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Suppose $X$ is a Banach space. What is the relationship between topology of pointwise convergence on $X$ and weak topology on $X$? Are they the same?

Recall that weak topology on $X$ is the coarsest topology such that every bounded linear functional on $X$ remains continuous.

Topology of pointwise convergence can be found here.

The motivation of the question comes from here.

In the paper given in the answer, the theorem requires a closed unit ball to be compact for weak topology. But after we have shown that the closed unit ball is compact for topology of pointwise convergence, we can apply the theorem. Why is this the case?

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Idonknow
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  • How do you define the topology of pointwise convergence (I assume of linear functionals, correct?). That is usually defined as the weak topology (on the dual of $X$) – Prahlad Vaidyanathan Nov 30 '15 at 16:08
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    related this http://math.stackexchange.com/questions/1455223/weak-topology-and-the-topology-of-pointwise-convergence and this http://math.stackexchange.com/questions/293004/what-is-the-topology-of-point-wise-convergence/293016#293016 (search MSE for pointwise convergence and weak topology, copied from the title of your question) – Mirko Nov 30 '15 at 16:11
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    In the link, topology of pointwise convergence is defined for a family of functions from a set $X$ to a space $Y$, but here you just have a space $X$. So how is the topology of pointwise convergence defined here? – This Is Me Dec 04 '15 at 06:17
  • @ThisIsMe: Actually this is also my concern. In this paper(http://kaltonmemorial.missouri.edu/docs/sm2003c.pdf), the author claimed that 'The closed unit ball of the space $Lip_0(X)$ is compact for the topology of pointwise convergence on $X$'. That's why I confused about topology of pointwise convergence and weak topology. – Idonknow Dec 04 '15 at 06:21
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    But $Lip_0(X)$ is a space of functions, so it makes sense to talk about this topology here. A sequence (or net) of functions $f_i$ in this unit ball converges to an $f$ iff $f_i(x)$ converges to $f(x)$ for every $x\in X$. That's I think what is meant with 'topology of pointwise convergence on $X$' there. – This Is Me Dec 04 '15 at 06:25
  • Okay, I understand now. But how is this topology of pointwise convergence the same as weak topology on the set of evaluation functionals on $Lip_0(X)$? I thought the weak topology on set of evaluation functionals gives us for all $f \in Lip_0(X)$, $f_i(x) \rightarrow f(x)$? – Idonknow Dec 04 '15 at 06:30
  • Weak topology here would mean that for $f_i,f\in Lip_0(X)$ we have $f_i$ converges to $f$ weakly iff $\varphi(f_i)\rightarrow\varphi(f)$ for all $\varphi$ in the dual of $Lip_0(X)$ (i.e. the continuous linear functionals). Note that the evaluation functionals $\varphi_x(f):=f(x)$ are examples, so you already get that weak convergence implies pointwise convergence. If you can show (I'm not immediately sure it is true) that these are are the only linear functionals, you would get equivalence. (Otherwise I think they should actually not be equivalent.) – This Is Me Dec 04 '15 at 21:41

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In the context you're talking about, the "weak topology" is not the weak topology of the Banach space (i.e., the coarsest topology making all bounded functionals continuous), but rather the topology defined by a specific set $E$ of functionals (i.e., the coarsest topology making every element of $E$ continuous, also known as "the weak topology with respect to $E$"). In this case, the set $E$ is just the set of all evaluations at points, so by definition the weak topology is the same as the topology of pointwise convergence.

For a general Banach space, "topology of pointwise convergence" is meaningless (what are the "points" you're evaluating at?) and even if it is meaningful, it usually is not the same as the weak topology.

Eric Wofsey
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