Topological Modules/Printable version


Topological Modules

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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  .

(On the condition of the axiom of choice.)

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 spaces edit

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.