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

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 $$h(x) = 0$$ $$?$$

• 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 $0 < \alpha x < x$). – 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.
• @megan - I've just edited my answer to address the issue of why $x$ can be taken to lie in the open interval $(0,1)$. – user4534 Jun 3 '19 at 16:20