Let $f$ be a periodic function. Prove that there exists a function $F$ such that $F'(x)=f(x)$ and $F$ is periodic

Question

Let $f$ be a periodic function. Prove that there exists a function $F$ such that $F'(x)=f(x)$ and $F$ is periodic.

Let $T\in\mathbb{R}$ be the period of $f$ (so $f(x)=f(x+T) \forall x\in\mathbb{R}$) I know that $\int_x^{x+T}f(t)dt=I$ for some constant $I\in \mathbb{R}$. If $I=0$ then we can take $F(x)=\int_0^xf(t)dt$ and then $\forall x\in\mathbb{R}:F(x+T)-F(x)=I=0 \implies F(x+T)=F(x)$ so $F$ is periodic.

But what if $I\neq0$?

I found that for $g(x)=f(x)+\frac{I}{T}$ I can get that $G(x)=\int_0^xg(t)dt$ that $G$ is periodic but $G'\neq f$. Anyone got any idea? Or a counterexample?

Answer 1

What you want to prove is false. Take $f(x)=1$. Then $f$ is periodic. But every primitive of $f$ is strictly increasing and therefore not periodic.

Answer 2

The assertion is simply false unless $I=0$.

Trivial example: Let $f=1$. Then $f$ is certainly periodic, but if $F'=f$ then $F(t)=t+c$, so $F$ is not periodic.

In fact it's not hard to show that if $F'=f$ and $f$ and $F$ are both periodic then $I=0$.

Answer 3

Prove that there exists a function $F$ such that $F'(x)=f(x)$

This does not hold true in general, even without the periodicity requirement.

Take for example $f(x)$ to be the Dirichlet function (the characteristic function of rationals). It is periodic (since every rational number is a period), but it does not have the Darboux (intermediate value) property, so it cannot be the derivative of another function by Darboux's theorem.

Answer 4

Your statement is false since $f(x)$ may be non-integrable at all e.g. take $f(x)=1$ for $x\in\Bbb Q\cap[0,1)$ and zero else where in $[0,1)$ and expand it with period $1$, but let's make it more precise. If a function is (Reimann) integrable with zero-DC (i.e. $\int_{T}f(x)dx=0$ where $T$ is the period) then such $F(x)=\int_{a}^{x}f(u)du$ exists (with some arbitrary constant $a$ which only affects the phase) and it is periodic since $$F(x+T){=\int_{a}^{x+T}f(u)du\\=\int_{a}^{x}f(u)du+\int_{x}^{x+T}f(u)du\\=\int_{a}^{x}f(u)du+\int_{0}^{T}f(u)du\\=\int_{a}^{x}f(u)du+0\\=\int_{a}^{x}f(u)du\\=F(x)}$$In fact the zero-DC condition is a necessary and sufficient condition.