Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 14

Problem

Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x \in I$.

Answer

In Rudin's book, the intermediate value theorem states that

Let $f$ be a continuous real function on the interval $[a,b]$. If $f(a) < f(b)$ and if $c$ is a number such that $f(a) < c < f(b)$, then there exists a point $x \in (a,b)$ such that $f(x)=c$.

Suppose $f(x) \neq x$ for all $x \in I$. Define a function $g(x) = f(x) -x$. Then $g$ is continuous and $g(0) = f(0) >0$, $g(1)=f(1) - 1 <0$. By the intermediate value theorem, there exists a point $c \in (0,1)$ such that $g(c)=0$  $\Rightarrow$  $g(c) = c$, which is a contradiction.

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