If $f_n \rightarrow f$ and $f$ is uniformly continuous, then does $f_n \rightarrow f$ uniformly? Let $f_n$ be a sequence of functions, $f_n : S\rightarrow T$, not necessarily continuous and suppose that $f_n \rightarrow  f$ as $n \rightarrow \infty$.  Let $f$ be uniformly continuous.  I.e. for all $\epsilon \gt 0, \exists \delta \gt 0$ such that $\forall x\in S, |f(x) - f(y)| \lt \epsilon$ whenever $|x - y| \lt \delta, y\in S$.
We want to show that for all $\epsilon \gt 0, \ \exists N$ such that for all $x \in S, n\gt N$, we have $|f_n(x) - f(x)|\lt \epsilon$.
If we need it, $S$ is compact and $T = \mathbb{C}$.
Since $S$ is compact, by the extreme value theorem there exists $c,d \in S$ such that $f(c) \leq f(x) \leq f(d), \ \forall x \in S$.
 A: This is false. For instance, consider the sequence of functions (which are even continuous)
$$f_n(x) = \begin{cases}{nx   \quad(0 \leq x \leq 1/2n) \\ 1/2-nx \quad (1/2n \leq x \leq 1/n) \\ 0 \quad ( 1/n \leq x  \leq 1) }\end{cases}$$
it converges to $0$ on the compact space $[0,1]$, but not uniformly.
A: No, e.g. Let $S=[0,1]$, and $f_n$ be the characteristic function of $\{1/n\}$, which converges to the zero function, but with $|f_n(1/n)-f(1/n)|=1$ for all $n$ showing that the convergence is not uniform.  You could modify this to make each $f_n$ continuous (differentiable, etc.) if desired.
A: People make examples pointlessly complicated.  How about this one:
$$
f_n(x) = \arctan(x-n).
$$
Then $f_n(x) \to-\pi/2$ as $n\to\infty$, and the function constantly equal to $-\pi/2$ is uniformly continuous, but the sequence does not converge uniformly.
A: Consider $f_n(x) = x^n$ on $(0,1)$. Then $f_n(x) \to 0$, and the function $x \mapsto 0$ is definitely uniformly continuous, but clearly the convergence is not uniform (since $f_n (\frac{1}{\sqrt[n]{2}}) = \frac{1}{2}$).
A: This is not going to work out well for you; sorry. For example, let $f_n\colon [0,1]\to [0,1]$ be given by
$$f_n(x)=\begin{cases}
0 & 0<x<\frac 1 n \\
1 & (x=0) \lor \left(\frac1n\le x\right).
\end {cases}
$$
Then $(f_n)$ converges pointwise to $x\mapsto 1$, but it does not converge uniformly.
