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# 3.6: Normed Linear Spaces

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By a normed linear space (briefly normed space) is meant a real or complex vector space $$E$$ in which every vector $$x$$ is associated with a real number $$|x|$$, called its absolute value or norm, in such a manner that the properties $$\left(\mathrm{a}^{\prime}\right)-\left(\mathrm{c}^{\prime}\right)$$ of §9 hold. That is, for any vectors $$x, y \in E$$ and scalar $$a,$$ we have

$$\left(\mathrm{i}\right) |x| \geq 0;$$

$$\left(\mathrm{i}^{\prime}\right)|x|=0$$ iff $$x=\overrightarrow{0};$$

$$\left(\mathrm{ii}\right) |a x|=|a||x| ;$$ and

$$\left(\mathrm{iii}\right)|x+y| \leq|x|+|y|\text{ (triangle inequality). }$$

Mathematically, the existence of absolute values in $$E$$ amounts to that of a map (called a norm map) $$x \rightarrow|x|$$ on $$E,$$ i.e., a map $$\varphi : E \rightarrow E^{1},$$ with function values $$\varphi(x)$$ written as $$|x|,$$ satisfying the laws (i)-(iii) above. Often such a map can be chosen in many ways (not necessarily via dot products, which may not exist in $$E$$ , thus giving rise to different norms on $$E .$$ Sometimes we write $$\|x\|$$ for $$|x|$$ or use other similar symbols.

Note 1. From (iii), we also obtain $$|x-y| \geq| | x|-| y| |$$ exactly as in $$E^{n}.$$

Example $$\PageIndex{1}$$

(A) Each Euclidean space (§9) such as $$E^{n}$$ or $$C^{n},$$ is a normed space, with norm defined by

$|x|=\sqrt{x \cdot x},$

as follows from formulas (a')-(c') in §9. In $$E^{n}$$ and $$C^{n},$$ one can also equivalently define

$|x|=\sqrt{\sum_{k=1}^{n}\left|x_{k}\right|^{2}},$

where $$x=\left(x_{1}, \ldots, x_{n}\right) .$$ This is the so-called standard norm, usually presupposed in $$E^{n}\left(C^{n}\right) .$$

(B) One can also define other, "nonstandard," norms on $$E^{n}$$ and $$C^{n} .$$ For example, fix some real $$p \geq 1$$ and put

$|x|_{p}=\left(\sum_{k=1}^{n}\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}.$

One can show that $$|x|_{p}$$ so defined satisfies $$(\mathrm{i})-($$ iii) and thus is a norm (see Problems 5-7 below).

(C) Let $$W$$ be the set of all bounded maps

$f : A \rightarrow E$

from a set $$A \neq \emptyset$$ into a normed space $$E,$$ i.e., such that

$(\forall t \in A) \quad|f(t)| \leq c$

for some real constant $$c>0$$ (dependent on $$f$$ but not on $$t ) .$$ Define $$f+g$$ and $$a f$$ as in Example (d) of §9 so that $$W$$ becomes a vector space. Also, put

$\|f\|=\sup _{t \in A}|f(t)|,$

i.e., the supremum of all $$|f(t)|,$$ with $$t \in A .$$ Due to boundedness, this supremum exists in $$E^{1},$$ so $$\|f\| \in E^{1}.$$

It is easy to show that $$\|f\|$$ is a norm on $$W .$$ For example, we verify (iii) as follows.

By definition, we have for $$f, g \in W$$ and $$x \in A,$$

\begin{aligned}|(f+g)(x)| &=|f(x)+g(x)| \\ & \leq|f(x)|+|g(x)| \\ & \leq \sup _{t \in A}|f(t)|+\sup _{t \in A}|g(t)| \\ &=\|f\|+\|g\|. \end{aligned}

(The first inequality is true because (iii) holds in the normed space $$E$$ to which $$f(x)$$ and $$g(x)$$ belong.) By (1), $$\|f\|+\|g\|+\|g\|$$ is an upper bound of all expressions $$|(f+g)(x)|, x \in A .$$ Thus

$\|f\|+\|g\| \geq \sup _{x \in A}|(f+g)(x)|=\|f+g\|.$

Note 2. Formula (1) also shows that the map $$f+g$$ is bounded and hence is a member of $$W .$$ Quite similarly we see that $$a f \in W$$ for any scalar $$a$$ and $$f \in W .$$ Thus we have the closure laws for $$W .$$ The rest is easy.

In every normed (in particular, in each Euclidean) space $$E,$$ we define distances by

$\rho(x, y)=|x-y| \quad\text{ for all x, y } \in E.$

Such distances depend, of course, on the norm chosen for $$E ;$$ thus we call them norm-induced distances. In particular, using the standard norm in $$E^{n}$$ and $$C^{n}$$ (Example (A)), we have

$\rho(x, y)=\sqrt{\sum_{k=1}^{n}\left|x_{k}-y_{k}\right|^{2}}.$

Using the norm of Example (B), we get

$\rho(x, y)=\left(\sum_{k=1}^{n}\left|x_{k}-y_{k}\right|^{p}\right)^{\frac{1}{p}}$

instead. In the space $$W$$ of Example (C) we have

$\rho(f, g)=\|f-g\|=\sup _{x \in A}|f(x)-g(x)|.$

Proceeding exactly as in the proof of Theorem 5 in §§1-3, we see that norm- induced distances obey the three laws stated there. (Verify!) Moreover, by definition,

$\rho(x+u, y+u)=|(x+u)-(y+u)|=|x-y|=\rho(x, y).$

Thus we have

$\rho(x, y)=\rho(x+u, y+u)\text{ for norm-induced distances;}$

i.e., the distance $$\rho(x, y)$$ does not change if both $$x$$ and $$y$$ are "translated" by one and the same vector $$u .$$ We call such distances translation-invariant.

A more general theory of distances will be given in §§11ff.