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Let me clarify my question. Say $\{T_n\}$ is a sequence of bounded linear operators from $X$ to itself, where $X$ is a Banach Space. There exists a bounded linear operator $T$, s.t., $\lim_{n\rightarrow \infty}T_n(x)=T(x)\qquad\text{for every $x\in X$}.$

Now, under what additional condition will the following convergence hold, $\lim_{n\rightarrow \infty} ||T_n-T||=0?$

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    @azarel You are right. Any other condition? I kind of need to use the convergence in norm to show that $T$ is compact.2012-03-18

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A sufficient condition is that $\{T_n\}$ is a Cauchy sequence with respect to the norm. If this holds, then the completeness of $B(X)$ implies that $T_n$ converge to something in the norm. But since $T_n\to T$ pointwise, it follows that something is in fact $T$.