Weak topology

The term is most commonly used for the initial topology of a topological vector space (such as a normed vector space) with respect to its continuous dual.

The remainder of this article will deal with this case, which is one of the concepts of functional analysis.

Starting in the early 1900s, David Hilbert and Marcel Riesz made extensive use of weak convergence.

The benefit of this more general construction is that any definition or result proved for it applies to both the weak topology and the weak* topology, thereby making redundant the need for many definitions, theorem statements, and proofs.

Suppose (X, Y, b) is a pairing of vector spaces over a topological field

The weak topology on Y is now automatically defined as described in the article Dual system.

has an absolute value |⋅|, then the weak topology 𝜎(X, Y, b) on X is induced by the family of seminorms, py : X →

If Y is a vector space of linear functionals on X, then the continuous dual of X with respect to the topology σ(X,Y) is precisely equal to Y.

[1](Rudin 1991, Theorem 3.10) Let X be a topological vector space (TVS) over

A subbase for the weak topology is the collection of sets of the form

In other words, a subset of X is open in the weak topology if and only if it can be written as a union of (possibly infinitely many) sets, each of which is an intersection of finitely many sets of the form

In this case, it is customary to write or, sometimes, If X is equipped with the weak topology, then addition and scalar multiplication remain continuous operations, and X is a locally convex topological vector space.

The uniform and strong topologies are generally different for other spaces of linear maps; see below.

In other words, it is the coarsest topology such that the maps Tx, defined by

Indeed, it coincides with the pointwise convergence of linear functionals.

An important fact about the weak* topology is the Banach–Alaoglu theorem: if X is normed, then the closed unit ball in

Moreover, the closed unit ball in a normed space X is compact in the weak topology if and only if X is reflexive.

Let X be a normed topological vector space over F, compatible with the absolute value in F. Then in

If X is a normed space, a version of the Heine-Borel theorem holds.

[1] This implies, in particular, that when X is an infinite-dimensional normed space then the closed unit ball at the origin in the dual space of X does not contain any weak* neighborhood of 0 (since any such neighborhood is norm-unbounded).

If X is a normed space, then X is separable if and only if the weak* topology on the closed unit ball of

[3] Consider, for example, the difference between strong and weak convergence of functions in the Hilbert space L2(

In contrast weak convergence only demands that for all functions f ∈ L2 (or, more typically, all f in a dense subset of L2 such as a space of test functions, if the sequence {ψk} is bounded).

For given test functions, the relevant notion of convergence only corresponds to the topology used in

For example, in the Hilbert space L2(0,π), the sequence of functions form an orthonormal basis.

One normally obtains spaces of distributions by forming the strong dual of a space of test functions (such as the compactly supported smooth functions on

If X is endowed with the weak topology induced by X# then the continuous dual space of X is X#, every bounded subset of X is contained in a finite-dimensional vector subspace of X, every vector subspace of X is closed and has a topological complement.

[4] If X and Y are topological vector spaces, the space L(X,Y) of continuous linear operators f : X → Y may carry a variety of different possible topologies.

There are, in general, a vast array of possible operator topologies on L(X,Y), whose naming is not entirely intuitive.

For instance, if Y is a normed space, then this topology is defined by the seminorms indexed by x ∈ X: More generally, if a family of seminorms Q defines the topology on Y, then the seminorms pq, x on L(X,Y) defining the strong topology are given by indexed by q ∈ Q and x ∈ X.