# Topological Modules/Printable version

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# Constructions

**Definition (quotient topological module)**:

Let be a topological module over the topological ring , and let a submodule with the subspace topology. Then the module together with the quotient topology, ie. the final topology induced by the quotient map , is called the **quotient module** of .

**Proposition (quotient map of topological quotient is open)**:

Let be a topological module and a submodule. Then the map is open.

**Proof:** Let be any open set. We have

which is open as the union of open sets.

**Proposition (quotient topological module is topological module)**:

Let be a topological module and a submodule. Then the quotient module is a topological module with the subspace topology.

**Proof:**

# Banach spaces

**Definition (Banach space)**:

A **Banach space** is a complete normed space.

TODO: Links

**Proposition (series criterion for Banach spaces)**:

Let be a normed space with norm . Then is a Banach space if and only if

- implies that exists in ,

whenever is a sequence in .

**Proof:** Suppose first that is a Banach space. Then suppose that converges, where is a sequence in . Then set ; we claim that is a Cauchy sequence. Indeed, for sufficiently large, we have

- .

Hence, also converges, because is a Banach space.

Now suppose that for all sequences the implication

holds. Let then be a Cauchy sequence in . By the Cauchy property, choose, for all , a number such that whenever . We may assume that , ie. is an ascending sequence of natural numbers. Then define and for set . Then

- .

Moreover,

- ,

so that

converges as a monotonely increasing, bounded sequence. By the assumption, the sequence converges, where

- .

Thus, is a Cauchy sequence that has a convergent subsequence and is hence convergent.

# Hahn–Banach theorems

**Theorem (geometrical Hahn–Banach theorem)**:

Let be a real topological vector space, and let be open and convex so that . Then there exists a hyperplane not intersecting .

**Proof:** The set of all vector subspaces of that do not intersect is inductive and also nonempty (because of the zero subspace). Hence, by Zorn's lemma, pick a maximal vector subspace that does not intersect . Claim that is a hyperplane. If not, has dimension . Now the canonical map is open, so that is an open, convex subset of . We consider the cone

and note that it has a nonzero boundary point; for otherwise would be clopen in which is path-connected (indeed by assumption , so that for any two points we find a 2-dimensional plane containing both, and by using a "corner point" when do lie on a line through the origin, we may connect them in , because a segment in a TVS yields a continuous path by continuity of addition and scalar multiplication), so that , which is impossible because for any in , we then have , for , so that by convexity, a contradiction. Hence, let . Then the line generated by does not intersect and hence not , and is a larger subspace of that does not intersect than in contradiction to the maximality of the latter.

# Barrelled spaces

**Proposition (pointwise limit of continuous linear functions from a barrelled LCTVS into a Hausdorff TVS is continuous and linear)**:

Let be a barrelled LCTVS over a field , let be a locally closed Hausdorff TVS over the same field, and suppose that a net () of linear and continuous functions is given. Suppose further that is a function such that

- .

Then is itself a linear and continuous functional.

**Proof:** First note that is linear, since whenever and , we have

since is a Hausdorff space, where limits are well-defined, and by continuity of addition. Then note that is continuous, since for all the set is bounded, so that the Banach—Steinhaus theorem applies and the family is uniformly bounded. Hence, suppose that is a closed neighbourhood of the origin. By uniform boundedness, select to be an open neighbourhood of the origin so that

- .

We conclude that , since closed sets contain their net limits. We conclude since is locally closed, so that represents a generic neighbourhood.

# Topological tensor products

## Tensor product of Hilbert spacesEdit

**Proposition (tensor product of orthonormal bases is orthonormal basis of tensor product)**:

Let be Hilbert spaces, and suppose that is an orthonormal basis of and is an orthonormal basis of . Then is an orthonormal basis of .

**Proof:** Let any element

of be given; by definition, each element of may be approximated by such elements. Let . Then by definition of an orthonormal basis, we find for and for and then resp. such that

- and .

Then note that by the triangle inequality,

- .

Now fix . Then by the triangle inequality,

In total, we obtain that

(assuming that the given sum approximates well enough) which is arbitrarily small, so that the span of tensors of the form is dense in . Now we claim that the basis is orthonormal. Indeed, suppose that . Then

- .

Similarly, the above expression evaluates to when and . Hence, does constitute an orthonormal basis of .

# Orthogonal projection

**Theorem (Von Neumann ergodic theorem)**:

Let be Hilbert space, and let be a unitary operator. Further, let the orthogonal projection onto the space be given by . Then

- ,

where the limit is taken with respect to the operator norm on , the space of bounded operators on . Moreover, the inequality

is a valid estimate for the convergence rate.

**Proof:** Suppose first that and . Then

Further, if we set

- ,

we obtain

If now the sequence is convergent, we see that its limit is indeed contained within . From the respective former consideration, we may hence infer that the sequence does in fact converge to . We are thus reduced to proving the convergence of the sequence in operator norm. Since is Hilbert space, proving that is a Cauchy sequence will be sufficient. But since

for this is the case; the gaps are closed using that

Taking in the next to last computation yields the desired rate of convergence. These computations also reveal the underlying cause of convergence: The sequence becomes more and more uniform, since applying to it does not change it by a large amount.