Weierstrass's Theorem/Lemma 2

Lemma for Weierstrass's Theorem
Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$.

Let $\norm {\,\cdot \,}_\infty$ denote the supremum norm on $C$.

Let $X$ consist of the $f \in C \closedint 0 1$ such that:
 * $\map f 0 = 0$
 * $\map f 1 = 1$
 * $\forall x \in \closedint 0 1: 0 \le \map f x \le 1$

For every $f \in X$, define $\hat f: \closedint 0 1 \to \R$ as follows:
 * $\map {\hat f} x = \begin {cases}

\dfrac 3 4 \map f {3 x} & : 0 \le x \le \dfrac 1 3 \\ \dfrac 1 4 + \dfrac 1 2 \map f {2 - 3 x} & : \dfrac 1 3 \le x \le \dfrac 2 3 \\ \dfrac 1 4 + \dfrac 3 4 \map f {3 x - 2} & : \dfrac 2 3 \le x \le 1 \end {cases}$

Then:
 * $\hat \cdot: X \to X$ is a contraction mapping.

Furthermore, we have the following inequality:
 * $\forall f, g \in X: \norm {\hat f - \hat g}_\infty \le \dfrac 3 4 \norm {f - g}_\infty$