# proof of that a continuous function has a fixed point

Consider $g(x)=f(x)-x$. $f(a)\ge a$ so $g(a)=f(a)-a\ge 0$. $f(b)\le b$ so $g(b)=f(b)-b\le 0$. By the Intermediate Value Theorem, since $g$ is continuous and $0\in[g(b),g(a)]$ there exists $c\in[a,b]$ such that $g(c)=f(c)-c=0$, so $f(c)=c$ for some $c\in[a,b]$."

My question is why $f(a)\ge a$ and $f(b)\le b$??

Thanks a lot!

Its a part of the assumption in the question, as far as I can tell. You assume $f(a) \geq a$ and $f(b) \leq b$, and then, based on the definition of $g$, conclude what the IVT allows you to conclude. You're also assuming implicitly that $f$ is continuous, at least on the domain that the question is concerned with, otherwise $g$ won't be continuous, as it depends on $f$, and one wouldn't be able to apply the IVT. So what the proof is basically saying is "if $f$ is a continuous function that satisfies these assumptions: $f(a) \geq a$ and $f(b) \leq b$, and if $g$ is defined such that... then by the IVT, we can conclude that ... " Also, this looks like its from Spivak. Am I right?
• Thank you! But in my exercise written "Prove that if $f$ is a continuous function $f:[a,b]\rightarrow [a,b]$, then exists $c\in[a,b]$, such that $f(c)=c$".No its not from Spivak. – Tina Dec 10 '12 at 12:36
• @Tina $f$ maps $[a,b]$ to $[a,b]$ so for any $x$ in the domain, $f(x)\in[a,b]$ in particular $f(a)\in[a,b]$, which implies $f(a)\ge a$. Similarly $f(b)\le b$. – David Mitra Dec 10 '12 at 12:48
• If $f$ maps into $[a,b]$, it is true for all $x\in[a,b]$ that $a \leq f(x) \leq b$, and, therefore, also for $a$ and $b$. – mkl Dec 10 '12 at 12:50