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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.

Victor Moore
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  • 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 07 '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 07 '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 07 '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 07 '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 07 '13 at 14:40

2 Answers2

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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 $f$).

Rhys
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  • How do you know that $f'$ is integrable? – 19021605 Feb 28 '25 at 04:09
  • Because it's a derivative. – Rhys Mar 15 '25 at 11:59
  • But derivative need not be integrable. https://math.stackexchange.com/questions/257069/what-is-an-example-that-a-function-is-differentiable-but-derivative-is-not-riema – 19021605 Mar 16 '25 at 07:52
  • Obviously that's a terrible outcome, and needs to be fixed (of course a derivative should be integrable, and the fundamental theory of calculus should hold!). Luckily somebody has already done the hard work for us: https://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral. Also, these subtleties hardly ever matter; you can almost always work with a nice-enough class of functions that you don't have to worry about such perverse behaviour. – Rhys Mar 21 '25 at 03:52
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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.

Thus, the derivative must have the same frequency as the original function.

Ben Grossmann
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