Characterization of the sinus

Solution 1:

This approach should do the job nicely. $f\in C^{\infty}(\mathbb{R})$ ensures $f\in L^2(-\pi,\pi)$, hence we may assume that

$$f(x) = k_0 + s_1 \sin(x)+c_1 \cos(x)+\sum_{n\geq 2}\left(c_n \cos(nx)+ s_n\sin(nx)\right) \tag{1}$$ holds for any $x\in(-\pi,\pi)$. Since $|f^{(n)}(x)|\leq 1$ we have $\int_{-\pi}^{\pi}f^{(n)}(x)^2\,dx \leq 2\pi$.
On the other hand, by Parseval's theorem, for any $m\geq 1$ we have

$$ \int_{-\pi}^{\pi}f^{(m)}(x)^2\,dx = \pi(s_1^2+c_1^2)+\pi\sum_{n\geq 2}n^{2m}(c_n^2+s_n^2) \tag{2}$$ hence in order that $\int_{-\pi}^{\pi}f^{(m)}(x)^2\,dx \leq 2\pi$ for any $m\geq 1$ we need to have $$ \forall n\geq 2,\qquad c_n=s_n=0.\tag{3} $$ This proves$^{(*)}$ the only functions fulfilling our constraints are functions of the form $f(x)=k_0+s_1 \sin(x)+c_1\cos(x)$. By imposing $f'(0)=1$ we get $s_1=1$ and by further imposing $\left|f(x)\right|\leq 1$ we get $k_0=c_1=0$ as wanted.

$(*)$ On the interval $(-\pi,\pi)$. On the other hand the constraints imply that $f$ is an entire function, hence once $f$ is defined on $(-\pi,\pi)$ it can be prolongated in a unique way to the whole real line.
Or you may apply the same argument over and over, first on $L^2(-\pi,\pi)$, then on $L^2(-2\pi,2\pi)$, then on $L^2(-4\pi,4\pi)$, $\ldots$, concluding that $f(x)=\sin(x)$ over any interval of the form $(-2^N\pi,2^N\pi)$, i.e. on the whole real line.

In the comments below Bob Pego makes an important observation: in order to grant that the Fourier series of $f^{(n)}(x)$ is the formal $n$-th derivative of the Fourier series of $f(x)$ some assumption on the decay of $s_n,c_n$ has to be made. Luckily, the Paley-Wiener theorem grants that they have an exponential decay. Unluckily, we are not really dodging that crucial step.

The idea is freely stolen from the Turàn power sums inequality.

_{Problem solving is really instructing. Before looking at a proof of this fact I was believing it was wrong, and I tried constructing (non-working) counter-examples like $f(x)=\frac{\text{Si}(\text{Si}(\pi x))}{\text{Si}(\pi)}$. After many failed attempts, I convinced myself the claim was correct. Then I saw the proof through Paley-Wiener, and believed it was the most efficient way for proving the statement. The last step back I took was to exhibit this simple proof through Fourier series. Thank you, David Hilbert, because reflexive spaces are a bless.}