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I was inspired by this question: it is quite easy to prove that for any positive odd number $2m+1$ there exists a function $f\in C^{\infty}(\mathbb{R})$ such that $$\left|\,f^{-1}(y)\right|=2m+1\tag{1}$$ for any $y\in\mathbb{R}$. My (double) question now is:

1) Is it possible that $(1)$ holds for a real-analytic function ?

2) Is it possible that $(1)$ holds for an entire real function ?

I am quite sure that the answer to the second point is negative. If we take $x_M$ as a point of local maximum ad $x_m$ as a point of local minimum, there must be $z_M,z_m\in\mathbb{R}$ with the properties that $f(z_M)=f(x_M),f(z_m)=f(x_m)$ and neither $z_m$ or $z_M$ are stationary points. If we prove that $f^{(k)}(z_m)=f^{(k)}(z_M)$ for any $k\geq 1$, then $g(z)=f'(z+z_m)-f'(z+z_M)\equiv 0$, so $f'$ is a $(z_M-z_m)$-periodic function. By translation, we can assume: $$z_m=-1,\quad z_M=1,\quad f(z_m)=-1,\quad f(z_M)=1,$$ and have a $2$-periodic function with the property that $f(\xi)\in\{-1,1\}$ for any $\xi\in[-1,1]$ such that $f'(\xi)=0$. So $f$ should be $(T_{2m+1}\circ\varphi)(x)$ with $T$ being a Chebyshev polynomial and $\varphi$ being a diffeomorphism of $[-1,1]$. But for such a function $f''(1)=-f''(1)\neq 0$.

However, there are too many "maybe" and "should" in this argument, and above all, it does not fit the real analytic case.

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Jack D'Aurizio
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  • Try $\sin(x) + a x$ with $a$ somewhere close to $0.22$ (exact value is difficult to solve for). Does that work? Also, what's the difference between real-analytic and real-entire? – Arthur Aug 13 '14 at 00:12
  • $f(x)=\frac{1}{x^2+1}$ is real analytic but not real- entire, since in $x=i$ we have a pole, hence the radius of convergence of the power series is never $+\infty$. – Jack D'Aurizio Aug 13 '14 at 00:34

1 Answers1

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Here I gave such an example for $2m+1=3$, same as in Arthur's comment above. The generalization is straightforward:

There is such a function of the form $f(x) = ax+\cos x$ with $a>0$.

We look for $a$ such that each maximal value matches the $(m+1)$th minimum to the right of it; on the pictures below $m=1$ and $m=2$. Since $f'(x) = a-\sin x$, the maxima are $\sin^{-1}a+2\pi n$ and the minima are $\pi-\sin^{-1}a+2\pi n$. This leads to equation $$a\,((2m+1)\pi -2\sin^{-1}a) =2$$ which cannot be solved explicitly; however it is easy to see that a solution exists. Indeed, the left side is a continuous function of $a$ which is equal to $0$ at $a=0$, and equal to $(2m-1)\pi>2$ at $a=1$. Therefore, there is $a\in (0,1)$ for which equality holds.

Plot for $m=1$

3-fiber

Plot for $m=2$

5-fiber

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