Suppose that $f:[0,1]\to \mathbb{R}$ is a continuous function

Question

Suppose that $f:[0,1]\to \mathbb{R}$ is a continuous function and that $f(c)=\min\{f(x):0 \le x \le 1\}$ for some $c\in(0,1)$ . Prove that $f(x)=f(y)$ for some $x$ and $y\in(0,1)$

Is any idea please thank you

Answer 1

First, notice that if $f(0)=f(c)$ or if $f(1)=f(c)$, there is nothing to do : you have to points of $[0,1]$ on which $f$ takes the same value.

So let's suppose $f(0)\neq f(c)$ and $f(1) \neq f(c)$. Because $f(c)$ is the minimum of $f$, you have $f(c) < f(0)$ et $f(c) < f(1)$. Consider $$ z = \frac{\min(f(0),f(1)) + f(c)}{2}$$

Of course you have $ f(c) <\min(f(0),f(1)) \leq f(0)$ and $ f(c) <\min(f(0),f(1)) \leq f(1)$, therefore you get $$z \in (f(c), f(0)) \quad \text{ and }\quad z \in (f(c), f(1))$$

Applying intermediate value theorem ($f$ is continuous) on the intervals $(0,c)$ and $(c,1)$, you get that there exists $x \in (0,c)$ such that $f(x)=z$, and $y \in (c,1)$ such that $f(y)=z$. Therefore you get $x\neq y$ such that $f(x)=f(y)$.

Answer 2

If $f(c)=f(0)$ or $f(c)=f(1)$ done. Otherwise $f(c)<f(0)$ and $f(c)<f(1).$ So $f$ takes on each real value between $f(c)$ and $\min(f(0),f(1)),$ once to the left of $c$ and once to the right. Apply intermediate value theorem.

Answer 3

"I don't understand what you saying – Inverse Problem". OK. I try to express his thought as best as I can here so you can understand his proof. To make it easier, put $p = f(c)$, and assume $q=f(0) < f(1) = r$. Since $p < q$, the intermediate value theorem says $f(x) = h$ for some $x \in (0,c)$, and $h \in (p,q)$, and also $f(y) = h$ for some $y \in (c,1)$ since $h \in (p,r)$ as well.Thus $f(x) = f(y)$ with $x \neq y$.