Integrating a nested periodic function

Question

Suppose $g(x)$ is a differentiable, real-valued, periodic function with period $a$ such that for all $u$, $\int_u^{u+a} g(x)dx=0$. Is it true, then, that $\int_u^{u+a}g(x+g(x))dx=0$ for all $u$?

Answer 1

No. Let $f$ be periodic of period $1$ such that $f(x)=x^2-1/3$ for $0\le x<1$. Then $$ \int_0^1f(x)\,dx=0,\text{ but }\int_0^1f(x+f(x))\,dx=\frac{1}{270}\, (339 + 10 \sqrt{21} - 50 \sqrt{57})=0.0271631. $$ Computations done with Mathematica.

The above example is not continuous, but it can be smoothed to provide a $C^1$ example. However, there are lots of other examples, like $f(x)=\sin x+\cos(2\, x)$; $$ \int_{0}^{2\pi}f(x+f(x))\,dx=0.874317\dots $$