Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, $$\lim_{n\rightarrow \infty} ||T_n-T||=0?$$

share|improve this question
1  
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
1  
@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
show 1 more comment

1 Answer

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 $T$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.