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I just want to make sure I am understanding this correctly. I want to show that operator norm convergence implies pointwise convergence.

Suppose we are looking at the space $L(X,Y)$ i.e the space of all linear operator from X to Y. Suppose that $S_m = \Sigma_{n = 1}^{n = m} T_n \rightarrow T$. Then the reason that operator norm convergence implies pointwise convergence is the following:

By definition we have $\|S_m - T\|_{op} \longrightarrow 0$ as $m \longrightarrow \infty$. So by definition $$| \Sigma_{n = 1}^m \ sup_{x \neq 0} \frac{T_n(x)}{\|x\|} - sup_{x \neq 0} \frac{T(x)}{\|x\|}| \longrightarrow 0$$

Thus we have that $$ \big|\Sigma_{n = 1}^m T_n(x) - T(x)| \big| = \big|\Sigma_{n = 1}^m T_n(\frac{x}{\|x\|})\|x\| - T(\frac{x}{\|x\|})\|x\|\big| \longrightarrow 0$$

Thus we are done. Is it good ?

2 Answers2

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Usually $L(X,Y)$ stands for the space of all continuous linear operators between normed spaces $X$ and $Y$. It it well-known that for $A\in L(X,Y)$ we have $\|A(x)\|\le \|A\|\cdot \|x\|$. Hence $\|T_n(x)-T(x)\|\le\|T_n-T\|\cdot \|x\|$. That is why operator convergence $T_n\to T$ implies pointwise convergence $T_n(x)\to T(x)$ for any $x$.

szw1710
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I think you may be mistaken as to what $\|S_m -T \|$ means. Things like $\sup_{x \neq 0} \frac{T_n(x)}{\|x\|}$ aren't well defined, because there's no order on the vector space $Y$. The definition of the norm of operator $S_m-T$ (at least one of the equivalent definitions) is actually $$ \| S_m -T \| = \sup_{x \neq 0} \frac{\| S_m(x) - T(x) \|}{\|x\|}. $$

So for any $x \neq 0$, you would necessarily have $\| S_m -T \| \geq \frac{\| S_m(x) -T(x) \|}{\|x\|}$. In other words $\| S_m( x) -T(x) \| \leq \| x\| \|S_m-T\|$ holds for any $x \in X$. This inequality gives you pointwise convergence if norm convergence holds.

Demophilus
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