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# Uniform Convergence of Function Sequences

For this discussion, we will only consider functions with $g n g n$ where $R→R$

Definition 1: Uniform Convergence
The sequence $g n | n =1∞ n 1 g n$ converges uniformly to function $gg$ if for every $ε>0 ε 0$ there is an integer $NN$ such that $n≥N n N$ implies
$| g n ⁢t−g⁢t|≤ε g n t g t ε$
1
for all $t∈R t$.
Obviously every uniformly convergent sequence is pointwise convergent. The difference between pointwise and uniform convergence is this: If $g n g n$ converges pointwise to $gg$, then for every $ε>0 ε 0$ and for every $t∈R t$ there is an integer $NN$ depending on $εε$ and $tt$ such that Equation 1 holds if $n≥N n N$. If $g n g n$ converges uniformly to $gg$, it is possible for each $ε>0 ε 0$ to find one integer $NN$ that will do for all $t∈R t$.

Example 1

$∀ t ,t∈R: g n ⁢t=1n t t g n t 1 n$ Let $ε>0 ε 0$ be given. Then choose $N=⌈1ε⌉ N 1 ε$. Obviously, $∀ n ,n≥N:| g n ⁢t−0|≤ε n n N g n t 0 ε$ for all $tt$. Thus, $g n ⁢t g n t$ converges uniformly to $00$.

Example 2

$∀ t ,t∈R: g n ⁢t=tn t t g n t t n$ Obviously for any $ε>0 ε 0$ we cannot find a single function $g n ⁢t g n t$ for which Equation 1 holds with $g⁢t=0 g t 0$ for all $tt$. Thus $g n g n$ is not uniformly convergent. However we do have: $g n ⁢t→g⁢t pointwise g n t g t pointwise$

Conclusion:
Uniform convergence always implies pointwise convergence, but pointwise convergence does not guarantee uniform convergence.