# Showing that a function is the sine function

How can I prove the following:

If $f: \mathbb{R} \rightarrow \mathbb{C}$ is a $2\pi$-periodic function of class $C^{\infty}$ such that $f'(0)=1$ and that for any $n\in \mathbb{N}, x\in\mathbb{R}$, $\vert f^{(n)}(x) \vert \leq 1$, then $f$ is the sine function?

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Perhaps consider its Fourier series and show that no coefficient for shorter periods than $2\pi$ can be nonzero? There are some useful estimates here. – Henning Makholm Feb 12 '12 at 21:17
You could also remove the asumption "$2\pi$ periodic" and add "bounded" and the result is the same. The proof will be more elaborate though. – Nathan Portland Feb 13 '12 at 1:33

Let $$\widehat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx.$$ Then, since $f$ is infinitely differentiable, we have $\widehat{f^{(k)}}(n)=\left(in\right)^{k}\widehat{f}(n).$ Parsevals theorem tells us that $$\sum_{n=-\infty}^{\infty}\left|\widehat{f}(n)\right|^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f(x)\right|^{2}dx,$$ and hence combining the fact that $|f^{(k)}(x)|\leq 1$ along with the formula for $\widehat{f^{(k)}}$ in terms of $\hat{f}(k)$ we have $$\sum_{n=-\infty}^{\infty}n^{2k}\left|\widehat{f}(n)\right|^{2}\leq1.$$ Taking $k$ to infinity implies the only non zero terms are $n=-1,0,1,$ and we have that $$f(x)=\widehat{f}(1)e^{ix}+\hat{f}(0)+\hat{f}(-1)e^{-ix}.$$ From here we just have to play around with the other givens until it works out. Since $f^'(0)=1$, and $f(x)\leq 1$, we can argue that $\hat{f}(0)=0$. This then allows us to solve for $f$.
Since $f$ is $\mathcal{C}^{\infty}$ we have $\forall k \in \mathbb{N}$ and $\forall x \in \mathbb{R}$ $$f^{(k)}(x) = \sum_{n\in \mathbb{Z}}{(in)^ka_n e^{inx}}$$ where $f(x) =\sum_{n\in \mathbb{Z}}{a_n e^{inx}}$
So $(in)^ka_n = \frac{1}{2\pi} \int_0^{2\pi}{f^{(k)}(t)e^{-int}dt}$ and since $|f^{(k)}(x) | \leq 1$ for all $x$ we get $|a_n| \leq \frac{1}{n^k}$ for all $k$.
So if $|n| \ne 1$ then $a_n = 0$ . $f'(0) = 1$ gives $f(x) = a_0 + b \sin(x)$ and using the fact that both $||f||_{\infty}$ and$||f'||_{\infty}$ are smaller than $1$ you get $f = \sin$