# Proof about Separating Points, Vanishing, and Uniform Closure

• Jul 13th 2010, 03:46 PM
ragnar
Proof about Separating Points, Vanishing, and Uniform Closure
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 $\displaystyle K =$ the unit circle in the complex plane, and $\displaystyle \mathscr{A}$ the algebra of functions of the form $\displaystyle f(e^{i \theta}) = \displaystyle \sum^{N}_{n = 0} c_{n}e^{in \theta}$ for $\displaystyle \theta \in \mathbb{R}$. Show that $\displaystyle A$ separates points on $\displaystyle K$.

So my thinking: I want to take two points $\displaystyle e^{ia} \ne e^{ib}$ and show that there is a function in $\displaystyle A$ such that $\displaystyle f(e^{ia}) \ne f(e^{ib})$. Well, assuming $\displaystyle N \geq M$, then $\displaystyle \displaystyle \sum^{N}_{n = 0} c_{n}e^{ina} = \sum^{M}_{m = 0} c_{m}e^{ima} \Leftrightarrow c_{1}(e^{ia} - e^{ib}) + ... + c_{N - M + 1}e^{i(N - M)a} + ... + c_{N}e^{iNa} = 0$. So I want to show that this last equality fails.
• Jul 13th 2010, 05:10 PM
ragnar
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, $\displaystyle \displaystyle \sum^{N}_{n = 0} c_{n}e^{ina} = \sum^{N}_{m = 0} c_{m}e^{ima} \Leftrightarrow c_{1}(e^{ia} - e^{ib}) + ... + c_{N}e^{iNa} = 0$. Well. We can go ahead and choose arbitrary $\displaystyle c_{2}, ... c_{n}$ so that everything after $\displaystyle c_{1}(e^{ia} - e^{ib})$ is now some constant, call it $\displaystyle c$. Since $\displaystyle e^{ia} \ne e^{ib}$, we can choose $\displaystyle c_{1}$ to be anything such that $\displaystyle c_{1}(e^{ia} - e^{ib}) \ne -c$, thereby failing the desired equality and proving that the values of the functions at the points are unequal.

No?