Proof check: Prove if $f'$ is periodic, then $f$ is periodic

Question

The proof $f(x + p) = f(x) \Rightarrow f'(x + p) = f'(x)$ can be done by simply applying the definition of the derivative and then using $f(x + p) = f(x)$. I am trying to prove the statement in the other direction, also by using the definition of the derivative.

Prove $f'(x + p) = f'(x) \Rightarrow f(x + p) = f(x)$

If $f'(x + p) = f'(x)$, then by the definition of the derivative

$\lim_{c\to0}\dfrac{f(x+p+c)-f(x+p)}{c}=\lim_{c\to0}\dfrac{f(x+c)-f(x)}{c}$ (1)

$\lim_{c\to0}\dfrac{f(x+p+c)-f(x+c)+f(x) - f(x+p)}{c}=0$ (2)

If $f(x) - f(x+p) = \lambda \ne \lambda(c) \ne 0$ then

$\lim_{c\to0}\dfrac{f(x)-f(x+p)}{c} = \pm\infty$ (3)

(3) inserted into (2) leads to

$\lim_{c\to0}\dfrac{f(x+p+c)-f(x+c)}{c} \pm \infty =0$ (4)

Whatever the result of the limit in (4), the left hand side will either be indetermined or it will invalidate (2), and therefore (1). So, $\lambda = 0 = f(x) - f(x+p) \Rightarrow f(x + p) = f(x)$.

I believe there is an error in the proof, is there one?

Answer 1

Let $f(t)=t$. Then $f'$ is periodic but $f$ is not.

Answer 2

The claim is false. For an example, consider $f(t)=\cos(t)+t$.

Answer 3

First, there's an easy counter example:$f(x)= \sin x + x$. And if you try your proof with this function, you will find what's wrong with your proof, that you assume (4) doesn't matter whatever it is. Two divergent limit might converge if they are summed.

Answer 4

Whatever the result of the limit in (4), the left hand side will either be indetermined or it will invalidate (2), and therefore (1). So, $\lambda = 0$ ... .

The conclusion that $\lambda = 0$ doesn't follow. It's true that if $\lambda \neq 0$ then (4) takes on the form $\pm\infty \pm\infty = 0,$ but so what? The point of calling something an indeterminate form is that you cannot draw conclusions from equations in which it appears.