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I was reading this question posted yesterday, and I learned through browsing linked answers and googling a bit, that the derivative of a periodic function is periodic. But is the converse true? I wasn’t able to find information about this. This is true for $\sin$ and $\cos$, but what about more general functions?

My attempt:

Let $f'(x)$ be a nice (infinitely differentiable) periodic function with period $T$. Then, let us see if $f(x)$ has the same period

$$ f(T) = f(0) + \int_0^Tf'(x) \,\mathrm dx \\ f(2T) = f(T) + \int_T^{2T}f'(x) \,\mathrm dx $$

If we want $f(0)=f(T)=f(2T)$, we must have

$$ \int_0^{2T}f'(x) \,\mathrm dx = 0 $$

Which seems harsh and restrictive and I don’t know what to do further with this condition. So, does periodic derivative imply a periodic function?

Random Math Enthusiast
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NNN
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    How about $f(x) = x$, which is not periodic but whose derivative $f'(x) = 1$ is periodic? I'm leaving this as a comment rather than an answer in case there's some interesting extra condition you could add which guarantees the antiderivative is also periodic. – Chris Grossack Mar 27 '25 at 03:47
  • Awesome. I should have thought simply and come up with your example. – NNN Mar 27 '25 at 03:49
  • @ChrisGrossack can't we say that the antiderivative of a periodic function is also periodic when the antiderivative evaluates out to be $0$ in that period? – Random Math Enthusiast Mar 27 '25 at 03:54
  • @RandomMathEnthusiast -- that sounds like exactly the kind of "interesting extra condition" I was alluding to! See also this question I just found, which shows that when the antiderivative is bounded with periodic derivative it must be periodic too. It does this by making exactly your observation about the definite integral over the period equaling $0$ – Chris Grossack Mar 27 '25 at 03:58
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    Related: An absolutely continuous function $f$ is periodic if and only if its derivative $f'$ is periodic and $\lim_{x\to\infty}f(x)/x=0$. – Brian Moehring Mar 27 '25 at 15:31

2 Answers2

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A periodic derivative doesn't always imply a periodic function

Take the function provided in the comments, for example. $f(x) = x$ is not periodic whereas $f'(x) = 1$ is periodic.

This brings us to the following question

So, how do we ensure that the function is periodic when the derivative is periodic?

The answer is quite simple. All it has to do is to satisfy the condition that the $\int_{x}^{x+T}f'(t). \operatorname{dt} = 0$ , where $T$ is the fundamental period. This is equivalent to saying that $f(x+T) - f(x)= 0$ or, $f(x+T) = f(x)$ which is exactly what we want.

Random Math Enthusiast
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    It's also useful that, given $f'$ is $T$-periodic, this condition only needs to be checked for a single value of $x$, eg $\int_0^T f'(t) \mathrm{d}t = 0$ (with $x = 0$). – ronno Mar 27 '25 at 12:29
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Does periodic derivative imply periodic function?

No.

For a differentiable, non-constant function to be periodic, it must have both increasing and decreasing regions. Its derivative will be positive in the increasing regions and negative in the decreasing regions. It follows, then, that if a function $f$'s derivative exists and is always non-negative (or always non-positive) but is not identically 0, then $f$ is not periodic.

There are lots of integrable, periodic functions other than 0 that are nowhere negative (or nowhere positive). Trivially, all constant functions other than 0, but also, for example, $f(x) = sin(x) + 1$. Any bounded periodic function gives rise to a whole family of periodic functions that do not take both positive and negative values. The indefinite integrals of these functions (where they exist) are examples of non-periodic functions with periodic derivatives, though not the only examples.

John Bollinger
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