For this discussion, we will only consider functions with ${g}_{n}$ where $$\mathbb{R}\to \mathbb{R}$$

Definition 1: Uniform Convergence

The sequence
$\left\{{g}_{n}\right\}{|}_{n=1}^{\infty}$
converges uniformly to function
$g$ if for every
$\epsilon >0$ there is an integer
$N$ such that
$n\ge N$ implies
for **all**
$t\in \mathbb{R}$.

$$\left|{g}_{n}\left(t\right)-g\left(t\right)\right|\le \epsilon $$

1

Example 1

$$\forall t,t\in \mathbb{R}:\left({g}_{n}\left(t\right)=\frac{1}{n}\right)$$ Let $\epsilon >0$ be given. Then choose $N=\lceil \frac{1}{\epsilon}\rceil $. Obviously, $$\forall n,n\ge N:\left(\left|{g}_{n}\left(t\right)-0\right|\le \epsilon \right)$$ for all $t$. Thus, ${g}_{n}\left(t\right)$ converges uniformly to $0$.

Example 2

$$\forall t,t\in \mathbb{R}:\left({g}_{n}\left(t\right)=\frac{t}{n}\right)$$ Obviously for any $\epsilon >0$ we cannot find a single function ${g}_{n}\left(t\right)$ for which Equation 1 holds with $g\left(t\right)=0$ for all $t$. Thus ${g}_{n}$ is not uniformly convergent. However we do have: $${g}_{n}\left(t\right)\to g\left(t\right)\text{pointwise}$$

Conclusion:

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