# Thread: Convergence of Sequences of Operator and Functionals

1. ## Convergence of Sequences of Operator and Functionals

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

2. Originally Posted by geezeela
Let $\displaystyle {T}_{n}\to T$, where $\displaystyle {T}_{n}\in B(X,Y).$. show that for every $\displaystyle \epsilon > 0$ and every closed ball $\displaystyle K\subset X$ there is an N such that [tex] $\displaystyle \left\|T_nx-Tx\right\| < \epsilon$ for all n > N and $\displaystyle x\in K$
I just did this somewhere. Where's the effort man? How about using the fact that $\displaystyle \|T_n(x)-T(x)\|\leqslant \|T_n-T\|_\text{op}\|x\|$.

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

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

5. Originally Posted by geezeela
What does that op stands for??
Operator norm? Isn't that the norm you're putting on $\displaystyle \mathcal{B}(X,Y)$.