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Let $ f:[a,b]→\mathbb R$ be a function with the following property:

For every $y\in f([a,b])$, you have exactly two $ x_1, x_2 $,such as $f(x_1)=f(x_2)=y$.

Prove that f can not be continuous everywhere.

Martund
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Chris Rafael
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  • @TheoBendit I think that the fact that the domain of $f$ is a closed interval in this question is relevant. – rafa11111 Dec 18 '18 at 23:35
  • @rafa11111 That's a good point. However, in the step where an $a \in \mathbb{R}$ is fixed, you can salvage the proof by ensuring that $a$ is not chosen to be the image of either end point. – Theo Bendit Dec 18 '18 at 23:39

2 Answers2

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By the extreme value theorem, $f$ attains its maximum at some point. Then it must attain its maximum at two points: let us name them $x_1 < x_2$. Now, $f$ restricted to the interval $[x_1, x_2]$ must attain its minimum at some point $ x_3 \in (x_1, x_2)$. And there must be some other point $x_4 \in [a, b]$ with $f(x_4)=f(x_3)$. If $x_4 \notin (x_1,x_2)$, then by the intermediate value theorem $f$ must attain the value $\frac{f(x_1)+f(x_3)}{2}$ at least three times. And if $x_4 \in (x_1, x_2)$, let us pick any $x_5$ between $x_3$ and $x_4$: then $f$ must attain $f(x_5)$ at least three times (because it is between $f(x_1)$ and $f(x_3)$).

Lucas Kenji Moori
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  • Where do we must end up to show that f isn't continuous everywhere ? Can you post the full answer because i am stuck ? – Chris Rafael Dec 19 '18 at 10:36
  • If I understood Lucas' answer, he showed in the end that there is a contradiction, because there will be three points with the same value instead of two. However, the summoning of the intermediate value theorem requires the hypothesis that $f$ is continuous. The contradiction arose due to an incorrect assumption, which, in the case, is the continuity of $f$. – rafa11111 Dec 19 '18 at 11:08
  • @rafa11111First , why must f attain its maximum at two points ? – Chris Rafael Dec 19 '18 at 11:17
  • @ChrisRafael because the function has each value at two different points, it's on the statement of your question. – rafa11111 Dec 19 '18 at 11:18
  • because f(x1)=f(x2) ? – Chris Rafael Dec 19 '18 at 11:18
  • ok yes . I didn't see yout anwer before i wrote the same thing – Chris Rafael Dec 19 '18 at 11:19
  • 2rafa11111 but the question didn't tell about maximum at x1 and x2 – Chris Rafael Dec 19 '18 at 11:20
  • i see.Ok . I understand it whu he take maximum at x1 and x2 – Chris Rafael Dec 19 '18 at 11:22
  • Why after this must f attain the value 9f(x1)+f(x3))/2 at least three times ? – Chris Rafael Dec 19 '18 at 11:27
  • We have $f(x_1)=f(x_2)>f(x_3)=f(x_4)$ (by choice of $x_1 ,..., x_4$); and $\frac{f(x_1)+f(x_3)}{2}$ is a number in between the two sides of this inequality. Suppose, for example, that $x_1 < x_3 < x_2 < x_4$: then, by the intermediate value theorem, $f$ (if it were continuous) would have to take the value $\frac{f(x_1)+f(x_3)}{2}$ somewhere in $(x_1, x_3)$, then again in $(x_3, x_2)$ and in $(x_2, x_4)$ (contradicting the hypothesis that $f$ takes every value exactly twice). – Lucas Kenji Moori Dec 19 '18 at 21:00
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Hint: assume there is such an f and then it has exactly two absolute minimum points and two absolute maximum points; if say one of the absolute minimum points is strictly inside the interval, show that there are small intervals to the right and left of it which are sent by f to the same interval and then you get a contradiction using the same property for the other absolute minimum point for which there is such a small interval either to the left or right (or both of course if it is inside); so the function must take its absolute minimum at the ends; in the same way it must take its absolute max at the ends and of course f is non-constant so that is impossible

Taking the absolute value or square functions shows that you cannot relax the condition to exclude one point in the interval

Conrad
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