TITLE: If $f$ is a smooth real valued function on real line such that $f'(0)=1$ and $|f^{(n)} (x)|$ is uniformly bounded by $1$ , then $f(x)=\sin x$?
QUESTION [26 upvotes]: Let $f : \mathbb R \to \mathbb R$ be a smooth ( infinitely differentiable everywhere ) function such that $f '(0)=1$ 
and $|f^{(n)} (x)| \le 1 , \forall x \in \mathbb R , \forall n \ge 0$ ( as usual denoting $f^{(0)}(x):=f(x)$) ; then is it true that 
$f(x)=\sin x , \forall x \in \mathbb R$ ?

REPLY [2 votes]: Here's my attempt, some of the details are missing. It is likely that an easier solution can be found.
Clarification: As David pointed out, there is no such Paley-Wiener type theorem. I will sketch what I think can be done to fix it in the end.
As David mentioned in the comments, your assumptions imply $f$ is an entire function (of exponential type $1$), and also give the following bound:
$$| f(z) | \le e^{|\Im z |}.$$
Suppose that a Paley-Wiener type theorem implies we can write $f$ as a Fourier transform:
$\begin{equation}f(z) = \int_{-1}^{1} e^{i t z} \, \rm{d} \nu(z), \tag{a}\label{a}\end{equation}$
where $\nu$ is a complex measure on $[-1,1]$. In particular,
$\begin{equation}f^{(n)}(z) = \int_{-1}^{1} (it)^n e^{i t z} \, \rm{d} \nu(z). \tag{b}\label{b}\end{equation}$
Consider odd $n>0$. Notice that the function $(it)^n$ is not continuous on $[-1,1]$, if we identify $-1$ and $1$. To fix it, note that $(it)^n e^{-\frac12 i\pi t}$ is continuous, and we have an absolutely convergent Fourier series,
$$ (it)^n e^{-\frac12 i\pi t} = \sum_{k \in \mathbb{Z}} C_n(k) e^{i \pi k t}, \quad  t\in [-1,1]. \tag{c}\label{c} $$
For example,
$$ it e^{-\frac12 i\pi t} = \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k + 1)^2} e^{i \pi k t}, \quad  t\in [-1,1]. \tag{d}\label{d}$$
Combining $\eqref{a}$, $\eqref{b}$, and $\eqref{d}$, we find
$$\begin{align*}
f^\prime (z) & = & \int_{-1}^1 i t e^{izt} \, \rm{d} \nu(t)
= \int_{-1}^1 i t e^{-\frac12 i \pi t} e^{i(z + \frac12 \pi)t} \, \rm{d} \nu(t)\\
&=& \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k+1)^2} \int_{-1}^1 e^{i t (z + \pi k + \frac12 \pi)} \, \rm{d} \nu(t) \\
&=& \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k+1)^2} \, f\left(z + \pi (k+\frac12)\right).
\end{align*}$$
In particular,
$$
1 = f^\prime(0) = \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{(-1)^k}{(2k+1)^2} \, f\left(\pi (k+\frac12)\right).
$$
Since $\frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{1}{(2k+1)^2} = 1$, and $|f|\le 1$, we find that $f\left(\pi (k+\frac12)\right) = (-1)^k$.
Notice also that $f^{\prime\prime}(0) = 0$, since otherwise this would contradict $|f^\prime(z)|\le 1$, for $z$ close to $0$.
Now (for odd $n>0$), using $\eqref{b}$, and $\eqref{c}$, we get in a similar way to the above
$$f^{(n)}(0) = \sum_{k\in\mathbb{Z}} C_n(k) \, f\left(\pi (k+\frac12)\right) = \sum_{k\in\mathbb{Z}} C_n(k) \, (-1)^k = i^n \cdot e^{-\frac12 \pi i} = (-1)^{\frac12 (n-1)}.$$
So for example, $f^{(3)}(0) = -1$, and this implies $f^{(4)}(0)=0$, and so on. Since $f$ is analytic and has the same derivatives as $\sin(z)$ at $z=0$, we find that $\sin(z) + c $ is the only function satisfying the requirements, for some real constant $c$. But $c=0$ is the only possibility, since $|f| \le 1$.

One can prove that $f$ can be approximated uniformly on compact sets by a sequence
$$\begin{equation}
g_j(z) = \int_{-1}^{1} e^{i t z} \, \rm{d} \nu_j(z),
\end{equation}$$
where $\nu_j$ are complex measures on $[-1,1]$. See Boas, Entire Functions, Theorem 6.8.14. We have to prove by induction that the first $n$ derivatives of $f$ and $\sin(z)$ at $z=0$ are the same. Then we need to repeat the above arguments, using $\varepsilon, \delta$ methods whenever we had equalities before. Since we have uniform convergence of entire functions, we can approximate (a fixed number of) the derivatives of $f$ on any compact set using the functions $g_j$.