Fixed points on functions mapping $[a,b]$ onto itself

Question

A function $f:[a,b] \to [a,b]$ is said to have a fixed point $c \in [a,b]$ if $f(c)=c$. Show that every continuous function $f$ mapping $[a,b]$ onto itself has at least one fixed point.

I came up with a basic proof of this. I don't think it's quite correct, but let me know what you think.

Suppose $f:[a,b] \to [a,b]$ is continuous on $[a,b]$

Since the domain and codomain of $f$ are the same, $f(x)=x$

Let $\varepsilon>0, c \in[a,b]$

Let $\delta= \varepsilon$

Suppose $|x-c|<\delta$

$\Rightarrow |x-c|<\varepsilon$

$\Rightarrow |f(x)-f(c)|<\varepsilon$

Thus, $\lim_{x \to c} f(x)=f(c)$ and therefore $f(x)$ is continuous for all $c \in [a,b]$

Then $f(c)=c$ for all $c \in [a,b]$

Thus, every continuous $f$ mapping $[a,b]$ onto itself has at least one fixed point

When you write that $f(x)=x$, what does that mean? What is that $x$? — José Carlos Santos, Nov 22 '20 at 15:13
$f(x)=x$ where $x \in [a,b]$. It's just a linear parent function — MartyrLogarius, Nov 22 '20 at 15:13
Let $f:[0,1]\to[0,1]$ then observe $f$ can be $x^n $ where $ n\geq 1$.So you can't assume $f(x)=x,,\forall x$ — PNDas, Nov 22 '20 at 15:16
So how should I create an arbitrary $f$ when there are countless counterexamples to my $f(x)=x$? — MartyrLogarius, Nov 22 '20 at 15:17
Observe that $f(a)\geq a$ and $f(b)\leq b$ So if I take $g(x)=f(x)-x$, then $g(a)\geq 0$ and $g(b)\leq 0$, so now you can apply IVT as both $f$ and $x$ are continuous, so $g$ is continuous. — PNDas, Nov 22 '20 at 15:20
You are not supposed to create a function. You are supposed to proved that there is a fixed point for every continuous function. — José Carlos Santos, Nov 22 '20 at 15:21

score 0 · Answer 1 · answered Nov 22 '20 at 15:15

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You can not assume $f(x)=x$. You need to prove this.

Hints: Let $g(x)=f(x)-x$. Can you see that $g$ is continuous? Use Intermediate value theorem. Good luck!

answered Nov 22 '20 at 15:15

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Fixed points on functions mapping $[a,b]$ onto itself

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