Integral to periodic function.

Question

I have this question. I would like to help me with this problem please . If $f'(x)$ is a periodic function, with period $a$, prove that $f(x)$ is a periodic function, if and only if $f(a)=f(0)$. I appreciate your help.

Answer 1

if $f(x)$ is periodic with period $a$ then $f(x) = f(x+a)$ for all x. $f(0) = f(a)$ and $f'(x) = \lim_\limits{x\to 0} \frac {f(x+h) + f(x)}{h} = \lim_\limits{x\to 0} \frac {f(x+a + h) + f(x+a)}{h} = f'(x+a)$

To go the other direction.

It is a necessary condition that $f(0) = f(a)$ for $f(x)$ to be periodic.

but is $f'(x)$ periodic and $f(0) = f(a)$ sufficient?

$f(x+a) - f(x) = \int_x^{x+a} f'(x) dx$

Given that $f'(x)$ is periodic $\int_x^{x+a} f'(x) dx$ is constant.

$f(0) - f(a) \implies\int_0^{a} f'(x) dx = 0\\ f(x+a) = f(x)$

Answer 2

\begin{equation} \int_0^xf^\prime(t)\,dt=\int_a^{a+x}f^\prime(t)\,dt \end{equation}

\begin{equation} f(x)-f(0)=f(a+x)-f(a) \end{equation}

So

\begin{equation} f(x)=f(x+a)+f(0)-f(a) \end{equation}