votes]: There are problems with your terminology.
In general, convergence in $\mathcal{L}(X)$ does not imply uniform convergence, i.e. $\|T - T_n\|_{op} \xrightarrow {n\to\infty} 0$ does not imply
$$\forall \varepsilon > 0 \,\,\exists n_0\in\mathbb{N} \text{ such that } \forall x \in X, \forall n \ge n_0 \text{ we have } \|Tx - T_nx\| < \varepsilon $$
However, the following holds:

$\|T - T_n\|_{op} \xrightarrow {n\to\infty} 0$ implies $T_n \xrightarrow{n\to\infty} T$ uniformly on bounded subsets $S \subseteq X$:
Let $M > 0$ be such that $\|y\| \le M, \forall y\in S$. For $x \in S$ we have:
  $$\|Tx - T_nx\| = \|(T - T_n)\,x\| \le \|T - T_n\|_{op}\cdot\|x\| \le \|T - T_n\|_{op}\cdot M \xrightarrow {n\to\infty} 0$$
As a consequence of this, we also have that $T_n \xrightarrow{n\to\infty} T$ poinwise.

What you are actually asking for is an example of a sequence of operators $(T_n)_{n=1}^\infty$ in $\mathcal{L}(X)$ which converges pointwise to $T \in\mathcal{L}(X)$, but not with respect to the operator norm $\|\cdot\|_{op}$, i.e. the sequence $(T_n)_{n=1}^\infty$ does not converge in $\mathcal{L}(X)$.
Edit:
$\|x\| < \infty$ is enough for pointwise convergence:
Let $x \in X$ and $\varepsilon > 0$. Since $\|T_n - T\|_{op} \xrightarrow{n\to\infty} 0$ there exists $n_0 \in \mathbb{N}$ such that $\|T_n - T\|_{op} < \frac{\varepsilon}{\|x\|}$ for $n \ge n_0$. We have:
$$\|Tx - T_nx\| = \|(T - T_n)\,x\| \le \|T - T_n\|_{op}\cdot\|x\| < \varepsilon$$
We need $\|x\| \le M$, $\forall x \in S$ for uniform convergence:
Let $\varepsilon > 0$. Since $\|T_n - T\|_{op} \xrightarrow{n\to\infty} 0$ there exists $n_0 \in \mathbb{N}$ such that $\|T_n - T\|_{op} < \frac{\varepsilon}{M}$ for $n \ge n_0$. For any $x \in S$ we have:
$$\|Tx - T_nx\| = \|(T - T_n)\,x\| \le \|T - T_n\|_{op}\cdot\|x\| \le \|T - T_n\|_{op}\cdot M <\varepsilon$$
Notice that a single $n_0 \in \mathbb{N}$ can be chosen to work for every $x \in S$.