# To find a $2\pi$-periodic function with a property

I recently came across the following question in my research, and I don't know how to proceed this problem.

Question: How to find a function $$g(x)$$ such that it satisfies

(1) $$2\pi$$ periodic

(2) odd

(3) $$g'(x)g'(y)+g(x)g(y)=g'(x-y)$$?

(4) $$\frac{dg(x)}{dx}$$, and $$\int_xg(x)$$ exists and continuous $$\in C^1$$

(5) $$g(x)\neq 0$$

Note: $$g(x)=\sin x$$ is one of the cases.

could we find something else?

What I tried

Since $$g$$ is odd, we assume it can be expanded as $$g(x)=\sum_{n=1}^\infty b_n\sin(nx)$$, Then $$g'(x)=\sum_{n=1}^\infty nb_n\cos(nx)$$. Thus $$g'(x)g'(y)=(\sum_{n=1}^\infty nb_n\cos(nx))(\sum_{n=1}^\infty nb_n\cos(ny))$$ and $$g'(x-y)=(\sum_{n=1}^\infty nb_n\cos(n(x-y)))$$ Then the condition becomes $$(\sum_{n=1}^\infty nb_n\cos(nx))(\sum_{n=1}^\infty nb_n\cos(ny))+ (\sum_{n=1}^\infty b_n\sin(nx))(\sum_{n=1}^\infty b_n\sin(ny))$$$$=(\sum_{n=1}^\infty nb_n\cos(n(x-y)))$$

The Right-hand side $$\sum_{n=1}^\infty nb_n\cos(n(x-y)))=\sum_{n=1}^\infty nb_n(\cos(nx)\cos(ny)+\sin(nx)\sin(ny))$$ But then I am stuck at how to proceed this, and whether this is correct way to go..

• What are you willing to assume about $g$, say $C^1$? Also you presumably want to exclude $g(x) = 0$? Commented Feb 26 at 2:48
• @AlekseiKulikov I hope at least: $\frac{dg(x)}{dx}$, and $\int_xg(x)$ exists and continuous $\in C$ (thus this implies $g\in C^1$). Yes, I want to exclude $g(x)=0$.
– tony
Commented Feb 26 at 2:53
• So, you do assume that $g'(x)$ exists and is continuous? Then I think I can prove that $g$ must be $\sin (x)$. Commented Feb 26 at 2:57
• @AlekseiKulikov yes exactly.. could you show me how, and how did you have this intuition.. Thank you :)
– tony
Commented Feb 26 at 3:02

It is enough to consider the equation for $$x = y$$ only, so the rest of the equation will be useless for us.

First, plugging $$x = y = 0$$ we get $$g'(0)^2 + g(0)^2 = g'(0)$$. Since $$g$$ is odd, $$g(0) = 0$$ so $$g'(0) = 1$$ or $$g'(0) = 0$$. If $$g'(0) = 0$$, then for all $$x$$ we have $$g'(x)^2 + g(x)^2 = 0$$, so $$g(x) = 0$$. So, $$g'(0) = 1$$ from now on.

Now, in some vicinity of $$0$$ we have $$g'(x) > 0$$ since $$g\in C^1$$. Then in this vicinity we have $$g'(x) = \sqrt{1 - g(x)^2}.$$

This is a differential equation, and until $$|g(x)| = 1$$ right-hand side is smooth, so existence and uniqueness theorem applies. Therefore, in the vicinity of $$0$$ our function must coincide with $$\sin(x)$$ (we used that $$g$$ is odd really only to have $$g(0) = 0$$). By the standard extension theorem this can go at least to the singularity of our differential equation, so $$g(x) = \sin(x), |x| \le \frac{\pi}{2}$$.

Note that after that $$g(x)$$ can be equal to $$1$$ for an arbitrarily long time (unless we also assume that $$g\in C^2$$ in which case we don't even need $$2\pi$$-periodicity and I think with some tedious work condition of $$g$$ being odd can also be removed). So, we have to use $$2\pi$$-periodicity somehow. We know that for $$\frac{3\pi}{2} \le x \le 2\pi$$ we have $$g(x) = \sin(x)$$. Consider the interval from $$x_1 =\frac{\pi}{2}$$ to $$x_2 = \frac{3\pi}{2}$$. We have $$g(x_1) = 1$$, $$g(x_2) = -1$$. So, there must exist $$x_1 < x_0 < x_2$$ such that $$g(x_0) = 0$$.

We have $$g'(x_0)^2 = 1$$, so $$g'(x_0) = \pm 1$$. By repeating the argument with differential equation from above (except we have to add an appropriate sign to it) we can see that in the interval $$[x_0 - \frac{\pi}{2}, x_0 + \frac{\pi}{2}]$$ our function must be $$\pm \sin(x-x_0)$$. However, if $$x_0\neq \pi$$ then this interval either contains $$x_1$$ or $$x_2$$ and the value at it would be smaller than $$1$$ in absolute value, which is a contradiction. So, $$x_0 = \pi$$. Finally, if $$g'(x_0) = 1$$ then we would still have contradiction for the value of $$g(x_1)$$, say (it would be $$-1$$ from the differential equation, while we already established that it is $$1$$). Thus, on $$[x_0 - \frac{\pi}{2}, x_0 + \frac{\pi}{2}]$$ we have $$g(x) = -\sin(x-x_0) = \sin(x)$$. So, $$g(x) = \sin(x)$$ on $$[0, \frac{\pi}{2}]$$, on $$[\frac{\pi}{2}, \frac{3\pi}{2}]$$ and on $$[\frac{3\pi}{2}, 2\pi]$$, so it is $$\sin(x)$$ everywhere by $$2\pi$$-periodicity.

As I said, must of the conditions can be likely relaxed or removed, and only the equation for $$x = y$$ was used, but the condition $$g\in C^1$$ does seem to be essential, because otherwise the sign of the derivative can jump and it is not clear to me how to proceed.