$$\mathbb{N}$$: natural number (e.g. $$1,2,3,4$$)

$$\mathbb{Z}$$: integer (e.g. $$0, \pm1,\pm2,\pm3,\pm4$$)

$$\mathbb{Q}$$: rational number (e.g. $$\frac{1}{3},2,3.1,4.2$$)

$$\mathbb{P}$$: irrational number (e.g. $$\sqrt{5},log 3, \pi$$)

$$\mathbb{R}$$: real number (e.g. $$\sqrt{5},\pi,e,2.3$$)

$$\exists!$$: Solution or results exists unique

$$\exists$$: Solution or results are exist

$$\nexists$$: Solution or result is not existed

$$\forall$$: for all cases, for all exmaples, or for all areas

$$\Rightarrow$$: A$$\Rightarrow$$B means that A implies B

$$A\equiv B(mod C)$$: If A mod C, residual is B

$$\mathbb{D}: $$Domain

f(x) := B: f(x) defined by B

[a,b] := {$$x\in\mathbb{R}$$ := $$a\leq x \leq b$$}: closed interval

(a,b) := {$$x \in \mathbb{R}$$ := a< x< b}: open interval

Assume [a,b] interval

$$a \neq b$$: Nondegenerate

$$a = b$$: Degenerate

f(x) := g(x), $$\forall x \in \mathbb{N}$$ $$\iff$$ f(1)=g(1) and assume f(k)=g(k) then f(k+1)=g(k+1)

f := $$\mathbb{R}$$ $$\rightarrow$$ $$\mathbb{I}$$

$$\mathbb{I}$$ = [a,b] or [a,b) or (a,b] or (a,b) $$\iff$$ f and $$\mathbb{I}$$ bounded on [a,b] or [a,b) or (a,b] or (a,b)

f unbounded $$\iff$$ f is not bounded

f := $$\mathbb{R} \rightarrow \mathbb{I}$$

$$\exists M \in \mathbb{R}$$ s.t. $$ x \leq M$$ for all $$x \in \mathbb{I}$$ $$\iff$$ f and $$\mathbb{I}$$ is bound above

$$\exists M \in \mathbb{R}$$ s.t. $$ x \geq M$$ for all $$x \in \mathbb{I}$$ $$\iff$$ f and $$\mathbb{I}$$ is bound below

f := $$\mathbb{R} \rightarrow \mathbb{I}$$

Let M be upper bound set

x $$\in M$$ s.t. $$S \leq y$$ for all y $$\in M$$ $$\iff$$ x is supremum of f or $$\mathbb{I}$$

f := $$\mathbb{R} \rightarrow \mathbb{I}$$

Let M be lower bound set

x $$\in M$$ s.t. $$S \geq y$$ for all y $$\in M$$ $$\iff$$ x is infimum of f or $$\mathbb{I}$$

Let f: X $$\rightarrow$$ Y and $$x_1, x_2$$ $$\in$$ X

f: 1-1 (injection) $$\iff$$ $$x_1$$ = $$x_2$$ implies f($$x_1$$) = f($$x_2$$)

f: onto (surjection) $$\iff$$ $$\forall y \in Y$$, $$\exists$$ $$\in$$ X s.t. y = f(x)

f: bijection $$\iff$$ f: 1-1 and onto $$\iff$$ $$x_1$$ = $$x_2$$ implies f($$x_1$$) = f($$x_2$$) and f($$x_1$$) = f($$x_2$$) implies $$x_1$$ = $$x_2$$

Let $$\mathbb{E}$$ be a set

$$\mathbb{E}$$ is countable $$\iff$$ $$\exists$$ a bijection f: {1,2,3,$$\cdots$$,n} $$\rightarrow$$ $$\mathbb{E}$$

$$\mathbb{E}$$ is countable $$\iff$$ $$\exists$$ a bijection f: $$\mathbb{N}$$ $$\rightarrow$$ $$\mathbb{E}$$

Function f: $$\mathbb{N}$$ $$\rightarrow$$ $$\mathbb{R}$$ where $$x_n$$ = f(n)

Notation: $$\{x_n\}_{n\in\mathbb{N}}$$, {$$x_n$$}

Let I be an interval as [a,b] and f: I $$\rightarrow$$ $$\mathbb{R}$$

f is convex $$\iff$$ f($$\lambda$$x + (1-$$\lambda$$)y) $$\leq$$ $$\lambda f(x) + (1-\lambda)f(y)$$

f is strict convex $$\iff$$ f($$\lambda$$x + (1-$$\lambda$$)y) < $$\lambda f(x) + (1-\lambda)f(y)$$

f is strong convex $$\iff$$ f($$\lambda$$x + (1-$$\lambda$$y) - $$\alpha$$$$\lVert$$$$\lambda$$x + (1-$$\lambda$$)y)$$\rVert^2$$ $$\leq$$ $$\lambda$$f(x) + (1-$$\lambda$$)f(y) -$$\lambda\alpha\lVert x\rVert^2$$ - (1 - $$\lambda$$)$$\alpha\lVert y \rVert^2$$

Let I be an interval as [a,b] and f: I $$\rightarrow$$ $$\mathbb{R}$$

f is convex $$\iff$$ f($$\lambda$$x + (1-$$\lambda$$)y) $$\geq$$ $$\lambda f(x) + (1-\lambda)f(y)$$

f is strict convex $$\iff$$ f($$\lambda$$x + (1-$$\lambda$$)y) > $$\lambda f(x) + (1-\lambda)f(y)$$

f is strong convex $$\iff$$ f($$\lambda$$x + (1-$$\lambda$$$y) - $$\alpha$$$$\lVert$$$$\lambda$$x + (1-$$\lambda$$)y)$$\rVert^2$$ $$\geq$$ $$\lambda$$f(x) + (1-$$\lambda$$)f(y) -$$\lambda\alpha\lVert x\rVert^2$$ - (1 - $$\lambda$$)$$\alpha\lVert y \rVert^2$$