Real Analysis, Continuous functions. Let $f : [0, 1]\to\mathbb R$ be a continuous function. 
If there exists $c\in(0, 1)$ such that $f(c)\leq f(x)$ for every
$x\in[0, 1]$, show that there exist $a,b\in[0, 1]$  such that $a\neq b$ and $f(a) = f(b)$.
 A: If $f(0)=f(1)$ we can take $a=0,b=1$. Otherwise we may assume that $f(0)<f(1)$ since a similar argument argument works when $f(0)>f(1)$. Let $t \in (f(c),f(0)])$. [Note that if $f(0)=f(c)$ the  there is nothing to prove].  Now apply intermediate value property to the intervals $[0,c]$ and $[c,1]$ to get points $a$ in the first interval and $b$ in the second such that $f(a)=f(b)=t$. [$t$ also lies between $f(c)$ and $f(1)$].  It is easy to see that $a=b$ is not possible because $t >f(c)$
A: Are you allowed to use following fact?


*

*An injective continuous real function $f: I \rightarrow \mathbb{R}$ on an interval $I$ is strictly monotone.


If yes you can reason as follows by contradiction:


*

*Assume $f$ is injective.

*$f(c) = \min_{x \in [0,1]} f(x) \stackrel{f \mbox{ strictly monotone}}{\Longrightarrow} c= 0$ or $c= 1$

*That's a contradiction to the assumed $c \in (0,1)$.

*It follows that $f$ cannot be injective, hence $\exists a,b \in [0,1]: a \neq b, f(a)=f(b)$

A: If there exists a,b such that f(a)=f(b)=f(c),then we are done. Therefore, we consider another case. Take a small neighborhood of c,$U=[c-\epsilon,c],V=[c,c+\epsilon]$,f has a maximum in $U$ ,denoted $f(d),d \in U $and f has a maximum in $V$,denoted $f(e),e\in V $since f is continued. We take $s=\min\{f(d),f(e)\}$,then we can use intermediate value theorem ,viz.,there exists $a\in U,b\in V$,such that $f(a)=f(b)=s$.
