# Fourier Analysis - Functions on a Circle

In general, if a continuous function $$g(x)$$ is defined on the interval $$[-\pi,\pi]$$, can I say it is definitely possible to extend this function to be a $$2\pi$$-period function? I think we need to make sure the endpoints match? Functions are $$2\pi$$-period on the real line are also called functions on a circle in Fourier analysis. But what about $$g(x) = x \\on\ [-\pi,\pi]$$

This is clearly continuous, but $$g(-\pi) \ne g(\pi)$$. How can I still extend this function to a continuous $$2\pi$$-period function? Or do I have some conceptual misunderstandings?

Thanks so much for your help.

• Hi Blazej, is that possible we can use change of variables to change the domain from $[-\pi, \pi]$ to $[-\pi/2, \pi/2]$ and then say we can extend this function to a new function $F(x)$ such that in $[-\pi/2, \pi/2]$ $F(x)$ is just equal to the same values and we require $F(x)$ to be continuous and $F(\pi) = F(-\pi)$. Then from there, we extend $F(x)$ to be $2\pi$-periodic? – xf16 Feb 2 '19 at 21:58
• Dear xf16, why do you want to insist on continuity?Given your input on $[-\pi,\pi]$ and assumption of periodicity you can define your function ambigously everywhere, except for $\pi, 3 \pi$ etc. Whatever value you choose at this point, Fourier series will be unchanged. Actually the natural choice would be the arithmetic average of the two values, which is $0$ in your case. Then Fourier series converges pointwise. – Blazej Feb 2 '19 at 22:20
If $$g(-\pi) \ne g(\pi)$$, then you cannot extend to a $$2\pi$$ periodic function because a $$2\pi$$ periodic function $$h$$ on $$\mathbb{R}$$ is required to satisfy $$h(x)=h(x+2\pi)$$ for all $$x$$, and your $$g$$ does not satisfy that for $$x=-\pi$$. You can extend any function defined on $$[-\pi,\pi)$$ or on $$(-\pi,\pi]$$ to a $$2\pi$$ periodic function on $$\mathbb{R}$$, but not if it is defined on $$[-\pi,\pi]$$ with $$h(-\pi)\ne h(\pi)$$.