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Prove that a differentiable function $f$ with $f(x+1)=f(x)$ has at least two points in $[0,1]$ such that $f ' (x) =0$.

I used Mean value theorem to obtain one such point in $[0,1]$ , but i am not sure how to go any further.

Thanks

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    $\begingroup$ It will help to use the fact that it should also hold that $f'(x+1)=f'(x)$ $\endgroup$
    – Mangomath
    Jun 6, 2014 at 11:08
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    $\begingroup$ If $c$ is the first point you found, consider the interval $[c,c+1]$. Find a stationary point in this interval and use the periodicity of $f$ to find another in $[0,1]$. $\endgroup$ Jun 6, 2014 at 11:08
  • $\begingroup$ @Mangomath: are you sure it's not $|f'(x+1)| = |f'(x)|$ $\endgroup$
    – Alex
    Jun 6, 2014 at 11:15
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    $\begingroup$ Yes. Upon finding the first point c i applied MVT on the interval [c,c+1] to find another point say z. Then z is an element of (c,c+1). There are two cases z is an element of (c,1) or [1,c+1). If z is an element of (c,1) then we are done otherwise z is an element of [1,c+1). Since f '(x+1)=f '(x) , thus there exists z-1 in [0,c) such that f ' (z-1) =0 . Is this legitimate ? $\endgroup$
    – user155561
    Jun 6, 2014 at 11:16
  • $\begingroup$ Yes, that looks good. $\endgroup$ Jun 6, 2014 at 11:22

2 Answers 2

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Case 1: $f'(x)$ has at least two roots in $(0,1)$. We are done.

Case 2: $f'(x)$ has exactly one root in $(0,1)$. Lets call this root $c$. Then $f$ is strictly monotonic on $(0,c)$ and $(c,1)$, and as $f(0)=f(1)$, the monotony on the two intervals must be different.

But then $f$ has the same monotony on $(1,1+c)$ as on $(0,c)$, therefor $f(1)$ is a local extrema. This implies $f'(1)=0$.

In this case you can also prove $f'(0)=0$, but it is not needed.

Case 3: $f'(x)$ has no root in $(0,1)$. Then, $f$ must be strictly monotonic on $(0,1)$, which is impossible as $f$ continuous and $f(0)=f(1)$.

Note: The proof uses the fact that $f'$ has the intermediate value property, in the form that if $f'$ has no root on an interval, it cannot change sign on that interval.

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Case 1: $f$ is constant. Every point is such as $f'(x)=0$.

Case 2: $\sup_{[0,1]} f > \inf_{[0,1]} f$. Use the fact that $f$ is bounded and that $f(\Bbb R) = f([0,1])$, and using the Bolzano Weierstrass theorem, you get $(x,y)\in [0,1]^2$ such as $$ \max_{[0,1]} f = f(x) \\ \min_{[0,1]} f = f(y) $$

Then show that $f'(x) = f'(y) = 0$.

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  • $\begingroup$ You might want to add something about how you use $f(x) = f(x+1)$. $\endgroup$
    – J. J.
    Jun 6, 2014 at 13:10
  • $\begingroup$ yes, thanks.{}{} $\endgroup$
    – mookid
    Jun 6, 2014 at 13:13

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