# When will weak convergence imply strong convergence?

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. – t.b. Mar 18 '12 at 3:05
@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. – henryforever14 Mar 18 '12 at 4:08
@henryforever14 I guess $T$ is compact? – azarel Mar 18 '12 at 4:18
@azarel why is that? – henryforever14 Mar 18 '12 at 14:29
@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. – azarel Mar 18 '12 at 19:59

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 $T$.