Suppose $f(x)$ is a continuous function on $[0,1]$ with $f(0) = f(1)$. Let $α ∈ (0,1)$. Prove that there exists an $x ∈ (0,1)$ such that $f(x) = f(αx)$.

I tried $h(x) = f(x) - f(αx)$ and Intermediate Value Theorem. $h(0) = f(0) - f(0) = 0$, $h(1) = f(1) - f(α)$, how can I prove there exists x in the open interval such that $h(x) = 0$ $?$

  • $\begingroup$ For intuition: What you are asking is that if the function ends up where it started ($f(0)=f(1)$), then inbetween there is an $x$ such that the function takes a value it already has taken before (because $0 < \alpha x < x$). $\endgroup$ – 77and33is100 Jun 3 '19 at 14:44

Hint: By the Extreme Value Theorem there exist $x_{max}, x_{min}\in [0,1]$ such that $$f(x_{max}) \geq f(x) \geq f(x_{min})$$ for all $x\in[0,1]$. Why don't you try applying the Intermediate Value Theorem to the function $h(x)=f(x)-f(\alpha x)$, not by evaluating $h(x)$ at the endpoints of your interval but rather at $x_{max}$ and $x_{min}$?

The above argument will show that $h(x_{min}) \leq 0$ and $h(x_{max})\geq 0$. Therefore there exists some number $x$ between $x_{min}$ and $x_{max}$ for which $h(x)=0$, which is what you wanted to show. Now how can we guarantee that $x\in (0,1)$? Well, for starters, if neither $x_{min}$ nor $x_{max}$ is in $\{0,1\}$ then $x$ will clearly lie in $(0,1)$. Moreover, if both $x_{min}$ and $x_{max}$ are in $\{0,1\}$ then the fact that $f(0)=f(1)$ implies that $f$ is constant on $[0,1]$, in which case the problem is trivial. So assume without loss of generality that $x_{min}=1$ and $x_{max}\not\in \{0,1\}$. Then $h(1)\leq 0$. If $h(1)< 0$ then $$0< x_{max} \leq x < 1=x_{min},$$ which means that $x\in (0,1)$. If $h(1)=0$ then $f(1)-f(\alpha)=0$ and consequently, $f(1)=f(\alpha)$. Therefore $\alpha$ is also a minimum of $f$ on $[0,1]$, so we could have taken $x_{min}=\alpha$ above. But this means that $\alpha=x_{min}$ and $x_{max}$ both lie in $(0,1)$, hence so will the zero $x$ of $h$ given to us by the Intermediate Value Theorem.

  • $\begingroup$ Thanks! How to prove x is in open interval? $\endgroup$ – sjh Jun 3 '19 at 15:52
  • $\begingroup$ @megan - I've just edited my answer to address the issue of why $x$ can be taken to lie in the open interval $(0,1)$. $\endgroup$ – user4534 Jun 3 '19 at 16:20

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