Let $f$ be bounded and Riemann integrable function on any bounded interval and $$F(x)=\int_0^xf(x)\,dx$$
And let $F$ and $f$ both be periodic functions Where $a$ is the period of $F$. Does it imply that the $a$ is also the period (not necessary the lowest) of $f$
To show it I thought that:
Meanwhile the conditions let us to apply the Fundamental theorem of calculus
We have $$(F(x)-F(x+a))'= f(x)-f(x+a)=0$$
$$f(x)=f(x+a)$$
Is it a right reasoning?
Best Answer
Basing on the comments it seems natural to assume that $f$ is continuous. What follows is not a direct answer to the OP question, but is closely related to it, in my opinion.
Let $F$ and $f$ be periodic, where $F$ is an antiderivative of $f.$ With no loss of generality we may assume that $F(0)=0.$ Let $a$ be the minimal period of $F.$ We will restrict to the nontrivial case $a>0.$ Then $f$ has period $a.$ The question arises, whether $a$ is the minimal period of $f.$ Assume by contradicton that $b$ is the minimal period of $f$ and $0<b<a.$ Then $a=nb$ for a natural number $n\ge 2.$
We have $$\displaylines{0=F(0)=F(a)=\int\limits_0^af(t)\,dt = \sum_{k=1}^n \int\limits_{(k-1)b}^{kb}f(t)\,dt\\ =\sum_{k=1}^n \int\limits_{0}^{b}f(t)\,dt=n\int\limits_{0}^{b}f(t)\,dt}$$ Therefore $$\int\limits_0^bf(t)\,dt =0$$ As $f$ has period $b$ we get $$0=\int\limits_0^bf(t)\,dt=\int\limits_x^{x+b}f(t)\,dt =F(x+b)-F(x)$$ which implies that $F$ has period $b,$ a contradiction.