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 $f(x)=\sin x$? 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 
$f(x)=\sin x , \forall x \in \mathbb R$ ?
 A: Yes, $f(x)=\sin(x)$. The proof below is not entirely elementary. There's probably a more elementary proof, although nobody's come up with one yet; the proof below does tell us a few interesting things about the class of functions satisfying the given condition.
Say $f\in X$ if $f$ is smooth and $$ \Vert f \Vert = \Vert f \Vert _X=\sup_{k\ge0}\sup_{x\in\Bbb R}|f^{(k)}(x)|<\infty.$$ (An example of one of the interesting things mentioned above: We will see below that if $f\in X$ then $ \Vert f \Vert _X= \Vert f \Vert _\infty$.)
As has been noted, if $f\in X$ then Taylor's Theorem with a suitable form of the remainder shows that $f$ is equal to its Maclaurin series on all of $\Bbb R$; thus $f$ is (the restriction to $\Bbb R$ of) an entire function. Simply noting $|\sum a_nz^n|\le\sum|a_n|\,|z|^n$ shows that $|f(z)|\le \Vert f \Vert e^{|z|}$. In particular $|f(iy)|\le \Vert f \Vert e^{|y|}$; now since $X$ is translation-invariant it follows that $$|f(x+iy)|\le \Vert f \Vert e^{|y|}.$$
Functions in $X$ enjoy a certain magical property:

Magical Property If $f\in X$ then $$f'(x)=\sum_{k=-\infty}^\infty
\frac{4(-1)^k}{(2k+1)^2\pi^2}f(x+\pi(k+1/2)).$$

(Hence $ \Vert f' \Vert _\infty\le \Vert f \Vert _\infty$ as noted above.)
The proof of the Magical Property is the not quite so elementary part of the argument (maybe it follows by some elementary complex analysis, I don't see how). First we show how the result follows:
Suppose as in the OP that $ \Vert f \Vert _X\le 1$, $f$ is real-valued on $\Bbb R$, and $f'(0)=1$. Then $$1=f'(0)
=\sum_{k=-\infty}^\infty
\frac{4(-1)^k}{(2k+1)^2\pi^2}f(\pi(k+1/2)).$$Now since $$\sum_{k=-\infty}^\infty
\frac{4}{(2k+1)^2\pi^2}=1$$and $-1\le f\le 1$ we cannot have any cancellation here; we must have $$f(\pi(k+1/2))=(-1)^k=\sin(\pi(k+1/2)).$$
So if we let $$g(z)=\frac{f(z)-\sin(z)}{\cos(z)}$$then $g$ is an entire function. Since $|f(x+iy)|\le \Vert f \Vert e^{|y|}$ it is clear that $g$ is bounded in the complement of the union of small disks centered at $\pi(k+1/2)$. Hence $g$ is bounded and so $g$ is constant. So $$f(z)=\sin(z)+c\cos(z).$$Since $f$ is real-valued on $\Bbb R$ it follows that $c\in\Bbb R$, and now it's easy to show that $|f|\le1$ implies $c=0$.

Proof of the Magical Property
Here we assume the reader is familiar with the basic properties of tempered distributions and their Fourier transforms. If you don't know that stuff probably you shouldn't bother asking about the stuff in this section, it would take a lot of space to explain. Of course if you do know that stuff and something below seems  problematic please say so.
Suppose $f\in X$.
Since $f\in L^\infty(\Bbb R)$, $f$ is a tempered distribution, and as such it has a Fourier transform, that being another tempered distribution. A version of the Paley-Wiener Theorem for tempered distributions, in for example Rudin Functional Analysis, shows that $$supp(\hat f)\subset[-1,1].$$(This special case is easier than the theorem in Rudin; I may add a few words about that if it looks like anyone's reading this.)
The Magical Property is really just an explicit version of Bernstein's inequality. Recall: If $g\in L^p(\Bbb R)$ and $supp(\hat g)\subset[-1,1]$ then $ \Vert g' \Vert _p\le \Vert g \Vert _p$. The proof proceeds by showing that there exists a complex measure $\mu$ with $ \Vert \mu \Vert =1$ and $\hat\mu(\xi)=i\xi$ for $-1\le\xi\le 1$; then considering the Fourier transform shows that $g'=g*\mu$, hence $ \Vert g' \Vert _p\le \Vert g \Vert _p \Vert \mu \Vert = \Vert g \Vert _p$. Modulo technicalities the Magic Property for $f\in X$ follows by the same argument, except we need to say exactly what $\mu$ is.
It's straightforward, if tedious, to verify that $$\sum_{k=-\infty}^\infty\frac{4(-1)^k}{(2k+1)^2\pi^2}e^{i\pi(k+1/2)t}=
\begin{cases}it,&(-1\le t\le 1),
\\i(2-t),&(1\le t\le 3).\end{cases}$$(Simply extend the function on the right to a function of period $4$ and calculate the Fourier coefficients.)
Hence if $$\mu=\sum_{k=-\infty}^\infty\frac{4(-1)^k}{(2k+1)^2\pi^2}\delta_{-\pi(k+1/2)}$$then $$\hat\mu(\xi)=i\xi\quad(-1\le\xi\le 1).$$Since $supp(\hat f)\subset[-1,1]$ this shows that $f'=f*\mu$, which is exactly what the Magic Property asserts.
A Technicality Asserting that $f'=f*\mu$ because $\hat\mu(\xi)=i\xi$ on the support of $\hat f$ is a little glib. The problem is that $\hat f$ is just a distribution, and $\hat\mu$ is not smooth on a neighborhood of $[-1,1]$. (This really is a problem; for example if $f$ is an arbitrary tempered distibution with $\hat f$ supported in $[-1,1]$ the convolution $f*\mu$ need not even exist.)
This is not hard to fix. For $0<\lambda<1$ let $$f_\lambda(x)=f(\lambda x).$$ It's enough to show that $f_\lambda'=f_\lambda*\mu$. This is straightforward, since $\hat\mu(\xi)=i\xi$ on a neighborhood of the support of $\hat f_\lambda$. Again, I may add details if it seems worthwhile.

Amuusing Note We've actually shown this:
Theorem Suppose $f\in L^\infty(\Bbb R)$. TFAE:

*

*$f\in X$.


*$f$ extends to an entire function with $|f(x+iy)|\le ce^{|y|}$.


*$\hat f$ is supported in $[-1,1]$.
(We've shown explicitly that (1) implies (2) and (2) implies (3). To show (3) implies (1), note that (3) implies that $ \Vert f' \Vert _\infty\le \Vert f \Vert _\infty$.)
A: 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 $g_j$.
