Weierstrass's Theorem/Lemma 3

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}$

The Contraction Mapping Theorem assures existence of a unique $h \in X$ with $\hat h = h$.

For every $n \in \N$ and $k \in \set {1, 2, 3, 4, \ldots, 3^n}$, the following inequality holds:
 * $\size {\map h {\dfrac {k - 1} {3^n} } - \map h {\dfrac k {3^n} } } \ge 2^{-n}$

Proof
For all $n \in \N$ and $k \in \set {1, 2, 3, \ldots, 3^n}$:
 * $1 \le k \le 3^n \implies 0 \le \dfrac{k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1} } \le \dfrac 1 3$
 * $3^n < k \le 2 \cdot 3^n \implies \dfrac 1 3 \le \dfrac {k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1}} \le \dfrac 2 3$
 * $2 \cdot 3^n < k \le 3^{n + 1} \implies \dfrac 2 3 \le \dfrac {k - 1} {3^{n + 1} } < \dfrac k {3^{n + 1} } \le 1$