Surjectivity of a an endomorphism such that $\varphi\left(f\right) = f \circ g -f$ Let $E$ the set of functions $f$ from $f :\left[0;1\right] \rightarrow\mathbb{R}$ and for all $x \in \left[0;1\right]$, there exists $C \geq 0$ such that $\left|f\left(x\right)\right| \leq C \left|x\right|$. Let $\varphi$ the endomorphism such that for all $f \in E$:
$$
\varphi\left(f\right) = f \circ g-f
$$
where $g : x \mapsto \frac{x}{2}$.
I need to show that $\varphi$ is an automorphism. I've shown that
$$
\varphi\left(f\right) = 0 \Rightarrow\forall x \in \left[0;1\right], \ f\left(\frac{x}{2}\right)=f\left(x\right) \Rightarrow \forall n \in \mathbb{N}, \ f\left(\frac{x}{2^n}\right)=f\left(x\right)
$$
And
$$
\left|f\left(\frac{x}{2^n}\right)\right| \leq \frac{x}{2^n} \underset{n \rightarrow +\infty}{\rightarrow}0
$$
Hence
$$
\left|f\left(x\right)\right|=0
$$
Hence $f=0$ and I have $\varphi$ injective. However, I'm stuck with surjectivity. I need to prove that for all $h \in E$, I can find $f$ such that
$$
\varphi\left(f\right) = f \circ g-f=h
$$
which means that for all $x \in \left[0;1\right]$, it should fulfill
$$
f\left(\frac{x}{2}\right)-f\left(x\right)=h\left(x\right)
$$
All my ideas are gone since we dont know about $f$ being differentiable. I've tried to express $f$ as a function of $h$ but in vain. Any hint would be much appreciated
 A: Hint(s): You've shown that we have:
$$\begin{align*} \forall x \in [0,1],\,\forall n \in \mathbb{N},\quad\left|f\left(\frac{x}{2^n}\right)\right| \leq\, C\frac{x}{2^n} \xrightarrow[n \to \infty]{} 0 \tag{1}\\
{}\end{align*}$$
On the other hand, the relation $h(x) = \displaystyle f\left(\frac{x}{2}\right) - f(x)$ is true for all $x \in [0,1]$, so for example, say, $\displaystyle \frac{x}{2}$...
${}$
${}$
${}$
Full solution below (tried to find a way to spoiler tag all of it and failed, so try to not read the following if you want to only use the hints at first):
First, let's find a candidate using our for now formal equation $h(x) = \displaystyle f\left(\frac{x}{2}\right) - f(x)$ with some telescoping, for all $n \in \mathbb{N}^*$:
$$\sum_{k=0}^{n-1} h\left(\frac{x}{2^k}\right) = \sum_{k=0}^{n-1} \left(f\left(\frac{x}{2^{k+1}}\right) - f\left(\frac{x}{2^k}\right)\right) = f\left(\frac{x}{2^n}\right) - f(x) \xrightarrow[n \to \infty]{} -f(x)$$
This gives us a possible candidate $x \mapsto \displaystyle -\sum_{n=0}^{\infty} h\left(\frac{x}{2^n}\right)$ for our $f$. Let's check that this function is well-defined, belongs to $E$, and does satisfy $\varphi(f) = h$.

*

*Let $C \geq 0$ be such that $|h(x)| \leq Cx$ for all $x \in [0,1]$ ($C$ exists since $h \in E$).
$h$ satsifies $(1)$ for that $C$, and thus, since the series $\displaystyle\sum_{n\geq 0} \frac{x}{2^n}$ converges, so is $\displaystyle\sum_{n \geq 0} \left|h\left(\frac{x}{2^n}\right)\right|$, which grants that $f$ is well-defined on $[0,1]$ since absolute convergence implies convergence for real-valued series.

*Using $(1)$ in more detail, we have, since $(1)$ holds for all $n$:
$$\forall x \in [0,1],\quad|f(x)| = \left|-\sum_{n \geq 0} h\left(\frac{x}{2^n}\right)\right| \leq \sum_{n \geq 0} \left|h\left(\frac{x}{2^n}\right)\right| \leq \sum_{n\geq 0} C\frac{x}{2^n} = Cx \cdot \sum_{n \geq 0} \frac{1}{2^n} = 2Cx$$
Therefore, $f$ indeed belongs in $E$.

*There only remains one thing to check. Let $x \in [0,1]$:
$$\varphi(f)(x) = -\sum_{n \geq 0} h\left(\frac{\frac{x}{2}}{2^n}\right) - \left(-\sum_{n \geq 0} h\left(\frac{x}{2^n}\right)\right) = h(x) + \sum_{n \geq 1} h\left(\frac{x}{2^n}\right) - \sum_{n \geq 1} h\left(\frac{x}{2^n}\right) = h(x)$$
Hence $\varphi : E \to E$ is indeed surjective.
