1
$\begingroup$

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 $h(x) = 0$ $?$

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.