Proof about Separating Points, Vanishing, and Uniform Closure

**ragnar** · July 13th 2010, 03:46 PM

I'm wondering if I'm getting stumped here because of my limited familiarity with complex numbers. So if anybody sees what's going on, if you could answer with a mild hint/nudge rather than a full answer, I would appreciate it.

The problem:

Let $K =$ the unit circle in the complex plane, and $\mathscr{A}$ the algebra of functions of the form $f(e^{i \theta}) = \displaystyle \sum^{N}_{n = 0} c_{n}e^{in \theta}$ for $\theta \in \mathbb{R}$ . Show that $A$ separates points on $K$ .

So my thinking: I want to take two points $e^{ia} \ne e^{ib}$ and show that there is a function in $A$ such that $f(e^{ia}) \ne f(e^{ib})$ . Well, assuming $N \geq M$ , then $\displaystyle \sum^{N}_{n = 0} c_{n}e^{ina} = \sum^{M}_{m = 0} c_{m}e^{ima} \Leftrightarrow c_{1}(e^{ia} - e^{ib}) + ... + c_{N - M + 1}e^{i(N - M)a} + ... + c_{N}e^{iNa} = 0$ . So I want to show that this last equality fails.

**ragnar** · July 13th 2010, 05:10 PM

Actually, I think I see what I was missing: Since I am keeping the function fixed, what I want is to violate the last equality, $\displaystyle \sum^{N}_{n = 0} c_{n}e^{ina} = \sum^{N}_{m = 0} c_{m}e^{ima} \Leftrightarrow c_{1}(e^{ia} - e^{ib}) + ... + c_{N}e^{iNa} = 0$ . Well. We can go ahead and choose arbitrary $c_{2}, ... c_{n}$ so that everything after $c_{1}(e^{ia} - e^{ib})$ is now some constant, call it $c$ . Since $e^{ia} \ne e^{ib}$ , we can choose $c_{1}$ to be anything such that $c_{1}(e^{ia} - e^{ib}) \ne -c$ , thereby failing the desired equality and proving that the values of the functions at the points are unequal.

No?

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