Showing |f(x)-f(y)| < $(\frac1\epsilon)^2*|x-y|$ for x,y on [$\epsilon^3/4,1$]

Question

f(x) in this case equals $x^\frac13$

So far I've tried setting $|x-y| < \delta$, with $\delta$ = 2$\epsilon^3$, therefore making $(1/\epsilon^2)|x-y| < 2\epsilon$, but this doesn't show that the right hand side of the title is greater than the left.

Can anyone help me with this?

Answer 1

Use the mean value theorem. If (without loss of generality) $x < y$ then there is a point $c \in (x,y)$ satisfying $$x^{1/3} - y^{1/3} = \dfrac 13 c^{-2/3} (x-y).$$ The best you can say about $c$ is that $c > \dfrac{\epsilon^3}{4}$ so that $0 < c^{-2/3} <\dfrac{\sqrt[3]{16}}{\epsilon^2}$ and thus $$|x^{1/3} - y^{1/3}| \le \frac{\sqrt[3]{16}}{3} \frac{1}{\epsilon^2} |x-y| \le\frac{1}{\epsilon^2} |x-y|. $$

Answer 2

WLOG, assume $x \geq y$.

Let's make a substitution! $$ w := x^{1/3} $$ $$ u := y^{1/3} $$ $$ |f(x) - f(y)| - 1/\epsilon^2|x-y| = (w - u) - 1/\epsilon^2 (w^3 - u^3) = (w - u)(1 - 1/\epsilon^2(w^2 +wu+u^2)) $$ Since $w-u > 0$, we need to prove that $$ 1 < 1/\epsilon^2(x^2 +xy+y^2) \Leftrightarrow \epsilon^2 < w^2 +wu+u^2 $$ $w, u \in [(\epsilon^3/4)^{1/3}, 1] = [\epsilon \cdot 4^{-1/3},1]$. Just make a bound by infimum.