Banach Spaces, Normed Vector Space

Normed Vector Space

A normed vector space, also called a normed linear space, is a real vector space S with a norm function denoted |x|. The norm has the following properties.

The norm maps S into the nonnegative reals.
The norm of x is 0 iff x = 0.
For any real number c, |cx| = |c|×|x|.
|x+y| ≤ |x|+|y|.

Sometimes the norm is derived from a dot product, but it need not be.

Turning the Norm into a Metric

Let the distance d(x,y) be the norm of x-y. Since x-y is -1 times y-x, property 3 above tells us this is well defined.

By property 2, d(x,y) = 0 iff x = y.

Let a triangle have vertices z, z+x, and z+x+y. Subtract z and apply property 4. This establishes the triangular inequality. Thus d becomes a distance metric, and S is a metric space, with the open ball topology.

Banach Space

A banach space is a normed vector space that forms a complete metric space. Every cauchy sequence in S converges to a limit point in S.

Continuous Module

Let u and v be points in S, and put a ball of radius ε around u+v. Points within ½ε of u, plus points within ½ε of v, wind up within ε of u+v. Plus is a continuous operator, and it turns S into a continuous group.

How about scaling by c? Keep points within ε/c of u, and the image is within ε of cu. (Treat c = 0 as a special case.) Thus scaling by c is a continuous function from R cross S into S, and S is a continuous R module.

Equivalent Norms

Two norms f(S) and g(S) are equivalent if there are positive constants b and c such that bf ≤ g ≤ cf.

Divide by b and c and obtain g/c ≤ f ≤ g/b. The relation is symmetric. Set b = c = 1 to show the relation is reflexive. If bf ≤ g and dg ≤ h, then bdf ≤ h. The relation is transitive, and we have an equivalence relation on norms. Norms clump together in equivalence classes, as they should, since they are called "equivalent norms".

Let's cover an open set in f with open balls in g. A point p in our open set is a certain distance d from the nearest edge, as measured by f. The points within bd of p, measured by g, are all within d of p, measured by f. So p is contained in an open g ball inside our open set. Open sets in f remain open in g, and by symmetry, open sets in g are open in f. The topologies are the same.

Note that the identity map on S, from f to g, is uniformly bicontinuous.

As we move from f to g, cauchy sequences remain cauchy, and the limit point of our sequence becomes the limit point of the same sequence under g. If S is complete under f, it is complete under g, i.e. still a banach space.

Pseudo Norm

A pseudo norm (without property 2) produces a pseudo metric. Collapse points that are 0 distance apart to build a new metric space. But this time S is a vector space, so there's more to the story.

If x and y are 0 distance apart, then u+x and u+y are 0 distance apart. Addition on equivalence classes is well defined. Also, addition remains commutative, and associative, and continuous, so we still have a topological group. Make similar observations for scaling, and the quotient space is a continuous module. Merge the inseparable points of a pseudo normed vector space and get a normed vector space.

Subspace

The word subspace has become ambiguous. If W is a subspace of S, is it a subset of S that inherits the open ball topology, or is it a vector space contained in S? Sometimes you have to infer the correct definition from context. The term linear subspace refers to a sub vector space. This is also called a linear manifold, and that's unfortunate, since a manifold is also a space that is locally homeomorphic to Rⁿ. I'll stick with linear subspace.

A finite dimensional (finitely generated) linear subspace is closed. For example, a line is a closed set in the plane. Choose a basis b₁ through b_n for this subspace, and let p be a point in the closure. Let q be a sequence of points in the subspace that monotonically approaches p. Describe each q_i as a linear combination of basis elements. As i approaches infinity, the coefficients on b_j form a cauchy sequence that approaches a limit c_j. Let r be the sum over c_jb_j. The distance from r to p has to be 0, else the sequence q could be bounded away from p. Therefore r = p, and p is part of our subspace.

Translate

The translate, or translation, or shift, of a set W in S, by a vector x, is the set x+y for all y in W. We're just sliding the set W along in S.