Si es uniformemente continua en $f$, $\mathbb{R}$ $f(x) \ge a >0$ $g(x) = 1/f(x)^2$, entonces el $g(x)$ es uniformemente continua

Question

Llegar a este punto $$ | g(x)-g(y) | = \left|\frac{1}{f (x) ^ 2}-\frac{1}{f (y) ^ 2} \right| = \left|\frac{f (x) ^ 2 - f (y) ^ 2} {f (x) ^ 2f (y) ^ 2} \right| \leq \frac{1}{a^4} | f(x)-f(y)|\,|f(x)+f(y) |. $$ quiero usar mi $|x-y|

Answer 1

Sólo has sido un poco demasiado rápido con el uso del límite:

$$\begin{align} \left\lvert \frac{f(x)^2-f(y)^2}{f(x)^2f(y)^2}\right\rvert &= \left\lvert f(x) - f(y)\right\rvert\left\lvert \frac{f(x) + f(y)}{f(x)^2f(y)^2}\right\rvert\ &= \left\lvert f(x) - f(y)\right\rvert\left\lvert \frac{1}{f(x)f(y)^2} + \frac{1}{f(x)^2f(y)}\right\rvert\ &\leqslant \left\lvert f(x) - f(y)\right\rvert\frac{2}{a^3} \end {Alinee el} $$

te lleva a casa.

Answer 2

Un límite superior que no es tan ingenioso sino que también sirve es el siguiente:\begin{align} |g(x)-g(y)| =& \left|\frac{1}{f(x)^2} - \frac{1}{f(y)^2}\right| \ =& \left|\frac{f(x)^2 - f(y)^2}{f(x)^2f(y)^2}\right| \ =& \left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right|\cdot \left|\color{red}{f(x)\cdot f(x)} - f(y)\cdot f(y)\right| \ = & \left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right| \cdot \left|\Big[ \color{red}{\big(f(x)-f(y)\big)}+ \color{blue}{f(y)} \Big] \cdot \Big[ \color{red}{\big(f(x)-f(y)\big)}+ \color{blue}{f(y) }\Big] - f(y)\cdot f(y)\right| \ = & \left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right| \cdot \bigg| \color{red}{\Big(f(x)-f(y)\Big)\cdot \Big(f(x)-f(y)\Big)} +2\cdot\color{blue}{f(y)}\cdot \color{red}{\Big(f(x)-f(y)\Big)} \bigg| \end{align} debido a la desigualdad del triángulo, $\cfrac{1}{f(x)}\leq \cfrac{1}{a}$ y $ \cfrac{1}{f(y)}\leq \cfrac{1}{a}$:\begin{align} |g(x)-g(y)| \leq & \left|\frac{1}{a^4} \right| \cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big| \ & \qquad\qquad + \left|\frac{1}{a^3} \right| \cdot \left|\frac{1}{f(y)} \right| \cdot2\cdot\big|\color{blue}{f(y)}\big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big| \ \leq & \left|\frac{1}{a^4} \right| \cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big| +\left|\frac{2}{a^3} \right| \cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big| \ \end{align}