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:

$\forall f, g \in X: \norm {\hat f - \hat g}_\infty \le \dfrac 3 4 \norm {f - g}_\infty$

Proof

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Sources

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• 1997: Gerard Buskes and Arnoud van Rooij: Topological Systems: From Distance to Neighborhood: $8.10$ (for the core proof)