1.3 Normed vector spaces

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While vector spaces have additional structure compared to a metric space, a general vector space has no notion of “length” or “distance.”

Definition 1

Let $V$ be a vector space over $K$ . A norm is a function $∥·∥:V\to \mathbb{R}$ such that

• $∥x∥\ge 0\forall x\in V$
• $∥x∥=0$ iff $x=0$
• $∥\alpha ,x∥=|\alpha |∥x∥\forall x\in V$ , $\alpha \in K$
• $∥x,+,y∥\le ∥x∥+∥y∥\forall x,y\in V$

A vector space together with a norm is called a normed vector space (or normed linear space ).

• $V={\mathbb{R}}^{N}$ : ${∥x∥}_{2}=\sqrt{{\sum }_{i=1}^{N}{|{x}_{i}|}^{2}}$
• $V={\mathbb{R}}^{N}$ : ${∥x∥}_{1}={\sum }_{i=1}^{N}|{x}_{i}|$ (“Taxicab”/“Manhattan” norm)
• $V={\mathbb{R}}^{N}$ : ${∥x∥}_{\infty }=\underset{i=1,...,N}{max}|{x}_{i}|$
• $V={L}_{p}\left[a,b\right]$ , $p\in \left[1,\infty \right)$ : ${∥x,\left(,t,\right)∥}_{p}={\left({\int }_{a}^{b},{|x\left(t\right)|}^{p},d,t\right)}^{1/p}$ (The notation ${L}_{p}\left[a,b\right]$ denotes the set of all functions defined on the interval $\left[a,b\right]$ such that this norm exists, i.e., ${\parallel x\left(t\right)\parallel }_{p}<\infty$ .)

Note that any normed vector space is a metric space with induced metric $d\left(x,y\right)=∥x,-,y∥$ . (This follows since $∥x,-,y∥=∥x,-,z,+,z,-,y∥\in ∥x,-,z∥+∥y,-,z∥$ .) While a normed vector space “feels like” a metric space, it is important to remember that it actually satisfies a great deal of additional structure.

Technical Note: In a normed vector space we must have (from N2) that $x=y$ if $∥x,-,y∥=0$ . This can lead to a curious phenomenon when dealing with continuous-time functions. For example, in ${L}_{2}\left(\left[a,b\right]\right)$ , we can consider a pair of functions like $x\left(t\right)$ and $y\left(t\right)$ illustrated below. These functions differ only at a single point, and thus ${\parallel x\left(t\right)-y\left(t\right)\parallel }_{2}=0$ (since a single point cannot contribute anything to the value of the integral.) Thus, in order for our norm to be consistent with the axioms of a norm, we must say that $x=y$ whenever $x\left(t\right)$ and $y\left(t\right)$ differ only on a set of measure zero. To reiterate $x=y⇎x\left(t\right)=y\left(t\right)\forall t\in \left[a,b\right]$ , i.e., when we treat functions as vectors, we will not interpret $x=y$ as pointwise equality, but rather as equality almost everywhere .

Questions & Answers

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Source:  OpenStax, Digital signal processing. OpenStax CNX. Dec 16, 2011 Download for free at http://cnx.org/content/col11172/1.4
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