3.6: Normed Linear Spaces
- Page ID
- 19040
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 \(E^{n}.\)
(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.