Ex3-28

Chapter 3, Question 28

Question: Prove the following assertion that is made in the proof of Theorem 3.33: If $$T:\mathbb R^n\to \mathbb R^n$$ is a Lipschitz transformation, then there is a constant $$c'>0$$ such that $$|TI|\leq c'|I|$$ for every interval $$I$$.

Proof: First we prove when $$I=Q$$ is a bounded cube in $$\mathbb R^n$$. Let $$d$$ be the side length of $$Q$$, then $$\text{diam} Q=\sqrt{\sum_{k=1}^nd^2 }=\sqrt nd$$. Let $$c$$ be the Lipschitz constant of $$T$$, so that $$|Tx-Ty|\leq c|x-y|\quad\text{ for all }x, y\in \mathbb R^n. $$

Fixed $$x\in Q $$, then for any $$y\in Q $$ we have

$$|Tx-Ty|\leq c|x-y|\leq c\cdot\text{diam }Q $$

Thus $$TQ $$ is contained in an $$n $$- dimensional sphere centered at $$Tx  $$ with radius $$c\cdot\text{diam }Q $$. Therefore, if we let $$Q' $$ be a cube with side length $$2c\cdot\text{diam }Q $$, centered at $$Tx $$, we will have $$TQ\subset Q' $$, hence

$$|TQ|\leq |Q'|=(2c\cdot \text{diam }Q)^n=(2c)^n n^{n/2} d^n =c' |Q| $$

Where $$c'=(2c)^n n^{n/2} $$. Note that this constant $$c'$$is independent of the choice of cube, even if $$|Q|=+\infty $$the inequality is still true, no matter $$|TQ|$$ is finite or not. Thus

$$|TQ|\leq c'|Q|\quad\text{ for each cube }Q\subset \mathbb R^n.$$

''This next part isn't necessary, since the book defines "interval" as "closed interval". But I suppose there is no harm in leaving it up.''

For a general interval $$I\subset \mathbb R^n$$, given a positive $$\varepsilon$$ we can choose an interval $$J $$such that $$I\subset J^\circ$$and $$|J|\leq (1+\varepsilon)|I|$$. Using Theorem 1.11, which states that every open set in $$\mathbb R^n$$ can be written as a union of nonoverlapping closed cubes, we can write $$J^\circ = \bigcup_{k=1}^\infty I_k$$ such that each $$I_k $$ is a closed cube; by Corollary 3.24 we can further write $$|J^\circ| = |J| = \sum_{k=1}^\infty |I_k|$$. Thus,

$$\begin{align}|TI|&\leq |T(J^\circ)|_e\\ &=\left|T\left(\bigcup_{k=1}^\infty I_k\right)\right|_e\\ &=\left|\bigcup_{k=1}^\infty TI_k\right|_e\\ &\leq \sum_{k=1}^\infty|TI_k|\\ &\leq c'\sum_{k=1}^\infty |I_k|\\ &=c'|J|\\ &\leq c'(1+\varepsilon)|I|.\end{align}$$Since $$\varepsilon$$is arbitrary, we conclude $$|TI|\leq c'|I|$$ for every interval $$I\subset \mathbb R^n$$.