# 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.