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Let $\mathbb{T}:=\lbrace z\in\mathbb{C}:|z|=1\rbrace$ and $A$ be the subalgebra of $C(\mathbb{T})=C(\mathbb{T};\mathbb{C})$ consisting of the Laurent polynomials $\lbrace\sum_{k=-n}^na_kz^k:a_k\in\mathbb{C}\rbrace$.

I have established that $A$ is dense in $C(\mathbb{T})$. I assume from this that $C(\mathbb{T})$ must be the space of convergent Laurent series? How can I prove that the closure of $A$ is the space of convergent Laurent series?

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  • $\begingroup$ this is a very confusing statement with something incorrect somewhere as on one hand continuous functions on the circle are indeed uniform limits of trigonometric polynomials (Feijer means) but on the other, they are not generally expandable in convergent Laurent series on the circle as the Fourier series (which must be that as easily seen) may not converge etc $\endgroup$
    – Conrad
    Oct 5, 2020 at 16:55

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If $f\in C(\mathbb T),$ does there exist a Laurent series $\sum_{-\infty}^{\infty}a_nz^n$ such that $\sum_{-N}^{N}a_nz^n$ converges to $f$ uniformly on $\mathbb T$ as $N\to \infty?$ If this is your question, then the answer is no. If it were true, then the Laurent series would equal the Fourier series of $f.$ But it is well known there are functions in $C(\mathbb T)$ whose Fourier series diverge at "lots of points".

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  • $\begingroup$ @Conrad But the OP already knows $A$ is dense in $C(\mathbb T).$ $\endgroup$
    – zhw.
    Oct 5, 2020 at 16:56
  • $\begingroup$ good point - still not sure what exactly the OP means - anyway I agree with you that if it is just to prove that $A$ dense in $C(\mathbb T)$ implies $C(\mathbb T)$ consists of functions with convergent Laurent series, that is flat out wrong $\endgroup$
    – Conrad
    Oct 5, 2020 at 17:00

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