$\{-\infty\} \cup \mathbb{R} \cup \{\infty\}.$

The supremum of a set $S \subset \mathbb{R}$ is a value $a \in \mathbb{R}_e$ such that $\forall s\in S,\ s \leq a$ and if $b \in \mathbb{R}_e$ such that $\forall s\in S,\ s \leq b$, then $a \leq b$.

Supremum is essentially the “least upper bound” in a set. It always exists, and is called $\sup S$. The opposite of supremum is the infinimum.

The infinimum of a set $S \subset \mathbb{R}$ is a value $a \in \mathbb{R}_e$ such that $\forall s\in S,\ s \geq a$ and if $b \in \mathbb{R}_e$ such that $\forall s\in S,\ s \geq b$, then $a \geq b$.

The infinimum is the “greatest upper bound”. Like the supremum, it always exists, and it is denoted $\inf S$. Supremum and Infinimum can be applied to scalar function $f: S\to \mathbb{R}$ by letting

$\sup_{x\in S} f(x) = \sup \{f(x) | x\in S \}.$

Let $V$ be a vector space of $\mathbb{R}$, then $\|\cdot\|: V \to \mathbb{R}$ is a norm if $\forall \boldsymbol{x},\boldsymbol{y}\in V, \alpha \in \mathbb{R}$,

$\|\boldsymbol{x}\| \geq 0, \qquad \boldsymbol{x} = 0 \Leftrightarrow \|\boldsymbol{x}\| = 0, \qquad \|\alpha \boldsymbol{x}\| = |\alpha|\|\boldsymbol{x}\|, \qquad \|\boldsymbol{x} + \boldsymbol{y}\| \leq \|\boldsymbol{x}\| + \|\boldsymbol{y}\|.$

A normed space $(V, \|\cdot\|)$ is a vector space which is equipped with a norm $\|\cdot\|: V \to \mathbb{R}$.

If we have an operator $A$ which takes vectors from normed space $(X, \|\cdot\|_X)$ and outputs vectors in normed space $(Y, \|\cdot\|_Y)$, then we can define another norm on the vector space of operators from $X\to Y$.

Let $A:X\to Y$ be an operator between normed spaces $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$, then the induced norm of $A$ is

$\|A\|_i = \sup_{\|\boldsymbol{x}\|_X \neq 0} \frac{\|A\boldsymbol{x}\|_Y}{\|\boldsymbol{x}\|_X}$

Two norms $\|\cdot\|$ and $|||\cdot|||$ on a vector space $V$ are said to be equivalent if $\exists k_1, k_2 > 0$ such that

$\forall \boldsymbol{x}\in V,\ k_1\|\boldsymbol{x}\| \leq |||\boldsymbol{x}||| \leq k_2\|\boldsymbol{x}\|$

If $V$ is a finite dimensional vector space if and only if all norms of $V$ are equivalent.

Let $(V, \|\cdot\|)$ be a normed space, $a\in \mathbb{R}$, $a > 0$, $\boldsymbol{x}_0\in V$, then the open ball of radius $a$ centered around $x_0$ is given by

$B_a(\boldsymbol{x}_0) = \{ \boldsymbol{x} \in V \ | \ \|\boldsymbol{x} - \boldsymbol{x}_0\| < a \}$

A set $S\subset V$ is open if $\forall \boldsymbol{s}_0\in S,\ \exists \epsilon > 0$ such that $B_\epsilon(\boldsymbol{s}_0) \subset S$.

A set $S$ is closed if $\sim S$is open.

A sequence of points $\boldsymbol{x}_k$ in normed space $(V, \|\cdot\|)$ converges to a point $\bar{\boldsymbol{x}}$ if

$\forall \epsilon > 0,\ \exists N < \infty,\ \text{ such that } \forall k \geq N, \|\boldsymbol{x}_k - \bar{\boldsymbol{x}}\| < \epsilon$

A sequence $\boldsymbol{x}_k$ is cauchy if

$\forall \epsilon > 0,\ \exists N < \infty \text{ such that } \forall n,m \geq N, \|\boldsymbol{x}_m - \boldsymbol{x}_n\| < \epsilon$

If $\boldsymbol{x}_n$ is a convergent sequence, then $\boldsymbol{x}_n$is a also a Cauchy sequence.

A normed space $(V, \|\cdot\|)$ is complete if every Cauchy sequence converges to a point in $V$.

$(C[a,b], \|\cdot\|_\infty)$, the set of continuously differentiable functions on the closed interval $[a,b]$ equipped with the infinity norm.

A point $\boldsymbol{x}^*$ is a fixed point of a function $P:X\to X$ if $P(\boldsymbol{x}^*)=\boldsymbol{x}^*$.

A function $P:X\to X$ is a contraction if $\exists c\in\mathbb{R}, 0 \leq c < 1$ such that

$\forall \boldsymbol{x},\boldsymbol{y}\in X,\ \|P(\boldsymbol{x}) - P(\boldsymbol{y})\| \leq c \|\boldsymbol{x}-\boldsymbol{y}\|$

$\boldsymbol{x}_{k+1} = P(\boldsymbol{x}_k)$

where $P$ is a function $P:X\to X$. When does this sequence converge, and to what point will it converge?

If $P:X\to X$ is a contraction on the Banach space $(X, \|\cdot\|)$, then there is a unique $\boldsymbol{x}^*\in X$ such that $P(\boldsymbol{x}^*) = \boldsymbol{x}^*$ and $\forall \boldsymbol{x}_0\in X$, the sequence $\boldsymbol{x}_{n+1} = P(\boldsymbol{x}_n)$ converges to $\boldsymbol{x}^*$.

A function $h:V\to W$ on normed spaces $(V, \|\cdot\|_V)$ and $(W, \|\cdot\|_W)$ is continuous at a point $\boldsymbol{x}_0$ if $\forall \epsilon > 0, \exists \delta > 0$ such that

$\|\boldsymbol{x}-\boldsymbol{x}_0\|_V < \delta \implies \|h(\boldsymbol{x}) - h(\boldsymbol{x_0})\|_W < \epsilon$

Continuity essentially means that given an $\epsilon-$ball in $W$, we can find a $\delta-$ball in $V$ which is mapped to the ball in $W$. If a function is continuous at all points $\boldsymbol{x}_0$, then we say the function is continuous.

A function $h:V\to W$ on normed spaces $(V, \|\cdot\|_V)$ and $(W, \|\cdot\|_W)$ is Lipschitz continuous at $\boldsymbol{x}_0\in V$ if $\exists r > 0$ and $L < \infty$ such that

$\forall \boldsymbol{x}, \boldsymbol{y}\in B_r(\boldsymbol{x}_0),\ \|h(\boldsymbol{x}) - h(\boldsymbol{y})\|_W \leq L \|\boldsymbol{x} - \boldsymbol{y}\|_V$

A good interpretation of Lipschitz Continuity is that given two points in a ball around $\boldsymbol{x}_0$, the slope of the line connecting those two points is less than $L$. It means that the function is growing slower than linear for some region around $\boldsymbol{x}_0$. Lipschitz continuity implies continuity. If a function is lipschitz continuous with respect to one norm, it is also lipschitz continuous with respect to all equivalent norms.

When the function $h$ is a function on $\mathbb{R}^n$ and is also differentiable, then Lipschitz continuity is easy to determine.

For a differentiable function $h:\mathbb{R}^n\to\mathbb{R}^n$,

$\exists r>0, L < \infty, \boldsymbol{x}_0\in\mathbb{R}^n,\ \forall \boldsymbol{x}\in B_r(\boldsymbol{x}_0), \left\lvert\left\lvert\frac{\partial h}{\partial \boldsymbol{x}}\right\rvert\right\rvert_2 \leq L$

implies Lipschitz Continuity at $\boldsymbol{x}_0$.

A function $h:\mathbb{R}\to V$ is piecewise continuous if $\forall k\in \mathbb{Z}$, $h:[-k, k] \to V$ is continuous except at a possibly finite number of points, and at the points of discontinuity $t_i$, $\lim_{s\to0^+} h(t_i+s)$ and $\lim_{s\to0^-}h(t_i+s)$exist and are finite.