# Prove that there is a number $y \in \mathbb{R}$ in $[0,1]$ st $f(y) = y$

$f$ is continuous on $[0,1]$ and the range of $f$ is a subset of $[0,1]$. Prove that there is a number $y \in \mathbb{R}~$ in $[0,1]$ such that $f(y) = y$ using $~h(x) = f(x) - x$

I wasn't sure how to approach this as I couldn't seem to determine whether the $~h(x) = f(x) - x$ was an identity of continuous functions or just part of the question. It seemed to be close to the definition of continuity at a point $\mu$:

If $\lim\limits_{x \to \mu} f(x) = f(\mu)$

but I wasn't sure how to use it (if I should at all).

What is a good direction to take with this?

-
I would not use $n$ to denote a real number... – Henrik Mar 25 '12 at 8:33
@user22705: How about $y$? : ) – stariz77 Mar 25 '12 at 11:07

Well, if there is a point $x$ such that $f(x) = x$, then we would have $h(x) = f(x) - x = 0$. Thus, we need only show that $h(x)$ has a zero.

A continuous function enjoys the intermediate value theorem. If we can show that $h(x)$ is greater than zero at some point in $[0,1]$, and less than zero at another point in $[0,1]$, it will have a zero. Try using the size of the domain and the size of the range to see that such points exist.

-

You are supposed to show that $h(x)=0$ for some $x\in [0,1]$. In order to do so, you can use the fact that $h$ is a continuous function and apply the intermediate value theorem. Note that $h(0)=f(0)\geq 0$ and $h(1)=f(1)-1\leq 0$, so by the IVT we have some $x\in [0,1]$ such that $h(x)=0$, so $f(x)=x$.

Edit: A more general approach using more advanced techniques allows us to show that for any continuous function $f:[a,b]\to [a,b]$ and non-intersection continuous curve $\gamma:[0,1]\to [a,b]\times [a,b]$ such that $\gamma(0)=(a,a),\gamma(1)=(b,b)$, there exists some $t\in [0,1],x\in [a,b]$ such that $(x,f(x))=\gamma(t)$. This is a proper generalization of the original question, which is equivalent to letting $a=0,b=1$ and defining $\gamma(t)=(t,t)$. Assume $f(a)> a$ as otherwise we are done. By the Jordan curve theorem, the regions $A$ above and $B$ below the curve $\gamma$ are connected yet $A\cup B$ is disconnected. Let $\Gamma(f)$ denote the graph of $f$. Note that $\Gamma(f)\cup A\cup B$ is connected, as each term is connected, $(a,f(0))\in \Gamma(f)\cap A$ and either $f(b)<b$ in which case $(b,f(b))\in \Gamma(f)\cap B$ or $f(b)=b$ in which case $(b,b)\in \Gamma(f)\cap (B\cup (b,b))$ and it is easy to verify that $B\cup (b,b)$ is connected. Thus $\Gamma(f)\cup A\cup B\neq A\cup B$ so we have some $(x,f(x))\in ([a,b]\times [a,b])\setminus (A\cup B)=\gamma$.

-
Is it better to leave Daniel's hint or give the answer? – Neil G Mar 25 '12 at 14:28
@NeilG In terms of Math.SE practice? There are differing philosophies. I tend to give answers because until I've worked things out I'm not certain of my hints. – Alex Becker Mar 25 '12 at 14:58
Out of curiosity, why the downvote? – Alex Becker Mar 25 '12 at 19:05
(It wasn't me…) – Neil G Mar 25 '12 at 19:39