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Suppose that $f:[0,1]→\mathbb{R}$ is a continuous function on the interval $[0,1]$, and that $f(0) =f(1)$. Use the Intermediate Value Theorem to show that there exists $c∈[0,1/2]$ such that $f(c) =f(c+1/2)$.

I understand the Intermediate Value Theorem and I know how it is used to prove an equation has a root within an interval. I just don't understand how to approach this question/

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    $\begingroup$ $f(c)=f\left(c+\frac12\right)\iff g(c)=f\left(c+\frac12\right)-f(c)=0$. Study $g$. $\endgroup$ – Nicolas FRANCOIS Aug 8 '18 at 19:00
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We want $f(c) = f(c + \frac12)$, i.e., $f(c)-f(c+\frac12)=0$. This already looks more like proving a function has a root. In fact, if we define a new function $g(c)=f(c)-f(c+\frac12)$, then our job is exactly to prove $g$ has a root in $[0,\frac12]$.

We have to verify that $g$ is continuous in order to use the IVT, but once you do this, it becomes a much more standard IVT question, which you indicated you are familiar with.

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Define $g(x)=f(x)-f(x+1/2)$. If $f(0)=f(1/2)$, we're done; otherwise, WLOG $f(1/2)<f(0)$. Thus $g(0)>0$ and $g(1/2)=f(1/2)-f(1)=f(1/2)-f(0)<0$; in particular, since $g$ is continuous (as a sum of continuous functions), we have that there exists a $c\in[0,1/2]$ such that $g(c)=0$. Equivalently, $f(c)=f(c+1/2)$.

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  • $\begingroup$ How do we know that $f(1/2)$ is not greater than $f(0)$? Although I realise it does not matter either way. $\endgroup$ – user499701 Aug 8 '18 at 19:28
  • $\begingroup$ WLOG means without loss of generality. If $f(1/2)>f(0)$, the consider $f_1(x)=-f(x)$ and apply the argument to $f_1(x)$. $\endgroup$ – Clayton Aug 8 '18 at 20:34

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