# Uniform Continuous and Uniform Convergence

• Oct 28th 2007, 02:28 PM
tbyou87
Uniform Continuous and Uniform Convergence
24.13 Prove that if (fn) is a sequence of uniformly continuous functions on an interval (a,b), and if fn -> f uniformly on (a,b), the f is also uniformly continuous on (a,b). Hint use eps/3 argument.

I don't see what I need to change from the "limit of continuous functions is continuous" theorem. At the end of that proof it concludes f is continous at x0. I guess I'm not sure how to expand that to uniform continuity.

Thanks
• Oct 28th 2007, 02:51 PM
Plato
If $\varepsilon > 0$ then $\left( {\exists N} \right)\left( {\forall x \in (a,b)} \right)\left[ {n \ge N \Rightarrow \quad \left| {f_N (x) - f(x)} \right| < \frac{\varepsilon }{3}} \right]$ from uniform convergence.

Now from uniform continuity $\left( {\exists \delta > 0} \right)\left[ {\left| {x - y} \right| < \delta \Rightarrow \quad \left| {f_N (x) - f_N (y)} \right| < \frac{\varepsilon }{3}} \right]$.

$\begin{array}{l}
\\
\left| {f(x) - f(y)} \right| \le \left| {f(x) - f_N (x)} \right| + \left| {f_N (x) - f_N (y)} \right| + \left| {f_N (y) - f(y)} \right| \\
\end{array}$