# Is this (classical?) exercice missing a hypothesis?

A friend just told me about an exercice he was given quite a few years ago, but he wasn't sure wether he remembered all the hypothesis correctly. Does anybody recognize this?

Let $f$ be a smooth real-valued function on $\Bbb R$ such that for all $n\in\Bbb N,~\sup_{\Bbb R}|f^{(n)}|=1$ and $f(0)=1$, then $f=\cos$.

He wasn't sure wether theose were the exact hypothesis needed for this to work. He noticed that $f$ is automatically a power series with infinite radius of convergence, but wondered wether one ought to impose $f$ to be even, and wether the condition on the suprermums of the derivatives is correct.

NOTE. If you know the hypothesis, please post them as an answer, and if you want to post a solution, please hide the text so that one needs to scroll over it to reveal it, thanks!

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Hum... if this is true, it is quite striking, +1. And I wonder in which spheres this is classical. Not mine. –  1015 Mar 4 '13 at 20:56
@Julien I agree :D I have tried a bit, and tried to prove that $f''(0)=-1$ which would be enough, but I don't make any progress, and doubt I'm on the right path. –  Olivier Bégassat Mar 4 '13 at 21:17
I understand that if you can prove $f''(0)=-1$, then you apply this to $-f''$ to find $f^{(4)}(0)=1$ and so on. But you are probably also assuming $f$ even, in which case the result follows clearly. Otherwise, how do deduce it? –  1015 Mar 4 '13 at 21:27
@Julien I'm not assuming it's even, its just that $f'(0)$ must be equal to $0$ for $f$ not to exceed $1$ in a neighborhood of $0$, and if we can show that $f''(0)=-1$ we would be done. –  Olivier Bégassat Mar 4 '13 at 22:50
@julien : Since $f(0) = 1$, then $f(x) - 1$ has the sign of $f'(0)x$ in a neighborhood of $0$ (but it must remain $\le 0$). –  Joel Cohen Mar 4 '13 at 23:25

If you have a journal subscription, a link to that paper is there. The essence of the approach is to investigate the Fourier transform of $f(x)$. An exponential dampening factor is introduced to make the function $L^1$. –  Christopher A. Wong Mar 4 '13 at 21:47