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(a) Show that there must exist $x,y \in [0,1] $ satisfying $|x-y| = \frac{1} {2}$ and $f(x) = f(y)$

I can start by defining a function $g(x) = f(x + \frac{1} {2}) - f(x)$ to guarantee an $x,y$ so that $|x-y| = \frac{1} {2}$ But how do I show that $f(x) = f(y)$?

(b) Show that for each $n \in \mathbb{N}$ $\exists x_n ,y_n \in [0,1]$ with $|x_n - y_n| = \frac{1} {n}$, and $f(x_n) = f(y_n)$

Actually I'm not sure where to start here. Any help is greatly appreciated. Thanks!

Jabernet
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    The key is the intermediate value theorem. – Daniel Fischer May 05 '16 at 18:51
  • @DanielFischer, I know that, it is the subject of the chapter. I am having trouble reasoning how to use it here. – Jabernet May 05 '16 at 18:52
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    Okay. So in part (a), you want to show that $g$ has a zero. Does $g$ satisfy the conditions under which the IVT guarantees a zero? – Daniel Fischer May 05 '16 at 18:54
  • @MartinR Not quite a duplicate, my question has a part b. Thanks – Jabernet May 05 '16 at 18:59
  • @Jabernet: ... and that is asked (and answered) in the referenced question as well: "In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$." – Martin R May 05 '16 at 19:00
  • @MartinR for future reference, I used the search function to try and find the questiion before asking and none of the suggestions that came up matched what I was asking. How am I supposed to know that it is the Universal Chord Theorem if my book did not call it that? – Jabernet May 05 '16 at 19:09
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    @Jabernet: Actually I found it by searching for "g(x) = f(x + \frac{1} {2}) - f(x)" and then following the links. But anyway: The question has been asked and answered before (and more than once, if you look at the Linked section of the referenced Q&A), therefore it is a duplicate. That does not mean that your question is bad. Closing as a duplicate is not a down vote. – Martin R May 05 '16 at 19:12

1 Answers1

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(a)

You are on the right track. If $g(0)=0$ or $g(0.5)=0$ then you are ready. If not then we still have $g(0)+g(0.5)=f(1)-f(0)=0$ so one of $\{g(0),g(0.5)\}$ is positive and the other negative. Now the intermediate theorem.

(b)

What you did for $n=2$ in the first part can also be done for $n=3,4,\dots$.

drhab
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