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A Fixed point of $f$ is a point $r$ such that $f(r)=r$.

I am not really convinced by this theorem holds for any function (even discontinuous functions). For example, I sketched this function:

enter image description here

This function has $f(0)>0, f(1)<1$, it is increasing and has no value $r$ where $f(r)=r$. Am I missing something or did the question miss a condition?

AspiringMat
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    What is $f(r)$? Your conditions state that $f$ is defined everywhere. – mlc Apr 26 '17 at 09:33
  • Yeah, you should definitely explain more clearly what you think is a counterexample. – Did Apr 26 '17 at 09:39
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    @Any: the theorem remains true if you replace monotonicity with continuity, but continuity need not be added to monotonicity. – mlc Apr 26 '17 at 09:44
  • @mlc I think I got confused a bit by discontinuity and being undefined. I think I figured out the proof. Let $S={x:f(x)-x\geq 0}$. $S$ is bounded above by $1$ so $SupS$ exists. Similarly for $S'={x:f(x)-x\leq 0}$ and bounded below by $0$ so $InfS'$ exists. We must have $SupS=InfS'=r$ since otherwise it would contradict their definition. Also, we have by monotone function theorem $f(r-)\leq f(r) \leq f(r+)$ but $f(r+)\leq r$ and $f(r-)\geq r$ so $r\leq f(r) \leq r$ or $f(r)=r$ – AspiringMat Apr 26 '17 at 10:19

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The function you provided is not defined in $[0,1]$.

As $f(0)>0$, the function $f$ will be over the line $y=x$ when $x=0$. Similarly, $f$ will be below the line $y=x$ when $x=1$. Then, the key here is to find a point where the function happens to go from over the line to below the line without touching it. Let's suppose this point is $r$. This is:

$\forall x < r : f(x) > x$ And $\forall x > r : f(x) < x$

Also, as the function is defined for every point in the interval $[0,1]$ and it is always increasing, $\forall \eta \in [0,1] f(\eta)<f(\xi) ~\forall\xi>\eta$

So, if we assume that $f(r) > r$, then for $ r+\epsilon$, $f(r+\epsilon)<r+\epsilon$ but as the function is increasing, $f(r+\epsilon)>f(r)$. When doing $\epsilon \rightarrow 0$, we obtain that $f(r)$ is both $>r$ and $<r$. Which is a contradiction.

Similar result is obtained supposing $f(r) < r$ with $f(r-\epsilon)>r-\epsilon$. The only option to avoid the contradiction is by selecting $r$ such as $f(r)=r$.

Another approach could be analysing the function $g(x)= f(x)-x$ and proving that $g(r)=0$.

Gyojin139
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  • How to you go from "find a point where ..." to "Let's suppose this point is $r$"? In other words, how exactly is $r$ defined? Also note that there can be multiple intervals where the graph of $f$ lies below or above the diagonal, see the second image in https://math.stackexchange.com/a/791470/42969. – Martin R Apr 26 '17 at 12:24