If $f(x)$ is a differentiable function on the real line and periodic with period $T$ does it have at least two stationary points in $(0,T]$?
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Yes. Consider $g(x)=f(x)-f(x+T/2)$. Since $f$ is differentiable, it is continuous, and hence $g$ is continuous. Now $g(x+T/2)=f(x+T/2)-f(x+T)=f(x+T/2)-f(x)=-g(x)$. Thus there is at least one zero of $g$ by the intermediate value theorem. But $g(x)=0$ implies $f(x)=f(x+T/2)=f(x+T)$, and then by the mean value theorem there are two points $\xi_1,\xi_2$ with $x\lt\xi_1\lt x+T/2$ and $x+T/2\lt\xi_2\lt x+T$ with $f'(\xi_1)=f'(\xi_2)=0$.
Yes. If $f$ is constant, then every point is stationary. Otherwise the image of $[0,T]$ is a closed interval (since $f$ is in particular continuous), and therefore $f$ has at least one global maximum and at least one global minimum. We know that maxima and minima of a differentiable function are stationary points. The maximum and minimum are different points because $f$ is not constant. If one of them are $0$ (and therefore outside $(0,T]$), just consider $T$ instead -- that will also be a global extremum because $f$ is periodic.