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Suppose that the sequence of operators in a Hilbert space $H$, $\left(T_{n}\right)_{n}$, is Cauchy (with respect to the operator norm) and that there is an operator $L$, such that $Lx=\lim_{n\rightarrow\infty}T_{n}x$, for all $x\in H$ (i.e. the $T_{n}$ converge pointwise to $L$).

How can I prove then, that $\left(T_{n}\right)_{n}$ converges with respect to the operator norm to $L$ (i.e. $\left(T_{n}\right)_{n}$ converges uniformly to $L$)?

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Fix $\varepsilon>0$; there is $N=N(\varepsilon)$ such that if $m,n\geq \varepsilon$, $\lVert T_n-T_m\rVert_{B(H)}\leq \varepsilon$. In particular, for all $n\geq N$ and all $x\in H$, we have $$\forall m\geq N,\lVert T_nx-T_mx\rVert\leq \varepsilon\lVert x\rVert.$$ Take the limit $m\to +\infty$ to get $$\forall n\geq N,\forall x\in H,\lVert T_nx-Lx\rVert\leq \varepsilon\lVert x\rVert.$$ We get $\lVert T_n-L\rVert_{B(H)}\leq \varepsilon$. The fact that $L$ is continuous can be seen taking the particular value of $\varepsilon:=1$.

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But why do we need the fact that $L$ is continuous ? (Or have you just stated that, because it is a nice corollary?) – pink_pyjamas Jul 29 '12 at 15:16
I just stated that, since in fact what we show here is that the space of bounded linear operator endowed with the canonical norm is complete. – Davide Giraudo Jul 29 '12 at 15:21

The space of bounded linear operators under the operator norm is complete. Hence, since $\{T_n\}$ is Cauchy, there exists an operator $T$ such that $T_n \to T$ in operator norm. In particular, $T_n \to T$ pointwise, since $\|T_n x - T x\| \le \|T_n -T\| \|x\|$. But $T_n \to L$ pointwise also, so we must have $T=L$.

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