Originally Posted by

**fifthrapiers** Grr, I mis-read it. I proved that they are uniformly CONVERGENT, not uniformly CONTINUOUS!!

Well, I know how to prove f is unif. continuous but not f + g...

So proving f is unif. continuous...:

Let (f_n) be the sequence of functions where $\displaystyle f_n : A \rightarrow \mathbb{R}$ are unif. cont. on A and that [tex](f_n) converges unif. to f on A. Let $\displaystyle \epsilon > 0$ be given. Then, $\displaystyle \exists$ a natural number $\displaystyle N \in \mathbb{N} \ni \forall n \geq N, |f_{n}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A$ by unif. conv. of $\displaystyle f_n$ to f on A. Particularly, $\displaystyle |f_{N}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A$.

As $\displaystyle f_N$ is unif. cont on A, $\displaystyle \exists \delta > 0 \ni |f_{N}(x) - f_{N}(y)| < \frac{\epsilon}{3}$ whenever $\displaystyle x,y \in A$ satisfy $\displaystyle |x-y| < \delta$. Then we have for this $\displaystyle \delta$, suppose $\displaystyle x,y \in A$ satisfy $\displaystyle |x - y| < \delta$ and consider:

\begin{eqnarray}

|f(x) - f(y)| &=& |f(x) - [f_{N}(x) - f_{N}(y)] - [f_{N}(y) - f_{N}(y)] - f(y)|\\

&=& |[f(x) - f_{N}(x)] + [f_{N}(x) - f_{N}(y)] + [f_{N}(y) - f(y)]|\\

&\leq& |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(y)| + |f_{N}(y) - f(y)|\\

&<& \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{eqnarray}

Thus, f is unif. cont. on A.

(NOTE: I get LaTeX syntax error on the last part (for the eqnarray stuff, not sure why) so I left it without [tex] so you can see the general jist. I think it's easy to read just in latex format.

But, I have no idea how to show f + g will be unif cont...