Bounded: $|f(x)|\le M$ for all $x\in \mathbb E$.

Uniformly bounded: $|f_n(x)| for all $x\in\mathbb E$, $n=1, 2, 3,...$

Pointwise bounded: $|f_n(x)<\phi(x)|$ where $\phi$ is a finite valued function, and $x\in\mathbb E$, $n=1, 2, 3,...$

$K$ is compact, $f_n$ is pointwise bounded and equitcontinuous $\Rightarrow f_n$ is uniformly bounded, and $f_n$ contains a uniformly convergent subsequence.

Continuous: $f: \mathbb E\subset X\to Y, p\in\mathbb E, \forall \varepsilon>0, \exists \delta>0$ s.t. $\forall x\in\mathbb E, d_X(x, p)<\delta\Rightarrow d_Y(f(x), f(p))<\varepsilon$.

Uniformly continuous: $\forall \varepsilon>0, \exists \delta>0$ s.t. $d_X(p, q)<\delta\Rightarrow d_Y(f(p), f(q))<\varepsilon$.

Equicontinuous: $\forall \varepsilon>0, \exists \delta>0$ s.t. $d(x, y)<\delta\Rightarrow |f(x)-f(y)|<\varepsilon$.

Equicontinuous $\Rightarrow$ uniformly continuous.

Compact set, uniformly convergence $\Rightarrow$ equicontinuous.

Convergent: A sequence $\{p_n\}$ in a metric space $X, \exists p\in X$ s.t. $\forall \varepsilon>0, \exists N$ s.t. $n\ge N\Rightarrow d(p_n, p)<\varepsilon$.

Uniformly convergent (definition 1): a sequence of functions $\{f_n\}, n=1, 2, 3, ...$ converges uniformly on $\mathbb E$ to a function $f$ if $\forall \varepsilon>0, \exists N$ s.t. $n\ge N\Rightarrow |f_n(x)-f(x)|\le \varepsilon$ for all $x\in\mathbb E$.

Uniformly convergent (definition 2): $\varepsilon>0, \exists N$ s.t. $m\ge N, n\ge N, x\in\mathbb E\Rightarrow |f_n(x)-f_m(x)|\le \varepsilon$.

Pointwise convergent: $f(x)=\lim_{n\to\infty} f(x) (x\in\mathbb E)$.

If $\{f_n\}$ is uniformly convergent, then $\lim_{t\to x}\lim_{n\to\infty}f_n(t)=\lim_{n\to\infty}\lim_{t\to x}f_n(x)$.

$K$ is compact, $\{f_n\}$ is continuous and pointwise convergent, $f_n\ge f_{n+1} \Rightarrow f_n\to f$ uniformly.

$f_n\to f$ uniformly, where $f_n\in\mathcal R Rightarrow \int_a^b f d_\alpha=\lim_{n\to\infty}\int_a^b f_n d\alpha$.

$\{f_n\}$ is differentiable, $\{f_n'\}$ converge uniformly on $[a, b]\Rightarrow f_n$ converges uniformly, and $f'(x)=\lim_{n\to\infty}f_n'(x) (a\le x\le b)$.

$\{f_n\}$ is differentiable, $f'$ is uniformly bounded $\Rightarrow f_n$ is equicontinuous.