1. ## Uniform Convergence

(1) Assume that $(f_n)$ converges uniformly to $f$ on $A$ and that each $f_n$ is uniformly continuous on $A$. Prove that $f$ is uniformly continuous on $A$

(2) Assume $(f_n)$ converges uniformly to $f$ on a compact set $K$, and let $g$ be a continuous function on $K$ satisfying $g(x) \not= 0$. Show $f_n/g)$ converges uniformly on $K$ to $f/g$

(3) Assume $(f_n)$ and $(g_n)$ are uniformly convergent sequences of functions.

(a) Show that $(f_n + g_n)$ is a uniformly convergent sequence of functions.
(b)Give an example to show that the product $(f_ng_n)$ may not converge uniformly.
(c) Prove that if there exists an $M>0$ such that $|f_n| \leq M$ and $g_n \leq M$ for all $n \in \mathbb{N}$, then $(f_ng_n)$ does converge uniformly.

I need a walkthrough with these, thanks

2. (2)

Since $g(x)$ is continuous on A it has a min., let $m=\min_{x\in A} [|g(x)|]>0$ we have $0<\frac{1}{|g(x)|}\leq \frac{1}{m}$ (1)

By definition for all $\epsilon>0$ there is a natural number $n_{\epsilon}$ such that if $n\geq{n_{\epsilon}}$ we have: $|f_n(x)-f(x)| for all $x\in A$

Now divide by $|g(x)|$: $\left|\frac{f_n(x)}{g(x)}-\frac{f(x)}{g(x)}\right| by (1)

So by definition $\frac{f_n}{g}$ converges uniformly to $f/g$ on A

(3)

(a) Suppose $f_n$ converges uniformly to $f$ on a interval A and $g_n$ converges uniformly to $g$ also on A

By definition for all $\epsilon>0$ there is a natural number $n_{\epsilon1}$ such that if $n\geq{n_{\epsilon1}}$ we have: $|f_n(x)-f(x)|<\frac{\epsilon}{2}$ for all $x\in A$

Again for all $\epsilon>0$ there is a natural number $n_{\epsilon2}$ such that if $n\geq{n_{\epsilon2}}$ we have: $|g_n(x)-g(x)|<\frac{\epsilon}{2}$ for all $x\in A$

Set $n_{\epsilon}=\max[n_{\epsilon1},n_{\epsilon2}]$

Then for all $\epsilon>0$ there is a natural number $n_{\epsilon}$ such that if $n\geq{n_{\epsilon}}$ we have: $|f_n(x)-f(x)|<\frac{\epsilon}{2}$ and $|g_n(x)-g(x)|<\frac{\epsilon}{2}$ for all $x\in A$

Summing $|f_n(x)+g_n(x)-(f(x)+g(x))|\leq |f_n(x)-f(x)|+|g_n(x)-g(x)|<\epsilon$for all $x\in A$

Thus (definition) we have that $(f_n+g_n)$ converges uniformly to $(f+g)$ on A

(1) Assume that $(f_n)$ converges uniformly to $f$ on $A$ and that each $f_n$ is uniformly continuous on $A$. Prove that $f$ is uniformly continuous on $A$
You probably know the theorem that a uniform limit of continuous functions is continuous. (If not, look it up in a textbook.) You should be able to adapt that proof to show that a uniform limit of uniformly continuous functions is uniformly continuous.

(3) Assume $(f_n)$ and $(g_n)$ are uniformly convergent sequences of functions.
...
(b)Give an example to show that the product $(f_ng_n)$ may not converge uniformly.
Let $f_n(x) = g_n(x) = x+(1/n)\;\;(x\in\mathbb{R})$.

(1) Assume that $(f_n)$ converges uniformly to $f$ on $A$ and that each $f_n$ is uniformly continuous on $A$. Prove that $f$ is uniformly continuous on $A$
Let $\varepsilon >0$
$f_n$ unif cont on $A$ implies $\exists \delta >0$ such that for $u,v \in A$ then $|f_n(u) - f_n(v)| < \frac{\varepsilon}{3}$ whenever $|u - v| < \delta$

$(f_n)$ converges uniformly to $f$ on $A$ implies $\exists \, N$ such that for all $n \geq N$ and for all $x \in A$, $|f_n(x) - f(x)| < \frac{\varepsilon}{3}$

thus, if $|u - v| < \delta$, we have

$|f(u) - f(v)| = |f_n(v) + f(u) - f(v) - f_n(v) + f_n(u) - f_n(u)|$

$\leq |f_n(v) - f(v)| + |f(u) - f_n(u)| + |f_n(u) - f_n(v)|$

$< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$