Finding the Image of an endomorphism on periodic indefinitely differentiable functions

Question

Let $T$ be a strictly positive real number, and $E$ the set of $T$-periodic functions from $\mathbb{R}$ to $\mathbb{R}$ that are indefinitely differentiable (of class $C^{\infty}$), and $\phi$ an endomorphism of the vector space $(E,+,\cdot)$ such that $\phi(f)=f'$ for all $f\in E$. I have to find the Kernel and Image of $\phi$. I found that $\mathrm{Ker} \, \phi$ is the set of constant functions, however, I'm having trouble finding $\mathrm{Im} \, \phi$. I thought of showing that $\phi$ is surjective and concluding that $\mathrm{Im} \, \phi = E$, but I think this is wrong because if $f'$ is $T$-periodic, that doesn't necessarily imply that $f$ is $T$-periodic. I don't know where to start now, any hints will be welcome.

Answer 1

The image of $\phi$ is the set $\left\{f\in E\,\middle|\,\int_0^Tf(x)\,\mathrm dx=0\right\}$. In fact, if $f\in E$ is such taht $\int_0^Tf(x)\,\mathrm dx=0$ and if you define$$F(t)=\int_0^tf(x)\,\mathrm dx,$$then $F'=f$ and $F\in E$. So, $f\in\operatorname{Im}(\phi)$. And if $f\in\operatorname{Im}(\phi)$, then, if $F\in E$ is such that $F'=f$; then$$\int_0^Tf(x)\,\mathrm dx=F(T)-F(0)=0.$$