# Convergence of Sequences of Operator and Functionals

• June 5th 2011, 06:52 PM
geezeela
Convergence of Sequences of Operator and Functionals
Let ${T}_{n}\to T$, where ${T}_{n}\in B(X,Y).$. show that for every $\epsilon > 0$ and every closed ball $K\subset X$ there is an N such that [tex] $\left\|T_nx-Tx\right\| < \epsilon$ for all n > N and $x\in K$
• June 5th 2011, 07:16 PM
Drexel28
Quote:

Originally Posted by geezeela
Let ${T}_{n}\to T$, where ${T}_{n}\in B(X,Y).$. show that for every $\epsilon > 0$ and every closed ball $K\subset X$ there is an N such that [tex] $\left\|T_nx-Tx\right\| < \epsilon$ for all n > N and $x\in K$

I just did this somewhere. Where's the effort man? How about using the fact that $\|T_n(x)-T(x)\|\leqslant \|T_n-T\|_\text{op}\|x\|$.
• June 5th 2011, 07:41 PM
geezeela
Quote:

Originally Posted by Drexel28
I just did this somewhere. Where's the effort man? How about using the fact that $\|T_n(x)-T(x)\|\leqslant \|T_n-T\|_\text{op}\|x\|$.

what does the {op} stands for?..
• June 5th 2011, 07:42 PM
geezeela
Quote:

Originally Posted by Drexel28
I just did this somewhere. Where's the effort man? How about using the fact that $\|T_n(x)-T(x)\|\leqslant \|T_n-T\|_\text{op}\|x\|$.

What does that op stands for??
• June 5th 2011, 07:43 PM
Drexel28
Quote:

Originally Posted by geezeela
What does that op stands for??

Operator norm? Isn't that the norm you're putting on $\mathcal{B}(X,Y)$.