# Is $f(x)=\sin(x)$ the unique function satisfying $f'(0)=1$ and $f^{(n)}(\Bbb R)\subset [-1,1]$ for all $n=0,1,\ldots$?

Question.Is there a function $f:\Bbb R \to \Bbb R$ with $f'(0)=1$ and $f^{(n)}(x)\in [-1,1]$ for all $n=0,1,\ldots$ and $x\in \Bbb R$, other than $f(x)=\sin(x)$?

Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.

Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).

It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.

If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)

My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...

Edit: I just realised that the set of solutions can be expanded by adding an $a$ whose absolute value is smaller than $1-k$ (i.e. $a\in[k-1,1-k]$). As that doesn't change the derivative(s), it also doesn't change the argument for why only one function satisfies the additional constraint. I.e. none of these solutions are solutions to the extended problem.

## 1 answer

**This is not a complete answer, but only an attempt for it.**

It should not be surprising that considering the specific form $f(x)=k\sin(jx+y)$ , the condition $f'(0)=1$ fixes value of all the 3 paramters $k,j,y$:

- Since $1$ is the maximum value $f'(x)$ is allowed to attain, it must be that $y=0$ (other other value resulting in identical function $f$ due to its periodicity).
- The condition $jk=1$ together with $|k|\leq 1$ and implies that $k=j=1$. Indeed, otherwise $|j|>1$, and so $f^{(n)}(0)$ would be greater than $1$ in absolute value for some $n$ large.

What I actually find surprising is how hard it seems to find any function $f$ with $f'(x)=1$ and $|f^{(n)}(x)|\leq 1$ for all $n=0,1,\ldots$ and $x\in \Bbb R$. A possible approach to do so is to consider the functions in the form $$ f(x) = \sin(g(x)), $$ where $g(x)$ is some function with $g'(0)=1$. Then the derivatives of $f$ are \begin{align} f'(x) &= g'(x) \sin(g(x)) \\ f''(x) &= g''(x) \sin(g(x)) + [g'(x)]^2 \sin(g(x)) \\ f^{(3)}(x) &= g^{(3)}(x) \sin(g(x)) + 2g''(x)g'(x)^2 \sin(g(x))+ [g'(x)]^3 \sin(g(x)) \\ &= \left[\left(\frac{d}{dx}+g'(x)\right)^2 g'(x) \right] \sin(g(x)),\\ f^{(n)}(x) &= \left[\left(\frac{d}{dx}+g'(x)\right)^{n-1} g'(x) \right] \sin(g(x)), \end{align} where $\left(\frac{d}{dx}+g'(x)\right)^{n-1}$ is first expanded using the binomial formula, and finally applied to $f'(x)$ in the sense that $\left(\frac d {dx}\right)^n f'(x) = f^{(n+1)}(x)$.

Unfortunately, I don't know how to continue from here. I tried to consider functions $g$ with $g'(0)=1$ and whose derivatives would be vanishing as $|x|\to \infty$, but there is always some derivative that happens to be greater than 1.

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