# Period of derivative is the period of the original function

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.

• This is somewhat badly worded (because what you have already shown is precisely that $f'$ has period $T$). What you want to prove is that if $f$ is periodic, then every period of $f'$ is also a period of $f$; in other words, $f'$ has no additional periods. My answer below does this for you. – Rhys Jun 7 '13 at 14:16
• @Rhys Not really, $f$ is periodic and $T$ is the minimum such that $f(x+T) = f(x)$. That shouldn't mean that $T$ is the minimum period such that $f'(x+T)=f'(x)$. It is clear that for all $x\in I$: $f'(x+2T) = f'(x)$ but that doesn't mean $f'$ is $2T$ periodic, right? – Victor Moore Jun 7 '13 at 14:21
• Your inclusion of the word 'minimum' here is the point; you want to prove that they have the same minimal period (I checked Wikipedia, which calls this the 'prime period'). Strictly, '$f$ has period $T$' usually just means $f(x+T) = f(x)$. It's a minor point, but some functions won't have a prime period. The simplest example is a constant function. – Rhys Jun 7 '13 at 14:28
• @Rhys Thanks, that is clarifier. My question would be: if $f$ is $T$-periodic, then $f'$ is $T$-periodic?, meaning that they have the same prime period $T$. – Victor Moore Jun 7 '13 at 14:33
• Yes, I think that's clearer now (although you're missing a single prime: $T' < T$). In any case, my answer below should suffice; let me know if you think it doesn't! – Rhys Jun 7 '13 at 14:40

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 $f$).
One solution is to note that $f(x)$ has an associated Fourier series, and since the derivative of a sinusoid of any frequency is another sinusoid of the same frequency, we deduce that the Fourier series of the derivative will have all the same sinusoidal terms as the original.