Let $(f_n)$ be a sequence of continuous functions on $[a,b]$. Suppose that, for each $x\in[a,b]$, $(f_n(x))$ is a nonincreasing sequence of real numbers.
a) Prove that if $f_n\to 0$ pointwise on $[a,b]$, then $f_n\to 0$ uniformly on $[a,b]$.
b) Prove that if $f_n\to f$ pointwise on $[a,b]$, and if $f$ is continuous on $[a,b]$, then $f_n\to f$ uniformly on $[a,b]$.
(a) Let $ \epsilon > 0 $. For each $ x \in [0,1] $, let $ n_{x} \in \mathbb{N} $ be the smallest positive integer $ n $ such that $ 0 \leq {f_{n}}(x) < \dfrac{\epsilon}{2} $; by the continuity of $ f_{n_{x}} $, there exists an open neighborhood $ U_{x} $ of $ x $ in $ [0,1] $ such that $ f_{n_{x}}[U_{x}] \subseteq [0,\epsilon) $. As $ [0,1] $ is compact and $ \{ U_{x} ~|~ x \in [0,1] \} $ is an open cover of $ [0,1] $, we can find $ x_{1},\ldots,x_{k} \in [0,1] $ such that $ \{ U_{x_{i}} \}_{i=1}^{k} $ is a finite subcover. Then as $ (f_{n})_{n \in \mathbb{N}} $ is a monotonically decreasing sequence of functions, it follows that for all integers $ n \geq N := \max(n_{x_{1}},\ldots,n_{x_{k}}) $, we have $ {f_{n}}(x) \in [0,\epsilon) $ for all $ x \in U_{x_{i}} $, for each $ i \in \{ 1,\ldots,k \} $. In other words, for all $ n \in \mathbb{N}_{\geq N} $, we have $ \| f_{n} \|_{\infty} < \epsilon $. Therefore, as $ \epsilon $ is arbitrary, we conclude that $ (f_{n})_{n \in \mathbb{N}} $ converges uniformly to the zero-function.
(b) For this part, simply define $ g_{n} := f_{n} - f $ for each $ n \in \mathbb{N} $ and apply Part (a) to the sequence $ (g_{n})_{n \in \mathbb{N}} $.