Let $f:I\to\mathbb R$ be a differentiable and periodic function with prime/minimum period $T$ (it is $T$-periodic) that is, $f(x+T) = f(x)$ for all $x\in I$. It is clear that
$$
f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{f(x+T+h) – f(x+T)}{h} = f'(x+T),
$$
but how to prove that $f'$ has the same prime/minimum period $T$? I suppose that there exist $\tilde T < T$ such that $f'(x+\tilde T) = f'(x)$ for all $x\in I$ but can't find the way to get a contradiction.
[Math] Period of derivative is the period of the original function
derivativesperiodic functionsreal-analysis
Best Answer
To see what happens, simply integrate (I will use $\tilde{T}$ instead of $T'$, since prime is being used for derivatives): \begin{align*} f'(x+\tilde{T}) &= f'(x)\\ \Rightarrow \int^y f'(x+\tilde{T})\,dx &= \int^y f'(x)\,dx \\ \Rightarrow \int^{y+\tilde{T}} f'(\tilde x)\,d\tilde x &= \int^y f'(x)\,dx \\ \end{align*} where we have substituted $\tilde x = x+\tilde{T}$. So we get $$ f(y+\tilde{T}) = f(y) + C $$ for some constant $C$. But we already know $f$ is periodic, so we must have $C = 0$. Hence $f(y+\tilde{T}) = f(y)$, so $\tilde{T}$ is some integer multiple of $T$ (since by assumption, $T$ is the prime period of $f$).