If a sequence of bounded linear operator $T_n$ converges to $T$, is it true that $\lim\|T_n\|=\|T\|$? If a sequence of bounded linear operator $T_n$ converges to $T$, is it true that $\lim\|T_n\|=\|T\|$? Can anyone please explain?
 A: You have, by the (reverse) triangle inequality, 
$$
|\,\|T\|-\|T_n\|\,|\leq\|T-T_n\|.
$$

The reverse triangle inequality:
By the triangle inequality, 
$$
\|A\|=\|A-B+B\|\leq\|A-B\|+\|B\|.
$$
So
$$
\|A\|-\|B\|\leq\|A-B\|.
$$
Reversing the roles,
$$
-(\|A\|-\|B\|)\leq\|A-B\|.
$$
From the two equations together, 
$$
|\,\|A\|-\|B\|\,|\leq\|A-B\|.
$$
A: One way to think about this question is: you are asking if the norm function is a continuous function of the operator $T$.    So to begin with, you should carefully specify what topology on the space of operators you are using.  If you are using the topology induced by operator norm, then your question comes down to asking: if $X$ is a metric space with metric $d$, and $x_0$ is some fixed point of $X$, then is the function $x \mapsto d(x,x_0)$ a continuous function on $X$?  The answer is yes.  (In fact the
function $d: X \times X \to\mathbb R$ is continuous.)
(In the application, $X$ will be the space of linear operators, $0$ will be the base-point $x_0$, and $d$ will be the metric $d(T_1,T_2) := || T_1 - T_2||$.)
A: As others have mentioned, the answer is yes if "converge" is interpreted as "converge in the operator norm topology".  On the other hand, if "converge" is in the strong or weak operator topology, the answer can be no.  For example, let $\cal H$ be an infinite-dimensional separable Hilbert space with orthonormal basis $e_n$, $n \in \mathbb N$, and let $P_n$ be the orthogonal projection on the closed linear span of $e_n, e_{n+1}, \ldots$.  Then $P_n \to 0$ in the strong operator topology, i.e. $P_n x \to 0$ as $n \to \infty$ for any $x \in \cal H$, but $\|P_n\| = 1$ while $\|0\| = 0$.
