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I have proved that a continuous function $f(x): \mathbb{R} \rightarrow\mathbb{R}$ cannot take every value in its range exactly twice. How is the general case with an even number of times proved?

Here's my proof for the basic case.

Suppose continuous $f(x): \mathbb{R} \rightarrow \mathbb{R}$ takes every value in its range exactly twice.
Let $f(a) = f(b) = w$ with $a < b$.
Claim that $\forall x \in (a,b), \; f(x)>w$ or $f(x) < w$.

  • If $f(x) = f(a) = f(b) \; \forall x \in (a,b)$, that clearly contradicts the hypothesis.
  • If $\exists x_1,x_2 \in (a,b): \; f(x_1) < w < f(x_2)$, then by Intermediate Value Theorem, $\exists x_3 \in (x_1,x_2)$ or $(x_2,x_1)$ such that $f(x_3) = w$ contradicting the hypothesis.

Assume for the moment that $\forall x \in (a,b), \; f(x)>w$.
By Extreme Value Theorem, $f(x)$ attains maximum $f(c_1) > f(a)=f(b)$ at some $c_1 \in (a,b)$.
By the hypothesis, $\exists c_2 \in \mathbb{R}\setminus \{c_1\}$ such that $f(c_2) = f(c_1)$.
Claim that $c_2 \in (a,b)$.

  • If $c_2 > b$, for some value $u \in \Big(f(a),f(c_1)\Big) = \Big(f(b),f(c_1)\Big) = \Big(f(b),f(c_2)\Big)$, $f(x) = u$ at least three times, namely $x_1 \in (a,c_1), \;x_2 \in(c_1,b), \;\text{and } x_3 \in(b,c_2)$, contradicting the hypothesis.
  • By similar reasoning, $c_2 \nless a$.

Without loss of generality, assume $a < c_1 < c_2 < b$.
By Extreme Value Theorem, $f(x)$ attains minimum $f(d)$ at some $d \in (c_1, c_2)$.
Because $f(a) = f(b) < f(d) < f(c_1) = f(c_2)$, by Intermediate Value Theorem, $f(x) = f(d)$ at least three times, namely $x_1 \in (a,c_1), \; d, \text{ and } x_2 \in (c_2,b)$.
This is a contradiction to the hypothesis.

Now for the other case when $\forall x \in (a,b), \; f(x) < w$. Let $g(x) = -f(x)$ and apply the above proof to conclude that $g(x)$ cannot take every value in its range exactly twice, so neither $f(x)$.

Thanks.

Myath
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    How does your proof work? Maybe we can generalise it. – PhoemueX Oct 26 '14 at 20:12
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    Here's how I'd start: choose $2k$ points $a_1,\dots,a_{2k}$ at which $f(a_j)=0$ say. Look at the $2k-1$ intervals formed by these numbers: at least $k$ of them are the same in terms of the function being positive of negative. Without loss of generality, $k$ of them are positive; on each of these $k$ intervals there is a maximum value; and then any positive value less than the least of these maximum values is already taken $2k$ times by the intermediate value theorem. Go from there.... – Greg Martin Oct 26 '14 at 23:42
  • My answer here also answers this question (regardless of how you interpret it), so I'm voting to close as a duplicate. – Eric Wofsey Sep 18 '15 at 05:27

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