Chapter 10

Normed Linear SpacesIn addition to the algebraic aspects of linear algebra presented earlier, there are many an-alytical and geometrical concepts which are usually included. This material involves thespecial fields R and C instead of general fields. It is these things which are typically gener-alized in functional analysis. The main new idea is that the notion of distance is included.This allows one to consider continuity, compactness, and many other topics from calculus.First is a general treatment of the notion of distance which has nothing to do with linearalgebra but is a useful part of the vocabulary leading most efficiently to the inclusion ofanalytical topics.

10.1 Metric SpacesThis section is here to provide definitions and main theorems about fundamental analyticalideas and terminology. The first part is on metric spaces which really have absolutelynothing to do with linear algebra but they provide a convenient framework for discussionof the analytical aspects of linear algebra.

10.1.1 LimitsIt is most efficient to discus things in terms of abstract metric spaces to begin with.

Definition 10.1.1 A non empty set X is called a metric space if there is a function d :X×X → [0,∞) which satisfies the following axioms.

1. d (x,y) = d (y,x)

2. d (x,y)≥ 0 and equals 0 if and only if x = y

3. d (x,y)+d (y,z)≥ d (x,z)

This function d is called the metric. We often refer to it as the distance.

Definition 10.1.2 An open ball, denoted as B(x,r) is defined as follows.

B(x,r)≡ {y : d (x,y)< r}

A set U is said to be open if whenever x ∈ U, it follows that there is r > 0 such thatB(x,r) ⊆U. More generally, a point x is said to be an interior point of U if there existssuch a ball. In words, an open set is one for which every point is an interior point.

For example, you could have X be a subset of R and d (x,y) = |x− y|.Then the first thing to show is the following.

Proposition 10.1.3 An open ball is an open set.

Proof: Suppose y ∈ B(x,r) . We need to verify that y is an interior point of B(x,r). Letδ = r−d (x,y) . Then if z ∈ B(y,δ ) , it follows that

d (z,x)≤ d (z,y)+d (y,x)< δ +d (y,x) = r−d (x,y)+d (y,x) = r

Thus y ∈ B(y,δ )⊆ B(x,r). ■

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