2
$\begingroup$

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a nondecreasing function.

Let $a and $f(b). Prove that there is a $a such that $f(c)=c$.

My attempt at a proof is as follows. Let $c:=\sup\{x:a\leq x\leq b\text{, }x\leq f(x)\}$.

This is where I'm stuck. Since I can't use more powerful theorem such as the IVT I find this problem far more complex.

  • 0
    If you define $c$ in that way, what you can tell us about $c$, $f(c)$, and other values that aren't $c$?2012-09-13

2 Answers 2

1

HINT: Let $L=\{x\in[a,b]:x. Is $L$ non-empty? Is $L$ bounded above?

Added: In other words, your idea is reasonable, though I used a slightly different set from yours. $L$ is non-empty and bounded (why?), so we can let $c=\sup L$. What do you know about $f(x)$ for $x\in[c,b]$? What happens if $f(c)\ne c$? Remember, $f$ is non-decreasing.

  • 0
    And it is needed. If we were talking about $\mathbb Q$, we could select a sequence $(a_n)$ converging to $\sqrt 2$ from below and $(b_n)$ from above. Then let $f(x)$ be the minimal a_n>x for x<\sqrt 2 and the maximal b_n for x>\sqrt 2.2012-09-13
0

Define another function $g(x) = f(x) - x$. Then try to find out if this function crosses $0$.

  • 0
    Note that $f$ isn't continuous anyways.2012-09-13