1. ## uniform continous

I have the function f: A-R such that f is uniformly continuous. abs(f(x)) greater than or = to k greater than 0 for all x in A. How would I show that (1/f) is also uniformly continuous on A?

I also need to find an example where the conclusion fails for a function which only satisfies abs(g(x)) greater than 0 for all x in A. Would the function
g(x)=(x-1)/(x+1) where 1/g is (x+1)/(x-1) be an example since it would not be continuous at 1?

2. Here is a brief outline of what must be done.
$\displaystyle \left| {f(x)} \right| \ge k > 0\quad \Rightarrow \quad \frac{1}{{\left| {f(x)} \right|}} \le \frac{1}{k}.$

$\displaystyle \left| {\frac{1}{{f(x)}} - \frac{1}{{f(y)}}} \right| = \frac{{\left| {f(y) - f(x)} \right|}}{{\left| {f(x)} \right|\left| {f(y)} \right|}} \le \frac{{\left| {f(y) - f(x)} \right|}}{{k^2 }}.$

Now you have control over $\displaystyle {\left| {f(y) - f(x)} \right|}.$

3. Thank you. I'm just having trouble finding a suitable delta.

4. What epsilon and delta would work here?

Does delta/2 hold to make the function continuous?

5. By uniform continuity of f
$\displaystyle \left[ {\varepsilon > 0} \right]\left( {\exists \delta :\left| {x - y} \right| < \delta \Rightarrow \left| {f(x) - f(y)} \right| < \frac{{\varepsilon k^2 }}{2}} \right).$