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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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    What kind of conditions are you looking for? Note for instance, if $\varphi_n$ is a sequence of functionals converging weak$^\ast$ to zero but not in norm then $T_n(x) = \varphi_n(x) \cdot x \to 0$ but of course not $T_n \to 0$ in norm.2012-03-18
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    @t.b. I am pretty aware of the fact that weak convergence is weaker than strong convergence. I need to construct a sequence of finite rank operator to approach a bounded operator. Now I can construct such a sequence that it converges pointwise. But I need it to be strong convergence.2012-03-18
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    @henryforever14 I guess $T$ is compact?2012-03-18
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    @azarel why is that?2012-03-18
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    @henryforever14 If $T_n$ are finite rank operators and $T_n$ converges to $T$ in norm then $T$ must be compact. On the other hand, if you already know that $T$ is compact then in this case weak convergence implies strong convergence.2012-03-18
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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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