Mostrando que una función es la función seno

Question

¿Cómo puedo demostrar lo siguiente:

Si$ f: \mathbb{R} \rightarrow \mathbb{C} $ es un$2\pi$ - función periódica de la clase$C^{\infty}$ tal que$f'(0)=1$ y que para cualquier$n\in \mathbb{N}, x\in\mathbb{R}$,$ \vert f^{(n)}(x) \vert \leq 1 $, entonces$f$ es la función seno?

Answer 1

Deje$$\widehat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx.$$ Then, since $ f$ is infinitely differentiable, we have $ \ widehat {f ^ {(k)}} (n) = \ left (en \ 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.

Answer 2

Desde$f$ es$\mathcal{C}^{\infty}$ tenemos$\forall k \in \mathbb{N}$ y$\forall x \in \mathbb{R}$ $$f^{(k)}(x) = \sum_{n\in \mathbb{Z}}{(in)^ka_n e^{inx}}$$ where $ f (x) = \ sum_ {n \ in \ mathbb {Z }} {a_n e ^ {} $ inx}

Así$(in)^ka_n = \frac{1}{2\pi} \int_0^{2\pi}{f^{(k)}(t)e^{-int}dt}$ y desde$|f^{(k)}(x) | \leq 1$ para todos los$x$ obtenemos$|a_n| \leq \frac{1}{n^k}$ para todos los$k$.

Así que si$|n| \ne 1$ #%% luego #%. $a_n = 0$ Da$f'(0) = 1$ y utilizando el hecho de que tanto$ f(x) = a_0 + b \sin(x) $ y$||f||_{\infty}$ son menores que$||f'||_{\infty}$ se obtiene$1$