Section 35*: Problem 2 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

When we approximate a function $f_{0}:A\to[-r,r]$ , we construct a function $g_{0}:X\to[-ar,ar]$ such that on $A$ it differs from $f$ no more than by $\begin{align*} \max\{2ar,(1-a)r\} & =r\max\{2a,1-a\}\\ & :=rb. \end{align*}$

By taking the difference $f_{0}-g_{0}$ on $A$ we obtain a new function $f_{1}:A\to[-rb,rb]$ . And we approximate it on $A$ by $g_{1}:X\to[-rab,rab]$ such that the difference on $A$ between the two functions is not greater than $rb^{2}$ .

Continuing this way we need to ensure that

$g=\sum_{k\ge0}g_{k}$ is well-defined. This holds as long as $\begin{align*} \sum_{k\ge0}rab^{k} & <\infty, \end{align*}$ or $b<1$ . This means $0<a<\tfrac{1}{2}.$
$g=f$ on $A$ . This holds as long as $rb^{k}\to+0$ as $k\to+\infty$ , or $b<1$ . And we have the same condition.

Note that the choice $a=\tfrac{1}{3}$ is optimal in the sense that $b$ as a function of $a$ reaches its minimum at $a=\tfrac{1}{3}$ , which ensures the fastest convergence of the approximation by the partial sums (not that it is important for the result itself).

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Section 35*: Problem 2 Solution

Section 35*: Problem 2 Solution