Prove that a continuous function has a fixed point

Suppose that $a < b$ and $f: [a,b] \to\mathbb{R}$ is a continuous function such that the range of $f$ contains $[a,b]$. Prove that $f$ has a fixed point.

I've already proved the case of $[0,1]$ but I don't know how to do the general case. Help.

Proof for $[0,1]$:

Let $g(x)=f(x)-x$, then $g(0)=f(0)\geq 0$, and $g(1)=f(1)-1\leq 0$.

Otherwise, by using IVT, there must exist some $c$ in $[0,1]$ such that $g(c)=0$ meaning that $f(c)=c$.

• Maybe you could write out your proof for the $[0,1]$ case? The general case should be similar. Dec 8, 2015 at 5:22
• Okay, how can I generalize this Dec 8, 2015 at 5:26
• I'm not convinced that $g(0)\geq0$ and $g(1)\leq0$... Dec 8, 2015 at 5:32
• well the range is contained in [0,1] Dec 8, 2015 at 5:34
• Use g (x) = f (x-a/b-a) Dec 8, 2015 at 5:34

By simply assuming that the range of $f$ contains $[a,b]$ pick $x_1,x_2$ such that $f(x_1)=\min_{[a,b]}f(x)$ and $f(x_2)=\max_{[a,b]}f(x)$ (which exist by Extreme Value Theorem). Since $f(x_1)\leq a$ and $f(x_2)\geq b$, apply your reasoning with $g(x)=f(x)-x$ on $[x_1,x_2]$ (or $[x_2,x_1]$) with the IVT.

• Wouldn't one have to also prove f is bounded? If [a,b] in range why not chose x1 X2 such that f (x1) =a and f (x2) = b if x2 < x1 you can do minor tweaking. Dec 8, 2015 at 5:49
• @fleablood, $f$ is continuous on a compact interval $[a,b]$, so it is bounded, uniformly continuous, etc... Dec 8, 2015 at 5:51
• True, but It also has a fixed point as it maps from a closed interval to a range containing the closed interval. If were proving basic and well known theorems we should be careful we know what we are stating and why. Dec 8, 2015 at 6:45

I'm not sure why you are stuck on [a,b]. The IVT applies to [a,b] just as it does to [0,1].

Let $x_1$ be such that $f(x_1) = a$ (as [a,b] is in range of f we know such $x_1$ exists. Let $x_2$ be such that $f(x_2) = b$.

Let $g(x) = f(x) - x$. $g(x_1) = a - x \le 0$ $g(x_2) = b - x \ge 0$. (as $x_1$ and $x_2$ are between a and b).

So by IVT there is an x between $x_1$ and $x_2$ where g(x) = 0. And so f(x) = x.

• Arggg. This was supposed to be a comment! Not an answer. Dec 8, 2015 at 5:37
• You can delete this answer and repost it as a comment. Dec 8, 2015 at 6:21
• Well, it sort of answers. The OP said s/he could only to the [0,1] and wanted hints on how to do it for [a,b]. To point out there is utterly no difference between [0,1] and [a,b] at all counts as a hint. Dec 8, 2015 at 6:40

If you have proved it for the case $[0,1]$ it should be fairly straight forward to use that for the general case by transformation. Consider $\phi(\theta) = f(a+(b-a)\theta)/(b-a)$ which will be a function from $[0,1]$ to $[0,1]$.

However your proof has issues, but the idea is right. The trick here is that you should use the fact that the function must take it's extreme values on the interval. That is you have $\xi_a$ and $\xi_b$ such that $f(\xi_a) = a$ and $f(\xi_b)=b$ and consider $f(x)-x$ on the interval $[\xi_a, \xi_b]$ (which should be understood as the closed interval between the end points regardless of which is larger). For example you have $f(\xi_a)-\xi_a = a - \xi_a \le a-a = 0$ with equality only if $\xi_a=a$ (which would mean that you have your fix point anyway). Similar reasoning applies to $\xi_b$. Then it's just a matter of applying the theorem of intermediate values.