Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the space of all bounded continuous real-valued functions of $\mathbb{R}$. I am having trouble understanding how to find the closed subalgebra generated by sine and cosine.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Denote $\mathcal A$ the subalgebra generated by $x\mapsto \cos x$ and $x\mapsto \sin x$. $\mathcal A$ contains all the maps $\cos(nx)$ and $\sin(nx)$ (can be shown by induction and the formulas of $\cos(a+b)$, $\sin(a+b)$) and also the constants (because $\cos^2x+\sin^2x=1$).

If we take a continuous $2\pi$-periodic function then by Stone-Weierstrass theorem it's in the closure of $\operatorname{span}\{1,\cos(kx),\sin(kx),k\in\mathbb N\}$ so the closure of $\mathcal A$ contains all continuous $2\pi$-periodic functions. Conversely, every function in $\mathcal A$ is continuous and $2\pi$-periodic and so is an uniform limit of such functions.

We conclude that the closure of $\mathcal A$ is the space of all continuous $2\pi$-periodic functions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.