If $f$ is uniformly continuous on $\mathbb{R}$, $f(x) \ge a >0$ and $g(x) = 1/f(x)^2$, then $g(x)$ is uniformly continuous

Question

I get to the this point $$ |g(x)-g(y)| = \left|\frac{1}{f(x)^2} - \frac{1}{f(y)^2}\right| = \left|\frac{f(x)^2 - f(y)^2}{f(x)^2f(y)^2}\right| \leq \frac{1}{a^4} |f(x)-f(y)|\,|f(x)+f(y)|. $$ I want to use my assumption $|x-y| < \delta$, but I don't know how to do that given $|f(x)-f(y)|$.

Answer 1

You've just been a little too fast with using the bound:

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

takes you home.

Answer 2

An upper bound that is not as resourceful but also serves is as follows: \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*} Due to do the triangle inequality, $\cfrac{1}{f(x)}\leq \cfrac{1}{a}$ and$ \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*}