I need hints for this particular question:
Prove that if a function $f$ is uniformly continuous on $A\subseteq \mathbb{R}$ and $|f(x)|\geq k>0$ for all $x\in A$, then the function $\frac{1}{f(x)}$ is also uniformly continuous on $A$.
My attempt: From the rough work given that $f(x)$ is uniformly continuous on $A$ for all $\epsilon> 0$, there exists a $\delta$ such that $|f(x)-f(y)|<\epsilon k^2$ when $x\in A$ and $|x-y|< \delta$, which impiles $\frac{1}{k^2} |f(x)-f(y)|< \epsilon$ for all $\epsilon >0$ and $|x-y|<\delta$.
Taking the hint from Ayman Hourieh into consideration we have using the same $\delta$ from rough work,
$\left |\frac{1}{f(x)} – \frac{1}{f(y)} \right | $ = $\left |\frac{1}{f(x)f(y)} \right | \left | f(x) – f(y) \right |\leq \frac{1}{k^2} \left | f(x)-f(y) \right |< \epsilon$ when $\left | x-y \right |<\delta$.
Is this proof okay?
Best Answer
Let $x_1, x_2 \in A$. We have: $$ \left|\frac{1}{f(x_1)} - \frac{1}{f(x_2)}\right| = \frac{\left|f(x_2) - f(x_1)\right|}{\left|f(x_1)f(x_2)\right|} $$
Now use the inequality you have to make an estimate...