# How weak convergence and weak topology arises naturally from infinite dimensional vector spaces

Originally published at 狗和留美者不得入内. You can comment here or there.

Let be a vector space over endowed with an inner product . Assume that it has an orthonormal basis . This means any can be expressed as

where for all . For each , we take any arbitrary sequence of reals that converges to . Let . We can also require that for all and that , which means uniform convergence across all . If we want that a cofinite number of the sequences at index are at least a certain distance away from the value converged to, we additionally define and require that for all , . Obviously, for all . Setting obviously satisfies these requirements.

We notice that iff the associated series converges, which of course does not happen all the time, which means that that via is not well defined. Thus, the two-norm on finite dimensional vector spaces induced by the inner product cannot be defined similarly on infinite dimensional vector spaces.

Proposition 1 There exists an infinite dimensional normed vector space that is a Banach space.

Proof: Keep in mind that for all , we must have one of the properties of norm. For simplicity we shall require that for any such that ,

which is of course compatible with triangle inequality. We shall require that is a Banach space which means that if is convergent, it is in . Moreover, we require that if converges, then . We now show that this is well-defined. We note that by ,

Assume that the infinite series converges, by definition, for any ,

Then, that

implies that

Applying absolute homogeneity of the norm to gives us

If is convergent, then must of course be bounded based on our extension of norm to the infinite series. Assume it converges to . Then by , . Thus, we have

which verifies absolute homogeneity. Verifying triangle inequality is straightforward and we leave this to the reader.

For , it is necessary that for an arbitrary real sequence . By the aforementioned properties, if we prescribe values for for all . If we require that for any , the associated coefficients satisfy , then values for such that suffices. It is easy to see that for any sequence of absolutely bounded real coefficients, there is some upper bound , which implies that the norm of the vector in associated with it is upper bounded by .

Finally, one easily verifies closure under scalar multiplication and addition of this vector space, wherein the vector coefficients are guaranteed to have bounded supremum.

We note that the bounded supremum requirement is the more essential part here. There is a in fact simple stupid way to construct a norm which is given by . In order for this to be a norm, we must of course always have .

Example 1 Let . Then, .

Example 2 We can let .

Example 3 We can define a sequence of vectors converging to via sequences of coefficients , we indeed have with respect to this norm. More specifically, goes to as . Thus, sequence converges with respect to norm given in Example 2. It also does with respect to the norm given in Example 1.

Example 4 If we let for all , then we would need to restrict the vector space elements to correspond to sequences such that .

Example 5 We let . Using the norm given in Example 1, we have that for all , . The tells us that it also cannot converge to any value in other . Thus does not converge with respect to the Example 1 norm. It does converge to with respect to the Example 2 norm though.

Proposition 2 Any is uniquely defined by prescribing the values of over all . In order for to be well-defined, we must have .

Proof: Follows directly from definition of basis and linearity of .

Definition 1 We say that , or that converges to strongly with respect to some norm iff .

Definition 2 We say that , or that converges to weakly, iff for all , .

Proposition 3 The sequence in Example 5 converges weakly (in the Example 1 norm).

Proof: If , then by Proposition 2, we have . Let and , with of course monotonically from above.

Then, we have that . That this upper bound converges to from above completes our proof.

Definition 3 The coarsest topology on normed vector space with respect to which every is continuous, which we shall denote with , is the weak topology.

Proposition 4 iff every neighborhood of with respect to contains for some all such that .

Proof: Follows directly from the definition of and of neighborhood.

Proposition 5 is a neighborhood basis of . If we let denote an arbitrary finite subset of , then the every set of the aforementioned collection can be denoted as .

Proof: Satisfaction of the continuity requirement means that for any open interval in , for any . We thus take to be a subbase of . (For the definition of subbase, see Definition 1 of [2].) Every open interval in has a midpoint which we take to be , and we take to be half the interval length. The collection of sets given in the proposition are all finite intersections of elements of . Every is the arbitrary union of finite intersections of elements of , by definition of subbase. Thus, either is the empty set, or it contains some . This completes our proof.

Definition 4 The strong topology of a normed vector space is the coarsest topology such that the norm is a continuous function.

The norm on the dual space is defined as in Definition 8 of [3]. This norm gives rise to a strong topology on the dual space as well.

Lemma 1 For any , is closed in the strong topology on .

Proof: is a closed set and the norm is a continuous function. Thus, is also closed. Proposition 7 of [4] tells us that the preimage of a closed set, when the function is continuous is also closed.

Proposition 6 For any , .

Proof: We note that if not, then would not be a well-defined norm on . By definition of , is continuous. By Proposition 2 of [3], continuous implies bounded. The final equality in Definition 8 of [3] then gives us the desired result.

Proposition 7 For any normal vector space , the weak topology is a subset of the strong topology and strong convergence implies weak convergence.

Proof: It suffices to prove that strong convergence implies weak convergence. Assume . Then, by Proposition 6, for any , , with , which completes our proof.

References

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