Every continuous map of a closed interval into itself has a fixed point

The Question: Please show this theorem: Let $f: I=[a,b] \rightarrow \mathbb{R}$ be a continuous map such that $f(I) \supset I$. Then $f$ has a fixed point on I.

My Attempt: Suppose there is a function, $g(x)=f(x)-x$. Since $f(x)$ is a continuous mapping, this implies that $g(x)$ is also a continuous mapping. If $f(a) = a$ and/or $f(b) = b$, we are done. We are going to investigate the case when $f(a) > a$ and $f(b) <b$. We know that $f(I) \supset I$, at least one element of $f(I)$ is not in $I$ but, all elements of $I$ is in $f(I)$.

Now from the intermediate value theorem, there exists a $c \epsilon (a,b)$ such that for every r in I, f(c) = r. Since $f(a) > a$ and $f(b) <b$, this implies that $f(a)-a >0$ and $f(b) -b < 0$. By the immediate value theorem, there exists a c in (a,b) such that $g(c)= f(c)-c =0$. This means that $f(c) =c$. Therefore, there is at least one fixed point on I.

Am I on the right track on this one?

• The original title had "difference equation" in it, which didn't seem to fit. I guess you copy-pasted it. Please try to write more descriptive titles (example above), instead of your usual "can you help me with this...". It's understood that you'd like help, since you are posting a question.
– user147263
Jun 23 '15 at 4:09

Let $g(x)=f(x)-x$; since $f(x)$ is continuous, so is $g(x)$. If $f(a)=a$ or $f(b)=b$, we’re done, so assume that $f(a)\ne a$ and $f(b)\ne b$. We’re told that $f[I]\subseteq I$, so we must have $f(a)>a$ and $f(b)<b$, and hence $g(a)>0$ and $g(b)<0$. By the intermediate value theorem there is a $c\in(a,b)$ such that $g(c)=0$, and hence $f(c)=c$. Thus, $f$ has at least one fixed point on $I$.
Now from the intermediate value theorem, there exists a $c\in(a,b)$ such that for every $r$ in $I$, $f(c) = r$.
This is clearly impossible: $f(c)$ is a single real number and cannot be equal to $r$ for every $r\in I$ (unless $a=b$, so that $I$ consists of a single point). Moreover, $f$ is the wrong function – you’re interested in $g$ here – and the intermediate value theorem doesn’t even guarantee that for *each $r\in I$ there is some $c_r\in(a,b)$ such that $g(c_r)=r$. It does, however, guarantee that for each $r\in\big(g(b),g(a)\big)$ there is some $c_r\in(a,b)$ such that $g(c_r)=r$, and that’s what we need here.