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This comes from problem 21c of Chapter 7 of Spivak's Calculus 4th Edition.

21c) Find a continuous function $f$ which takes on every value exactly $n$ times, where $n$ is odd.

The solution I came up is $f_n(x)= [ (-1)^{⌊x⌋+⌊\frac{x}{n}⌋}~\cdot~(x-⌊x⌋-\frac{1}{2}) ]+ ⌊\frac{x}{n}⌋ + \frac{1}{2}$

Where $⌊x⌋$ is the floor function.

It seems to work, but I want to prove it works. Well atleast I want to prove it is continuous on $\mathbb{R}$. No idea where to start. Any suggestions?

1 Answers1

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Continuity of $f_n$

For $x \in [kn + i, kn+i + 1)$ with $(n,i) \in \mathbb Z \times \{0, \dots, n-1\}$ you have $\lfloor x \rfloor = kn+i$, $\lfloor \frac{x}{n} \rfloor = k$ and

$$\begin{aligned}f_n(x) &= \left[(-1)^{k(n+1)+i}\left(x-kn-i -\frac{1}{2}\right)\right] + k + \frac{1}{2}\\ &= \left[(-1)^{i}\left(x-kn-i -\frac{1}{2}\right)\right] + k + \frac{1}{2} \end{aligned}$$ and therefore for $x \in [kn + i, kn+i + 1)$

$$f_n(x) = \begin{cases} x-kn-i+k &\text{ for } i \text{ even}\\ -x+kn+i+k+1 &\text{ for } i \text{ odd} \end{cases}$$

Based on that we get for $i$ even

$$\lim\limits_{x \to (kn+i+1)^-} f_n(x)= k+1 = f_n(kn+i+1)$$ and for $i$ odd

$$\lim\limits_{x \to (kn+i+1)^-} f_n(x)= k= f_n(kn+i+1).$$

Knowing that $f_n$ is continuous on all intervals $(kn + i, kn+i + 1)$, this allows to conclude that $f_n$ is continuous on each interval $(kn, k(n+1))$ with $k \in \mathbb Z$. As

$$\lim\limits_{x \to kn^-} f_n(x)=k= f_n(kn)$$ we can finally conclude that $f_n$ is continuous on $\mathbb R$.

Values taken by $f_n$

  1. $g_n(x)=f_n(x) - ⌊\frac{x}{n}⌋$ is a periodic map with $n$ as a period.
  2. $g_n$ takes exactly $n$ times each value of $(0,1)$ on the interval $(0,n)$.
  3. The cardinal of $f^{-1}_n(\{0\})$ is equal to $n$.
  4. For all $x \in \mathbb R$ $f_n(x+n) = f_n(x)+1$.
  5. Conclude.

This can also be seen directly from the study of the continuity and the values taken on each interval on which $f_n$ is linear.

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