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Suppose $f$ is continuous on $[0,2]$ and $f(0) = f(2)$. For which $a\in(0,2)$ must there exist $x,y\in[0,2]$ so that $\lvert y − x\rvert = a$ and $f(x) = f(y)$

I'm really unsure how to approach this problem ...we did a similar problem where $a=1$, by defining $g(x) = f(x+1)-f(x)$ on $[0,1]$, then applying the IVT.

Is this problem approached in a similar way? If not, what's a good starting point? Thank you for any ideas!

ky370211
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    Starting point: You know the claim holds for $a=1$. Thus, there are $b<c$ with $c-b=1$ and $f(c)=f(b)$. By applying an analogous argument than for the proof for $a=1$ to $f|{[b,c]}$ (or by shifting and rescaling), you can show that the claim also holds for $a=1/2$. Inductively, it holds for $a =2^{-n}$ for arbitrary $n \in \Bbb{Z}{\geq 0}$. – PhoemueX Mar 09 '16 at 21:25
  • Could you check the intervals of x,y and a ? Are they correct in the question ? – Our Aug 01 '16 at 08:03
  • By $[0, 2]$ do you mean the interval or the point $[0, 2]$? – Obinna Nwakwue May 18 '17 at 00:19

1 Answers1

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The answer is $a \in \{\frac 2 n|n\gt 1\}$ .

Proof: Take some $a \in \{\frac 2 n|n\gt 1\}$ and suppose for purposes of contradiction that $f$ is continuous on $[0,2]$ with $f(0)=f(2)$ but there are no $x, y \in [0,2]$ such that $|x-y|=a$ and $f(x)=f(y)$.

Consider the function $g(x)=f(x+a)-f(x)$ which is well defined on $[0,2-a]$ .

Then $g(x) \neq 0$ throughout $[0,2-a]$ and due to the intermediate value theorem it must either be positive throughout $[0,2-a]$ or negative throughout $[0,2-a]$.

If $g(x)\gt0$ then $f(0)\lt f(a)\lt f(2a) …\lt f(n a)=f(2)$ which contradicts the condition $f(0)=f(2)$ . Similarly if $g(x)\lt0$ then $f(0)\gt f(a)\gt f(2a) …\gt f(n a)=f(2)$ .

So no such function $f$ can exist if $a=\frac2 n$ .

On the other hand suppose $a \not \in \{\frac 2 n|n\gt 1\}$ . To show the condition need not be satisfied in this case define $f(x) = \cos(\frac{2\pi x} a) +\frac x 2(1-\cos(\frac {4\pi} a))$

Then if $|x-y|=a$ , $\cos(\frac{2\pi x} a)=\cos(\frac{2\pi y}a)$ and $|f(x)-f(y)|=\frac a 2(1-\cos(\frac{4\pi} a) ) \ne 0$, since $\cos(\frac{4\pi} a)\ne1$

But $f(2)=\cos(\frac{4\pi} a) + (1-\cos(\frac{4\pi} a))=1=f(0)$ so $f$ is a counterexample

Arthur Champernowne
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