uniformly convergence on compact metric space Let $K$ be a compact metric space. Let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous functions on $K$ such that $f_n$ converges to a function $f$ pointwise on $K$. 
on Walt. Rudin's book Principles of mathematical analysis, 7.13, if we assume 
(1). $f$ is continuous; 
(2). $f_n(x)\geq f_{n+1}(x)$ for all $x\in K$ and all $n$; 
then it is proved that $f_n$ converges to $f$ uniformly on $K$. 
Is there counterexample satisfying (1) but not (2)? And satisfying (2) but not (1)? 
 A: $f_n(x)=x^n$ on $[0,1]$. Satisfies (2) but not (1).
$f_n(x)=nx(1-x)^n$  on $[0,1]$- satisfies 1 but not 2, and the convergence is not uniform.
A: Yes to both.
If we assume (2) but not (1), then let
$$
f_n(x) =
\begin{cases}
1 & \text{if } x = 0 \\
1-nx & \text{if } 0 \le x \le \frac{1}{n} \\
0 & \text{if } \frac{1}{n} \le x
\end{cases}
$$
$f_n$ is continuous, $f_n(x) \ge f_{n+1}(x)$, and $f_n$ and converges pointwise to the discontinuous characteristic function $\chi_{\{0\}}$.
However, the convergence is not uniform because if $\epsilon < 1$,
then for all $n$ there is a value of $x > 0$ with $f_n(x) > \epsilon$.  Alternatively, you know the convergence cannot be uniform because the uniform limit of continuous functions is continuous.
If we assume (1) but not (2), then for $n \ge 2$ let
$$
f_n(x) =
\begin{cases}
nx & \text{if } 0 \le x \le \frac{1}{n} \\
2 - nx & \text{if } \frac{1}{n} \le x \le \frac{2}{n} \\
0 & \text{if } \frac{2}{n} \le x
\end{cases}
$$
The functions are continuous and converge to the continuous $0$, but the convergence is again not uniform.
