1. ## Uniform Convergence

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

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

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

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

I need a walkthrough with these, thanks

2. (2)

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

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

Now divide by $\displaystyle |g(x)|$: $\displaystyle \left|\frac{f_n(x)}{g(x)}-\frac{f(x)}{g(x)}\right|<m\cdot{\epsilon}\cdot{\fr ac{1}{m}}=\epsilon$ by (1)

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

(3)

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

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

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

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

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

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

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

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

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

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

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

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

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

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