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This question was asked by myself on the math stackexchange a few days ago. I thought I'd repeat it here:

Let $X$ be a normed, real vector space of uncountable dimension. Let $X^*$ denote the set of all continuous linear functions from $X$ to $\mathbb{R}$. Does there exist a metric $d : X^* \times X^* \to \mathbb{R}$ such that a sequence $(\varphi_n)_{n=1}^\infty$ in $X^*$ converges to $\varphi$ with respect to $d$ if and only if $\varphi_n(x) \to \varphi(x)$ for all $x \in X$?

I believe the answer to this question is no, but I have been unable to show it.

I am aware that "the topology of pointwise convergence" -- the weak* topology -- is not metrizable. However, there exist distinct topologies which agree on their convergent sequences, so it is possible for such a metric to exist while still inducing a different topology from the weak* topology.

I have seen this question which asks a more general question about when the metric topology agrees with the weak topology induced by a family of functions. The answer to this question presents sufficient conditions; I am interested in necessary conditions.

Here's what I know:

  • If $X$ has countable dimension, then there does exist a metric which does this: namely, $d(\varphi, \psi) = \sum_{n = 1}^\infty 2^{-n} \frac{|\varphi(x_n) - \psi(x_n)|}{|\varphi(x_n) - \psi(x_n)|+1}$ where $\{x_n : n \in \mathbb{N}\}$ is a linear basis of $X$.
  • Pointwise convergence on every basis vector implies pointwise convergence everywhere. So, the question may equivalently be stated: "does there exist a metric $d : X^* \times X^* \to \mathbb{R}$ such that $\phi_n \to_d \phi$ if and only if $\phi_n(x_i) \to \phi(x_i)$ for all basis vectors $x_i$?"
  • When $X$ is separable, such a metric exists on the unit ball (with respect to the operator norm) of $X^*$.

I have tried (without success) to reduce the problem to that of the metrizability of the weak* topology by arguing that any metric which satisfies the desired property must in fact induce the weak* topology. In particular, I have tried to show that the family of open sets $S = \left\{ e_x^{-1}(U) : x \in X, U \subseteq \mathbb{R} \text{ is open} \right\}$ is a subbase for the topology induced by $d$ (here, $e_x$ denotes the evaluation map $f \mapsto f(x)$).

I have also tried to directly derive a contradiction by constructing a sequence of functionals which must converge to zero but do not converge pointwise.

Mustafa Motiwala
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    The subtle aspect of the question is that a metric with the same convergent sequences as the weak$^$ topology could still generate a different topology. For a Banach space $X$, there is no metric generating the weak$^$-uniformity: The uniform boundedness principle implies that $(X^,\sigma^)$ is sequentially complete and then $(X^,\sigma^)$ would be a Fréchet space. The open mapping theorem then implies that $\sigma^*$ coinides with the dual norm topology which implies that $X$ is finite dimensional. – Jochen Wengenroth Dec 07 '23 at 17:30
  • Thank you, Jochen, this is very helpful! I don't think I quite understand how the uniform boundedness principle implies completeness, and why the open mapping theorem implies equality of the $\sigma^*$ and dual norm topologies. If we have a Cauchy sequence $\varphi_n$ with respect to a metric $d$ satisfying our condition, how can we ensure that $\sup_{n \in \mathbb{N}} \varphi_n(x)$ exists for all $x \in X$? – Mustafa Motiwala Dec 08 '23 at 01:18
  • If $\varphi_n$ is Cauchy for the $\sigma^*$-uniformity, then $\varphi_n(x)$ is Cauchy in $\mathbb R$ and hence convergent and, in particular, bounded. The limits $\varphi(x)=\lim\limits_{n\to\infty}\varphi_n(x)$ define a linear functional which is continuous because of the uniform boundedness theorem. – Jochen Wengenroth Dec 08 '23 at 13:10
  • Thanks. I was thinking of that, but I couldn't figure out how to make the argument that $\varphi_n(x)$ is still Cauchy, since we don't know how the metric is actually defined. It certainly seems natural, though; I'll see if I can figure it out. Could you explain how the open mapping theorem is applied? – Mustafa Motiwala Dec 08 '23 at 16:53
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    The identity $(X,|\cdot|^) \to (X^,\sigma^)$ is continuous and hence has a continuous inverse if the range is a Fréchet space. Recall that this is a site for research problems. – Jochen Wengenroth Dec 08 '23 at 16:58

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