Norms: a short overview

Recall: A $\textit{norm}$ (in the sense of analysis) is a map from a vector space to the non-negative real numbers
$|| \cdot ||: V \longrightarrow [0,\infty)$ with the following 3 properties :
(i) $||x|| \geq 0 \quad \forall{x \in{V}}$ and $||x||=0 \Longleftrightarrow x=0$ (positive-definite)
(ii) $||\alpha x||=|\alpha| \cdot ||x||$ $\forall \alpha\in{F}$, $x\in{V}$ (homogeneity)
(iii) $||x+y|| \leq ||x||+||y||$ (triangle inequality)

Any vector space equipped with a mapping satisfying (i)-(iii) above is called a $\textit{normed space}$

Examples:
1. The real numbers as a vector space together with the absolute value function $(\mathbb{R},|\cdot |)$ form a classic example of a normed space. Many other mappings that are defined by taking the absolute value of a real number inherit (i)-(iii) above and are thus a norm.
2. Consider the vector space of real-valued continuous functions defined on a closed interval $[a,b]$ (denoted $C[a,b]$). The norm $$||f||_p = \left(\int_a^b|f(x)|^p dx \right)^{\frac{1}{p}} \quad p\in{[1,\infty)}$$ When $p=1$, the above defines both a norm and a linear functional by the linearity and triangle inequality of the integral. The triangle inequality for the cases $ 1 < p < \infty $ is called Minkowski's inequality. For the $ p= \infty $ case, $$||f||_\infty \equiv \inf_{a \leq x \leq b}\{M \geq 0 : |f(x)| \leq M \quad a.e. \}$$ Since real-valued continuous functions on a closed bounded interval attain their sup and inf, the above norm is equivalent to $$||f||_{max} \equiv \max_{a \leq x \leq b} |f(x)| $$ When $C[a,b]$ is equipped with the norm $||f||_{max} \equiv \max_{a \leq x \leq b} |f(x)|$ this normed space is $\textit{complete}$: Sequences which are Cauchy with respect to the max norm also converge to an element in $C[a,b]$.

3. The mapping $||f||_1 \equiv \int_X|f(x)|d\mu(x)$ forms a $\textit{pseudonorm}$ on the vector space of integrable functions $L^1(X, \mu)$. Regarding (i), we may have $||f||=0$ for some $f$ that is not identically zero as long as $f = 0$ $\mu$-almost everywhere. e.g. The Dirichlet function $\chi_{\mathbb{Q}}=0$ $m-a.e.$ where $m$ is the usual Lebesgue measure on $\mathbb{R}$. To make a norm, we define the equivalence relation $f$~$g$ if $\mu(\{x|f(x) \neq g(x)\})=0$. Then, the set of equivalence classes of $L^1$ functions together with $|| \cdot ||$ forms a normed space.

Regarding (ii) above, the homogeneity requirement of norms has important implications:
$\textit{The powers of the norms must agree} $: Consider the following inequality
$||f||_X^2 \leq c||f||_Y$ If this inequality is true for all $f$, then in particular it is true if we replace $f$ with $\alpha f$.
$||\alpha f||_X^2 \leq c ||\alpha f||_Y$
$(|\alpha| ||f||_X)^2 \leq c |\alpha| ||f||_Y$
$|\alpha|^2 ||f||^2 \leq c |\alpha| ||f||_Y$
$|\alpha| ||f||_X^2 \leq c||f||_Y$
Letting $\alpha \longrightarrow 0$, we get $0 \leq c||f||_Y$...which is trivial. Also,
$||f||_X^2 \leq \frac{c}{|\alpha|} ||f||_Y$. Letting $\alpha \longrightarrow + \infty$, we get $||f||_X^2 = 0 \Longleftrightarrow f = 0$...nonsense!
Additionally, $\textit{the powers of the internal constants must agree}$. Consider the inequality
$||\alpha f|| \leq c||f||$. If true for all $f$, then again replace $f$ with $\alpha f$
$||\alpha \alpha f|| \leq c|| \alpha f|| $
$ | \alpha|^2 ||f|| \leq c |\alpha| ||f||$
$||f|| \leq \frac{c}{|\alpha |} ||f||$. Letting $| \alpha | \longrightarrow + \infty $ , we get
$||f|| = 0$...nonsense!

Regarding (iii) norms also satisfy the reverse triangle inequality: $$\left| ||x||-||y|| \right| \leq ||x-y|| \quad \forall x,y \in{(V,|| \cdot ||)} $$ Let $x,y$ be in a normed space and let $||x|| > ||y||$ for if they are equal, then the reverse triangle inequality is trivial.
$||x+y|| \leq ||x|| + ||y||$
$ ||x+y|| - ||y|| \leq ||x||$. $\leftarrow$ since this is true forall $x,y$, it is also true if we replace $x$ with $x-y$.
$||(x-y)+y|| - ||y|| \leq ||x-y||$
$||x||-||y|| \leq ||x-y||$. With both sides being non-negative, we may put absolute values on the left hand side
$\left| ||x||-||y|| \right| \leq ||x-y||$
Note that if we started with the above line, we could have reversed each of the above steps to obtain the triangle inequality. So, these two inequalities are equivalent in any normed space.