Topological Modules/Printable version

Topological Modules

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Constructions

Definition (quotient topological module):

Let $M$ be a topological module over the topological ring $R$, and let $N\leq M$ a submodule with the subspace topology. Then the module $M/N$ together with the quotient topology, ie. the final topology induced by the quotient map $q:M\to M/N$, is called the quotient module of $M$.

Proposition (quotient map of topological quotient is open):

Let $M$ be a topological module and $N\leq M$ a submodule. Then the map $q:M\to M/N$ is open.

Proof: Let $U\subseteq M$ be any open set. We have

$p^{-1}(p(U))=\bigcup _{n\in \mathbb {N} }U+n$

which is open as the union of open sets. $\Box$

Proposition (quotient topological module is topological module):

Let $M$ be a topological module and $N\leq M$ a submodule. Then the quotient module $M/N$ is a topological module with the subspace topology.

Proof: $\Box$

Banach spaces

Definition (Banach space):

A Banach space is a complete normed space.

TODO: Links

Proposition (series criterion for Banach spaces):

Let $X$ be a normed space with norm $\|\cdot \|$. Then $X$ is a Banach space if and only if

$\sum _{n=1}^{\infty }\|x_{n}\|<\infty$ implies that $\lim _{N\to \infty }\sum _{n=1}^{N}x_{n}$ exists in $X$,

whenever $(x_{n})_{n\in \mathbb {N} }$ is a sequence in $X$.

Proof: Suppose first that $X$ is a Banach space. Then suppose that $\sum _{n=1}^{\infty }\|x_{n}\|$ converges, where $(x_{n})_{n\in \mathbb {N} }$ is a sequence in $X$. Then set $S_{N}:=\sum _{n=1}^{N}x_{n}$; we claim that $(S_{N})_{N\in \mathbb {N} }$ is a Cauchy sequence. Indeed, for $M>0$ sufficiently large, we have

$N\geq M\Rightarrow \|S_{M}-S_{N}\|=\left\|\sum _{n=M+1}^{N}x_{n}\right\|\leq \sum _{n=M+1}^{N}\|x_{n}\|\leq \sum _{n=M+1}^{\infty }\|x_{n}\|<\epsilon$.

Hence, $(S_{N})_{N\in \mathbb {N} }$ also converges, because $X$ is a Banach space.

Now suppose that for all sequences $(x_{n})_{n\in \mathbb {N} }$ the implication

$\sum _{n=1}^{\infty }\|x_{n}\|<\infty \Rightarrow \lim _{N\to \infty }\sum _{n=1}^{N}x_{n}$

holds. Let then $(y_{n})_{n\in \mathbb {N} }$ be a Cauchy sequence in $X$. By the Cauchy property, choose, for all $k\in \mathbb {N}$, a number $N_{k}\in \mathbb {N}$ such that $\|y_{m}-y_{n}\|<1/2^{k}$ whenever $m,n\geq N_{k}$. We may assume that $N_{1}\leq N_{2}\leq \cdots \leq N_{k}\leq \cdots$, ie. $(N_{k})_{k\in \mathbb {N} }$ is an ascending sequence of natural numbers. Then define $x_{1}:=y_{N_{1}}$ and for $k\geq 2$ set $x_{k}:=y_{N_{k}}-y_{N_{k-1}}$. Then

$\sum _{j=1}^{k}x_{j}=y_{N_{k}}$.

Moreover,

$\sum _{j=1}^{k}\|x_{j}\|\leq \|y_{N_{1}}\|+\sum _{j=2}^{k}2^{-(k-1)}$,

so that

$\sum _{j=1}^{\infty }\|x_{j}\|$

converges as a monotonely increasing, bounded sequence. By the assumption, the sequence $(S_{k})_{k\in \mathbb {N} }$ converges, where

$S_{k}=\sum _{j=1}^{k}x_{j}=y_{N_{k}}$.

Thus, $(y_{n})_{n\in \mathbb {N} }$ is a Cauchy sequence that has a convergent subsequence and is hence convergent. $\Box$

Hahn–Banach theorems

Theorem (geometrical Hahn–Banach theorem):

Let $V$ be a real topological vector space, and let $U\subseteq V$ be open and convex so that $0\notin V$. Then there exists a hyperplane $W\leq V$ not intersecting $U$.

(On the condition of the axiom of choice.)

Proof: The set of all vector subspaces of $V$ that do not intersect is inductive and also nonempty (because of the zero subspace). Hence, by Zorn's lemma, pick a maximal vector subspace $W\leq V$ that does not intersect $U$. Claim that $W$ is a hyperplane. If not, $V/W$ has dimension $\geq 2$. Now the canonical map $p:V\to V/W$ is open, so that $U':=p(U)$ is an open, convex subset of $V/W$. We consider the cone

$C:=\bigcup _{\lambda >0}\lambda U'$

and note that it has a nonzero boundary point; for otherwise $C$ would be clopen in $V/W\setminus \{0\}$ which is path-connected (indeed by assumption $\operatorname {dim} V/W\geq 2$, so that for any two points $x,y\in V/W$ we find a 2-dimensional plane containing both, and by using a "corner point" $z$ when $x,y$ do lie on a line through the origin, we may connect them in $V/W\setminus \{0\}$, because a segment in a TVS yields a continuous path by continuity of addition and scalar multiplication), so that $C=V/W\setminus \{0\}$, which is impossible because for any $x\neq 0$ in $V/W$, we then have $x\in \mu U'$, $-x\in \lambda U'$ for $\mu ,\lambda >0$, so that $0\in U'$ by convexity, a contradiction. Hence, let $w\in \partial C$. Then the line $L$ generated by $w$ does not intersect $C$ and hence not $U'$, and $p^{-1}(L)$ is a larger subspace of $V$ that does not intersect $U$ than $W$ in contradiction to the maximality of the latter. $\Box$

Barrelled spaces

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

Let $X$ be a barrelled LCTVS over a field $\mathbb {K}$, let $Y$ be a locally closed Hausdorff TVS over the same field, and suppose that a net $T_{\lambda }:X\to Y$ ($\lambda \in \Lambda$) of linear and continuous functions is given. Suppose further that $T:X\to Y$ is a function such that

$\lim _{\lambda \in \Lambda }T_{\lambda }(x)=T(x)$.

Then $T$ is itself a linear and continuous functional.

Proof: First note that $T$ is linear, since whenever $\alpha \in \mathbb {K}$ and $v,w\in X$, we have

$T(v+\alpha w)=\lim _{\lambda \in \Lambda }(T_{\lambda }(v)+\alpha T_{n}(w))=T(v)+\alpha T(w)$

since $Y$ is a Hausdorff space, where limits are well-defined, and by continuity of addition. Then note that $T$ is continuous, since for all $v\in X$ the set $\{T_{\lambda }|\lambda \in \Lambda \}$ is bounded, so that the Banach—Steinhaus theorem applies and the family $\{T_{\lambda }|\lambda \in \Lambda \}$ is uniformly bounded. Hence, suppose that $V\subseteq Y$ is a closed neighbourhood of the origin. By uniform boundedness, select $U\subseteq X$ to be an open neighbourhood of the origin so that

$\forall \lambda \in \Lambda :T_{n}(U)\subseteq V$.

We conclude that $T(U)\subseteq V$, since closed sets contain their net limits. We conclude since $Y$ is locally closed, so that $V$ represents a generic neighbourhood. $\Box$

Topological tensor products

Tensor product of Hilbert spaces

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

Let $H_{1},H_{2}$ be Hilbert spaces, and suppose that $(e_{\lambda })_{\lambda \in \Lambda }$ is an orthonormal basis of $H_{1}$ and $(f_{\mu })_{\mu \in \mathrm {M} }$ is an orthonormal basis of $H_{2}$. Then $(e_{\lambda }\otimes e_{\mu })_{(\lambda ,\mu )\in \Lambda \times \mathrm {M} }$ is an orthonormal basis of $H_{1}\otimes H_{2}$.

Proof: Let any element

$\sum _{j=1}^{n}f_{j}\otimes g_{j}$

of $H_{1}\otimes H_{2}$ be given; by definition, each element of $H_{1}\otimes H_{2}$ may be approximated by such elements. Let $\epsilon >0$. Then by definition of an orthonormal basis, we find $m_{j},l_{j}$ for $j\in [n]$ and $\alpha _{j,k},\beta _{j,k},\lambda _{j,k},\mu _{j,k}$ for $j\in [n]$ and then $k\in [m_{j}]$ resp. $[l_{j}]$ such that

$\left\|\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}-f_{j}\right\|<\epsilon$ and $\left\|\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}-g_{j}\right\|<\epsilon$.

Then note that by the triangle inequality,

$\left\|\sum _{j=1}^{n}f_{j}\otimes g_{j}-\sum _{j=1}^{n}\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\leq \sum _{j=1}^{n}\left\|f_{j}\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|$.

Now fix $j\in [n]$. Then by the triangle inequality,

${\begin{aligned}\left\|f_{j}\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|&\leq \left\|f_{j}\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes g_{j}\right\|+\left\|\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\\&=\left\|f_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\right\|\|g_{j}\|+\left\|\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right\|\left\|g_{j}-\left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\\&\leq \epsilon \left(\|g_{j}\|+\left\|\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right\|\right).\end{aligned}}$

In total, we obtain that

$\left\|\sum _{j=1}^{n}f_{j}\otimes g_{j}-\sum _{j=1}^{n}\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\leq \epsilon \sum _{j=1}^{n}\left(\|g_{j}\|+2\|f_{j}\|\right)$

(assuming that the given sum approximates $f_{j}$ well enough) which is arbitrarily small, so that the span of tensors of the form $e_{\lambda }\otimes f_{\mu }$ is dense in $H_{1}\otimes H_{2}$. Now we claim that the basis is orthonormal. Indeed, suppose that $(\lambda ,\mu )\neq (\lambda ',\mu ')$. Then

$\langle e_{\lambda }\otimes f_{\mu },e_{\lambda '}\otimes f_{\mu '}\rangle =\langle e_{\lambda },e_{\lambda '}\rangle \langle f_{\mu },f_{\mu '}\rangle =0$.

Similarly, the above expression evaluates to $1$ when $\lambda =\lambda '$ and $\mu =\mu '$. Hence, $(e_{\lambda }\otimes e_{\mu })_{(\lambda ,\mu )\in \Lambda \times \mathrm {M} }$ does constitute an orthonormal basis of $H_{1}\otimes H_{2}$. $\Box$

Orthogonal projection

Theorem (Von Neumann ergodic theorem):

Let $H$ be Hilbert space, and let $U:H\to H$ be a unitary operator. Further, let the orthogonal projection onto the space $W:=\{x\in H|Ux=x\}$ be given by $P_{W}:H\to H$. Then

$\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}U^{k}=P_{W}$,

where the limit is taken with respect to the operator norm on $B(H)$, the space of bounded operators on $H$. Moreover, the inequality

$\left\|{\frac {1}{n}}\sum _{k=1}^{n}U^{k}-P_{W}\right\|\leq {\frac {2}{n}}$

is a valid estimate for the convergence rate.

Proof: Suppose first that $x\in H$ and $z\in W$. Then

${\begin{aligned}\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-P_{W}(x),z\rangle &=\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-P_{W}(x),z\rangle +{\frac {n}{n}}\langle P_{W}(x)-x,z\rangle \\&=\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-x,z\rangle \\&={\frac {1}{n}}\sum _{k=1}^{n}\left(\langle U^{k}x,z\rangle -\langle x,z\rangle \right)\\&{\overset {z\in W}{=}}{\frac {1}{n}}\sum _{k=1}^{n}\left(\overbrace {\langle U^{k}x,U^{k}z\rangle } ^{=\langle x,z\rangle }-\langle x,z\rangle \right)=0.\end{aligned}}$

Further, if we set

$y_{n}:={\frac {1}{n}}\sum _{k=1}^{n}U^{k}x$,

we obtain

${\begin{aligned}\|Uy_{n}-y_{n}\|^{2}&=\langle y_{n}-Uy_{n},y_{n}-Uy_{n}\rangle \\&={\frac {1}{n^{2}}}\langle Ux-U^{n+1}x,Ux-U^{n+1}x\rangle \\&={\frac {1}{n^{2}}}\left(\langle Ux,Ux\rangle -\langle U^{n+1}x,Ux\rangle -\langle Ux,U^{n+1}x\rangle +\langle U^{n+1}x,U^{n+1}x\rangle \right)\\&{\overset {\text{Cauchy‒Schwarz}}{\leq }}{\frac {4\|x\|^{2}}{n^{2}}}.\end{aligned}}$

If now the sequence $(y_{n})_{n\in \mathbb {N} }$ is convergent, we see that its limit is indeed contained within $W$. From the respective former consideration, we may hence infer that the sequence $(y_{n})_{n\in \mathbb {N} }$ does in fact converge to $P_{W}(x)$. We are thus reduced to proving the convergence of the sequence in operator norm. Since $H$ is Hilbert space, proving that $\left({\frac {1}{n}}\sum _{k=1}^{n}U^{k}\right)_{n\in \mathbb {N} }$ is a Cauchy sequence will be sufficient. But since

${\begin{aligned}\|y_{n}-y_{mn}\|&=\left\|y_{n}-\sum _{k=1}^{m}{\frac {1}{m}}U^{nk}y_{n}\right\|\\&\leq \sum _{k=1}^{m}{\frac {1}{m}}\|y_{n}-U^{nk}y_{n}\|\\&\leq {\frac {2\|x\|}{n}}\end{aligned}}$

for $m\geq 1$ this is the case; the gaps are closed using that

${\begin{aligned}\|y_{n}-y_{m}\|&=\left\|{\frac {m-n}{mn}}\sum _{k=1}^{n}U^{k}x-{\frac {1}{m}}\sum _{k=n+1}^{m}U^{k}x\right\|\\&\leq {\frac {m-n}{mn}}\|x\|+{\frac {m-n}{m}}\|x\|.\end{aligned}}$

Taking $m\to \infty$ 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 $U^{m}$ to it does not change it by a large amount. $\Box$