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My homework: Let $A$ be a set. Endow $\mathbb{C}^A$ with the product topology. Then any continuous linear functional $\Lambda:\mathbb{C}^A\rightarrow\mathbb{C}$ is of the form $\Lambda(f)=\sum_{i=1}^n \alpha_i f(a_i)$ for some complex $\alpha_1,\dotsc,\alpha_n$ and elements $a_1,\dotsc,a_n \in A$.

My effort: I proved it under the assumption that $A$ is countable, but I don't know how to do the general case. Here's an outline of what I did:

  1. Proved that there's a finite subset $B \subset A$ such that $\Lambda(\delta_a)=0$ for any $a \in A\setminus B$.
  2. Proved that convergence in $\mathbb{C}^A$ is equivalent to pointwise convergence.
  3. Used the fact that $\sum_{i=1}^n f(a_i)\delta_{a_i}$ converges pointwise to $f$ as $n\rightarrow \infty$. Then used the continuity and linearity of $f$ and the fact that $B$ is finite.
Gils
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  • How did you prove 1. for the countably infinite case? – Berci Nov 13 '12 at 21:46
  • @Berci: I proved 1. for any set actually. I only used countability in (3). What I did in (1): I took the preimage of the open unit disc in $\mathbb{C}$. It's open because $\Lambda$ is continuous. I found a basic open set in the preimage containing the origin. It only limits a finite number of coordinates. For any unlimited coordinate $a\in A$ we must have $\Lambda(\delta_a)=0$ because scalar multiplication commutes with the action of the linear $\Lambda$. – Gils Nov 13 '12 at 22:18
  • I got it! My line of proof was not the best one probably. Changed it a bit to show that $\Lambda$ vanishes on functions vanishing on $B$. – Gils Nov 14 '12 at 00:33

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