Prove that $f$ Contains A Fixed Point Using The Intermediate Value Theorem

Question

Use the Intermediate Value Theorem to prove that any continuous function with domain $[0, 1]$ and range a subset of $[0, 1]$ must have a fixed point.

My approach:

I will call this continuous function in question $f$, and define $g(x) = x$.

What I will(attempt to) show is that $f$ will intersect the function $g$ at some point.

Let the range of $f$ be $[r_1, r_2]$ (which is a subset of $[0, 1]$)

Let $g(a) = r_1$ and $g(b) = r_2$ ($a$ and $b$ naturally are in the domain of $f$)

$r_1 \leq f(x) \leq r_2 \Leftrightarrow g(a) \leq f(x) \leq g(b)$ so $f$ must intersect $g$ at some point

So there exists a fixed point of $f$ between $a$ and $b$ or possibly at $a$ and/or $b$.

Please let me know what I'm doing right/wrong.

I'm quite sure this has been discussed before. Anyway, here's the hint: consider $g(x)=f(x)-x$; then $g(0)=f(0)-0$ and $g(1)=f(1)-1$. What can you say about $f(0)-0$ and $f(1)-1$? — egreg, Jul 29 '14 at 11:41
What I gather from that is $f(0) - 0$ is somewhere in the range of $[0, 1]$ and $f(1) - 1$ is somewhere in the range of $[0, -1]$. So that implies that somewhere between $0$ and $1$ $f(x)=x$? — Kermit the Hermit, Jul 29 '14 at 12:10
$f(0)-0\ge0$, while $f(1)-1\le0$. If one of them is zero, you're done, otherwise… — egreg, Jul 29 '14 at 12:34
Yes yes, that's exactly what I understood from your comment. Thanks a lot, egreg, I appreciate your assistance. — Kermit the Hermit, Jul 29 '14 at 12:40
Possible duplicate of Show that a continuous function has a fixed point — Jigao, Dec 16 '18 at 13:19

score 4 · Accepted Answer · answered Jul 29 '14 at 12:55

4

Hint: consider the continuous function $g(x)=f(x)-x$; then \begin{align} g(0)&=f(0)-0\ge0\\ g(1)&=f(1)-1\le0 \end{align} Your task is to argue that there exists $x_0\in[0,1]$ such that $f(x)=x$, or $g(x)=0$.

answered Jul 29 '14 at 12:55

egreg

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Prove that $f$ Contains A Fixed Point Using The Intermediate Value Theorem

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