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Is $f(x)=\sin(x)$ or $f(x)=\cos(x)$ the only function on $\Bbb R$ such that $f^{(n)}(\Bbb R) \subset [-1,1]$ for all $n=0,1,\ldots$ and $f^{(k)}(0)=1$ for some $k\in \Bbb N$?

I found this question for $k=1$ on Codidact. I originally though that it would be easy to find functions satisfying the criteria, but surprisingly all my attempts have failed.

Note that for $k=1$ it is easy to see that out of the parametric family $f(x)=a\sin(bx)$ only the function with $a=b=1$ satisfies the criteria. I though that there would be a way modify the sine to make it oscillate progressively slower as going away from $0$, but I would always get one of the higher-order derivatives out of the $[-1,1]$ range.

Pavel Kocourek
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    Two comments:
    1. I believe you want to exclude the constant functios.
    2. As you say $f(x)=asin(bx)$ can work, but in many more cases you say. The derivates are always of the form $\pm ab^n sen(bx)$ or $\pm ab^n cos(bx)$. In both cases, if $0 \leq a, b \leq 1$ then $0 \leq ab^n \leq 1$ for all $n$, as well as $cos(bx)$ and $sin(bx)$.
     – Brian Britos Simmari Feb 01 '23 at 16:33
  • Also think about the function: $f(x)=\frac{sen(x)}{x}$ if $x\neq 0$ and $f(0)=1$. – Brian Britos Simmari Feb 01 '23 at 16:34
  • Just to let you know: I'm the one who wrote that question on codidact.I think it was originally inspired by a question here, but I simply don't remember anymore. – Henrik supports the community Feb 01 '23 at 16:35
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    For periodic functions see https://math.stackexchange.com/q/108630/42969. It is also mentioned in a comment (but without proof) that the statement holds without assuming periodicity. – Martin R Feb 01 '23 at 16:37
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    Does this answer your question? https://math.stackexchange.com/q/1863239/42969 – Martin R Feb 01 '23 at 16:39
  • @MartinR Thanks a lot for the reference, it answers it for $k=1$, I can't see now if the other values of $k$ then follow. – Pavel Kocourek Feb 01 '23 at 17:05
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    @Henriksupportsthecommunity Thanks for sharing the question on Codidact, I hope that one day there will be a vibrant community there. – Pavel Kocourek Feb 01 '23 at 17:07

1 Answers1

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It follows from

that $f^{(k-1)}(x) = \sin(x)$, that is the difficult part and answers your question for $k=1$.

The general case now follows by repeated integration: $f^{(k-1)}(x) = \sin(x)$ implies that $$ f^{(k-2)}(x) = -\cos(x) + C $$ for some constant $C$, and $f^{(k-2)}(\Bbb R) \subset [-1,1]$ is only possible if $C=0$.

Continuing in that way we get $f^{(k-3)}(x) = -\sin(x)$ and so on, and finally that $f$ is one of the functions $\sin, \cos, -\sin$, or $-\cos$.

Martin R
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