Chapter 7 #11 : Intersection of Continuous Function f with Diagonal x On The Interval [0, 1]

11. Suppose that $f$ is a continuous function on $[0, 1]$ and that $f(x)$ is in $[0, 1]$ for each $x$. Prove that $f(x) = x$ for some number $x$.

Proof:

We know that $f(x) \in [0, 1]$ for all $x \in [0, 1]$. Then we have:

(1)

\begin{align} 0 \le f(0) \le 1 \;\; \text{and} \;\; 0 \le f(1) \le 1 \end{align}

Now consider the function $h = f - x$, which is certainly continuous on $[0, 1]$ (by Theorom 6.1 (1)), and consider these two values of $h(x)$:

(2)

\begin{align} h(0) = f(0) - 0 \ge 0 \quad \text{,from (1)} \\ h(1) = f(1) - 1 \le 0 \quad \text{,from (1)} \end{align}

Then we have $h(0) \le 0 \le h(1)$ and by Theorem 7.1 there exists some $x$ such that $h(x) = 0$. But this means that $f(x) = x$. $Q.E.D.$