Counter example: strong convergence does not imply convergence in operator norm 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 $n\to \infty$.

 A: Consider, in $X=l^2 = \lbrace (x_n) \in \mathbb{R}^\mathbb{N}, \sum x_n^2 < \infty \rbrace$.
Let $T_n (x)=(0,\dots,0,x_n,0,\dots)$ and $T\equiv 0$.
$\forall x \in X, \|T_n(x)-T(x) \|=\|(0,\dots,0,x_n,0,\dots)\|=|x_n| \to 0$ since $ \sum x_n^2 < \infty$.
But : $$\|T_n-T\|_{op}=\|T_n\|_{op}=1.$$
A: Let $\{e_n\}$ be an orthonormal basis for $X$, and let $P_n$ be the projections onto the subspace spanned by $e_1,\ldots,e_n$, that is,
$$P_nx=\langle x,e_1\rangle e_1+\cdots+\langle x,e_n\rangle e_n.$$
Then $P_nx\to x$ as $n\to\infty$ for all $x\in X$, i.e., $\{P_n\}$ converges strongly to the identity operator $I$.  But for each $n$, we have 
$$\|P_n-I\|\geq\|(P_n-I)e_{n+1}\|=\|e_{n+1}\|=1$$
and therefore $\{P_n\}$ does not converge to $I$ in norm.
A: 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 $x \in S$.
