Smooth function with derivatives of any order bounded

Question

Question
Let $f\in C^\infty(\mathbb R)$. It is given that $\forall x\in \mathbb R\ \forall n\in\mathbb Z_{\ge\color{red}0}$, $|f^{(n)}(x)|\le 1$ and that $f'(0)=1$. Is $f(x)\equiv\sin x$ the unique solution? (Note that $|f(x)|\le 1$ when we choose $n=0$)

Currently, I thought of two possible ways to solve this problem. One is by Taylor expansion and the other is Fourier expansion. It can be shown that the bounded derivative of any order and $C^\infty$ should leads to that the function is analytic as the remainder of Taylor series vanishes. Then, $f(x)=x+\sum_{n=2}^\infty a_nx^n$. Consider auxiliary function $g(x)=x-\sin x+\sum_{n=2}^\infty a_nx^n$, it should vanish everywhere (or at least in $(a,b)$ since it is proved that $f\in C^\omega$). $$g(x)=\sum_{n=2}^\infty\frac{f^{(n)}(0)-\Im i^n}{n!}x^n$$I have no idea how to prove $g$ vanishes.
I think it does not have much possibility to solve this problem via Fourier series as $f'(0)=1$ leads to an equation involves infinitely many variables.

Answer 1

How did you get $f(0)=0$? Anyways, the answer is negative.

Let $f$ be a solution with $f(0)=0$ and $f'(0)=1$. For any $|r|\leq1$, consider $h(x):=f(rx)/r=x+\sum_{n=2}^{\infty}a_nr^{n-1}x^n$. It isn't $\sin(x)$ if $r\neq1$ and still $|h^{(n)}(x)|=|r^{n-1}f^{n}(rx)|\leq1$

Edit：as pointed out in the comments，my proof didn‘t check the case $n=0$. Sorry for that.