# Unordered sums in Hilbert spaces

Jordan Bell
April 3, 2014

## 1 Preliminaries

Let $\mathbb{N}$ be the set of positive integers. We say that a set is countable if it is bijective with a subset of $\mathbb{N}$; thus a finite set is countable. In this note I do not presume unless I say so that any set is countable or that any topological space is separable. A neighborhood of a point in a topological space is a set that contains an open set that contains the point; one reason why it can be handy to speak about neighborhoods of a point rather than just open sets that contain the point is that the set of all neighborhoods of a point is a filter, whereas it is unlikely that the set of all open sets that contain a point is a filter.

## 2 Unordered sums in normed spaces

A partially ordered set is a set $J$ and a binary relation $\leq$ on $J$ that is reflexive ($\alpha\leq\alpha$), antisymmetric (if both $\alpha\leq\beta$ and $\beta\leq\alpha$ then $\alpha=\beta$), and transitive (if both $\alpha\leq\beta$ and $\beta\leq\gamma$ then $\alpha\leq\gamma$).11 1 Paul R. Halmos, Naive Set Theory, §14. A directed set is a partially ordered set $(J,\leq)$ such that if $\alpha,\beta\in J$ then there is some $\gamma\in J$ such that $\alpha\leq\gamma$ and $\beta\leq\gamma$. If $X$ is a topological space, a net in $X$ is a function from some directed set to $X$. If $z:J\to X$ is a net in $X$ and $N$ is a subset of $X$, we say that $z$ is eventually in $N$ if there is some $\alpha\in J$ such that $\alpha\leq\beta$ implies $z(\beta)\in N$. We say that the net $z$ converges to $x\in X$ if for every neighborhood of $x$ the net is eventually in that neighborhood. The importance of the notion of a net is that if $X$ and $Y$ are topological spaces and $f$ is a function $X\to Y$ then $f$ is continuous if and only if for every $x\in X$ and for every net $z:J\to X$ that converges to $x$, the net $f\circ z:J\to Y$ converges to $f(x)$.22 2 James R. Munkres, Topology, second ed., p. 188.

Let $X$ be a normed space, let $I$ be a set, and let $\mathscr{F}$ be the set of all finite subsets of $I$. $\mathscr{F}$ is a directed set ordered by set inclusion. Define $S:\mathscr{F}\to X$ by

 $S(F)=\sum_{i\in F}f(i)\in X,\qquad F\in\mathscr{F}.$

$S$ is a net in $X$, and if the net $S$ converges to $x\in X$, we say that the sum $\sum_{i\in I}f(i)$ converges to $x$, and write $\sum_{i\in I}f(i)=x$.

###### Theorem 1.

If $X$ is a normed space, $f:I\to X$ is a function, $x\in X$, and $I_{0}$ is a subset of $I$ such that if $i\in I\setminus I_{0}$ then $f(i)=0$, then $\sum_{i\in I}f(i)$ converges to $x$ if and only if $\sum_{i\in I_{0}}f(i)$ converges to $x$.

###### Proof.

Let $\mathscr{F}$ be the set of all finite subsets of $I$, let $\mathscr{F}_{0}$ be the set of all finite subsets of $I_{0}$, define $S:\mathscr{F}\to X$ by $S(F)=\sum_{i\in F}f(i)$, and let $S_{0}$ be the restriction of $S$ to $\mathscr{F}_{0}$. Suppose that $\sum_{i\in I}f(i)$ converges to $x$, and let $\epsilon>0$. There is some $F_{\epsilon}\in\mathscr{F}$ such that if $F_{\epsilon}\subseteq F\in\mathscr{F}$ then $\left\|S(F)-x\right\|<\epsilon$. Let $G_{\epsilon}=F_{\epsilon}\cap I_{0}$. If $G_{\epsilon}\subseteq G\in\mathscr{F}_{0}$, then

 $S_{0}(G)-x=\sum_{i\in G}f(i)-x=\sum_{i\in F}f(i)-x=S(F)-x,$

giving $\left\|S_{0}(G)-x\right\|=\left\|S(F)-x\right\|$. Hence $G_{\epsilon}\subseteq G\in\mathscr{F}_{0}$ implies that $\left\|S_{0}(G)-x\right\|<\epsilon$, showing that the net $S_{0}$ converges to $x$, i.e. that $\sum_{i\in I_{0}}f(i)$ converges to $x$.

Suppose that $\sum_{i\in I_{0}}f(i)$ converges to $x$, and let $\epsilon>0$. There is some $G_{\epsilon}\in\mathscr{F}_{0}$ such that if $G_{\epsilon}\subseteq G\in\mathscr{F}_{0}$ then $\left\|S_{0}(G)-x\right\|<\epsilon$. If $G_{\epsilon}\subseteq F\in\mathscr{F}$, then, with $G=F\cap I_{0}$,

 $S(F)-x=\sum_{i\in F}f(i)-x=\sum_{i\in G}f(i)-x=S_{0}(G)-x,$

so $G_{\epsilon}\subseteq F\in\mathscr{F}$ implies that $\left\|S(F)-x\right\|<\epsilon$. This shows that $S$ converges to $x$, that is, that $\sum_{i\in I}f(i)$ converges to $x$. ∎

###### Theorem 2.

If $X$ is a normed space, $f:I\to X$ is a function, and $\sum_{i\in I}f(i)$ converges, then $\{i\in I:f(i)\neq 0\}$ is countable.

###### Proof.

Suppose that $\sum_{i\in I}f(i)$ converges to $x$, let $\mathscr{F}$ be the set of all finite subsets of $I$, and let $S(F)=\sum_{i\in I}f(i)$, $F\in\mathscr{F}$. For each $n\in\mathbb{N}$, let $F_{n}\in\mathscr{F}$ be such that if $F_{n}\subseteq F\in\mathscr{F}$ then

 $\left\|S(F)-x\right\|<\frac{1}{n}.$

If $G\in\mathscr{F}$ and $G\cap F_{n}=\emptyset$, then

 $\left\|S(G)\right\|=\left\|S(G\cup F_{n})-S(F_{n})\right\|\leq\left\|S(G\cup F% _{n})-x\right\|+\left\|S(F_{n})-x\right\|<\frac{2}{n}.$

Let $J=\bigcup_{n\in\mathbb{N}}F_{n}$. If $i\in I\setminus J$, then for each $n\in\mathbb{N}$, we have $\{i\}\cap F_{n}=\emptyset$, whence $\left\|S(\{i\})\right\|<\frac{2}{n}$. That is, if $i\in I\setminus J$ then for each $n\in\mathbb{N}$ we have $\left\|f(i)\right\|<\frac{2}{n}$, which implies that if $i\in I\setminus J$ then $f(i)=0$. Therefore $\{i\in I:f(i)\neq 0\}\subseteq J$, and as $J$ is countable, the set $\{i\in I:f(i)\neq 0\}$ is countable. ∎

However, we already have a notion of infinite sums: a series is the limit of a sequence of partial sums.

###### Theorem 3.

If $X$ is a normed space, $x_{n}\in X$, and $\sum_{n\in\mathbb{N}}x_{n}$ converges to $x$, then $\sum_{n=1}^{N}x_{n}\to x$ as $N\to\infty$.

###### Proof.

Let $\epsilon>0$, let $\mathscr{F}$ be the set of all finite subsets of $\mathbb{N}$, and let $S:\mathscr{F}\to X$ be $S(F)=\sum_{n\in F}x_{n}$. The net $S$ converges to $x$, so there is some $F_{\epsilon}\in\mathscr{F}$ such that if $F_{\epsilon}\subseteq F$ then $\left\|S(F)-x\right\|<\epsilon$. Let $N_{\epsilon}=\max F_{\epsilon}$. If $N\geq N_{\epsilon}$, then for $F=\{1,\ldots,N\}$ we have $F_{\epsilon}\subseteq F$ and so

 $\left\|\sum_{n=1}^{N}x_{n}-x\right\|=\left\|S(F)-x\right\|<\epsilon,$

showing that $\sum_{n=1}^{N}x_{n}\to x$ as $N\to\infty$. ∎

When we talk about the sum $\sum_{i\in I}f(i)$, the set of all finite subsets of $I$ is ordered by set inclusion, but we don’t care about any ordering of the set $I$ itself. If the sum $\sum_{n\in\mathbb{N}}x_{n}$ converges then for any bijection $\sigma:\mathbb{N}\to\mathbb{N}$, $\sum_{n=1}^{\infty}x_{\sigma(n)}=\sum_{n\in\mathbb{N}}x_{n}$. If $x_{n}$ is a sequence in a normed space and for every bijection $\sigma:\mathbb{N}\to\mathbb{N}$ the series $\sum_{n=1}^{\infty}x_{\sigma(n)}$ converges, we say that the sequence $x_{n}$ is unconditionally summable. If an unordered sum converges, then it is unconditionally summable, and if a countable unordered sum is unconditionally summable the unordered sum converges.

###### Theorem 4.

If $X$ is a Banach space, $x_{n}\in X$, and $\sum_{n=1}^{\infty}\left\|x_{n}\right\|<\infty$, then $\sum_{n\in\mathbb{N}}x_{n}$ converges.

###### Proof.

For each $k\in\mathbb{N}$ there is some $K(k)$ such that

 $\sum_{n=K(k)+1}^{\infty}\left\|x_{n}\right\|<\frac{1}{k};$

suppose that if $j then $K(j). Define

 $v_{k}=\sum_{n=1}^{K(k)}x_{n}.$

For $\epsilon>0$, let $N>\frac{1}{\epsilon}$. If $k>j\geq N$, then

 $\left\|v_{k}-v_{j}\right\|=\left\|\sum_{n=1}^{K(k)}x_{n}-\sum_{n=1}^{K(j)}x_{n% }\right\|=\left\|\sum_{n=K(j)+1}^{K(k)}x_{n}\right\|\leq\sum_{n=K(j)+1}^{K(k)}% \left\|x_{n}\right\|\leq\sum_{n=K(j)+1}^{\infty}\left\|x_{n}\right\|,$

hence if $k>j\geq N$, then $\left\|v_{k}-v_{j}\right\|<\frac{1}{j}\leq\frac{1}{N}$. This shows that $v_{k}$ is a Cauchy sequence, and hence $v_{k}$ converges to some $x\in X$.

Let $\mathscr{F}$ be the set of all finite subsets of $\mathbb{N}$ and define $S:\mathscr{F}\to X$ by $S(F)=\sum_{n\in F}x_{n}$. Let $\epsilon>0$, and as $v_{k}\to x$ there is some $N_{1}$ such that if $k\geq N_{1}$ then $\left\|v_{k}-x\right\|<\epsilon$. Let $N_{2}>\frac{1}{\epsilon}$, put $N=\max\{N_{1},N_{2}\}$, and put $F_{\epsilon}=\{1,\ldots,K(N)\}$. If $F_{\epsilon}\subseteq F\in\mathscr{F}$, then

 $\displaystyle\left\|S(F)-x\right\|$ $\displaystyle=$ $\displaystyle\left\|\sum_{n\in F}x_{n}-x\right\|$ $\displaystyle\leq$ $\displaystyle\left\|\sum_{n\in F}x_{n}-\sum_{n\in F_{\epsilon}}x_{n}\right\|+% \left\|\sum_{n\in F_{\epsilon}}x_{n}-x\right\|$ $\displaystyle=$ $\displaystyle\left\|\sum_{n\in F\setminus F_{\epsilon}}x_{n}\right\|+\left\|v_% {N}-x\right\|$ $\displaystyle<$ $\displaystyle\sum_{n\in F\setminus F_{\epsilon}}\left\|x_{n}\right\|+\epsilon$ $\displaystyle\leq$ $\displaystyle\sum_{n=K(N)+1}^{\infty}\left\|x_{n}\right\|+\epsilon$ $\displaystyle<$ $\displaystyle\frac{1}{N}+\epsilon$ $\displaystyle<$ $\displaystyle 2\epsilon.$

Therefore the net $S$ converges to $x$, i.e. $\sum_{n\in\mathbb{N}}x_{n}$ converges to $x$. ∎

The following theorem shows us in particular that the converse of Theorem 3 is false. One direction of the following theorem is Theorem 4 with $X=\mathbb{C}$. The other direction follows from the Riemann rearrangement theorem.33 3 Walter Rudin, Principles of Mathematical Analysis, third ed., p. 76, Theorem 3.54.

###### Theorem 5.

If $\alpha_{n}\in\mathbb{C}$, then $\sum_{n\in\mathbb{N}}\alpha_{n}$ converges if and only if $\sum_{n=1}^{\infty}|\alpha_{n}|<\infty$.

Let $X$ be a normed space and $z:J\to X$ a net. We say that $z$ is Cauchy if for every $\epsilon>0$ there is some $\alpha\in J$ such that $\alpha\leq\beta$ and $\alpha\leq\gamma$ together imply that $\left\|z(\beta)-z(\gamma)\right\|<\epsilon$.44 4 Ronald G. Douglas, Banach Algebra Techniques in Operator Theory, second ed., p. 3, Proposition 1.7.

###### Theorem 6.

If $X$ is a Banach space and $z:J\to X$ is a Cauchy net, then there is some $x\in X$ such that $z$ converges to $x$.

###### Proof.

Let $\alpha_{1}\in J$ such that if $\alpha_{1}\leq\alpha$ then $\left\|z(\alpha)-z(\alpha_{1})\right\|<1$, and for $n>1$ let $\alpha_{n}\in J$ be such that if $\alpha_{n}\leq\alpha$ then $\left\|z(\alpha)-z(\alpha_{n})\right\|<\frac{1}{n}$ and such that $\alpha_{n-1}\leq\alpha_{n}$. Define $x_{n}=z(\alpha_{n})$. For $\epsilon>0$, let $N>\frac{1}{\epsilon}$. If $n\geq m\geq N$, then, as $\alpha_{n}\geq\alpha_{m}$,

 $\left\|x_{n}-x_{m}\right\|=\left\|z(\alpha_{n})-z(\alpha_{m})\right\|<\frac{1}% {m}\leq\frac{1}{N},$

showing that $x_{n}$ is a Cauchy sequence in $X$. Hence there is some $x\in X$ such that $x_{n}\to x$.

Let $\epsilon>0$, let $N_{1}>\frac{1}{\epsilon}$, let $N_{2}$ be such that if $n\geq N_{2}$ then $\left\|x_{N_{2}}-x\right\|<\epsilon$, and set $N=\max\{N_{1},N_{2}\}$. If $\alpha_{N}\leq\alpha$, then, by construction of the sequence $\alpha_{n}$,

 $\displaystyle\left\|z(\alpha)-x\right\|$ $\displaystyle\leq$ $\displaystyle\left\|z(\alpha)-z(\alpha_{N})\right\|+\left\|z(\alpha_{N})-x\right\|$ $\displaystyle=$ $\displaystyle\left\|z(\alpha)-z(\alpha_{N})\right\|+\left\|x_{N}-x\right\|$ $\displaystyle<$ $\displaystyle\frac{1}{N}+\epsilon$ $\displaystyle<$ $\displaystyle 2\epsilon,$

showing that the net $z$ converges to $x$. ∎

###### Theorem 7.

If $H$ is an infinite dimensional Hilbert space and $\{e_{n}:n\in\mathbb{N}\}$ is an orthonormal set in $H$, then $\sum_{n\in\mathbb{N}}\frac{1}{n}e_{n}$ converges.

###### Proof.

Let $\mathscr{F}$ be the set of finite subsets of $\mathbb{N}$ and let $S(F)=\sum_{n\in F}\frac{1}{n}e_{n}$, $F\in\mathscr{F}$. Define $v_{N}=\sum_{n=1}^{N}\frac{1}{n}e_{n}$. If $N_{1}>N_{2}\geq N$, then, as $e_{n}$ are orthonormal,

 $\left\|v_{N_{1}}-v_{N_{2}}\right\|^{2}=\left\|\sum_{n=N_{2}+1}^{N_{1}}\frac{1}% {n}e_{n}\right\|^{2}=\sum_{n=N_{2}+1}^{N_{1}}\frac{1}{n^{2}}<\sum_{n=N+1}^{% \infty}\frac{1}{n^{2}}<\sum_{n=N}^{\infty}\frac{1}{n(n+1)}=\frac{1}{N},$

so $v_{N}$ is a Cauchy sequence in $H$ and hence converges to some $h\in H$. For $\epsilon>0$, let $N_{1}>\frac{1}{\epsilon}$, let $\left\|v_{N_{2}}-h\right\|^{2}<\epsilon$, put $N=\max\{N_{1},N_{2}\}$, and put $F_{\epsilon}=\{1,\ldots,N\}$. If $F_{\epsilon}\subseteq F\in\mathscr{F}$, then, using that $e_{n}$ are orthonormal and $0\leq(a-b)^{2}=a^{2}-2ab+b^{2}$,

 $\displaystyle\left\|S(F)-h\right\|^{2}$ $\displaystyle\leq$ $\displaystyle\left(\left\|S(F)-S(F_{\epsilon})\right\|+\left\|S(F_{\epsilon})-% h\right\|\right)^{2}$ $\displaystyle\leq$ $\displaystyle 2\left\|S(F)-S(F_{\epsilon})\right\|^{2}+2\left\|S(F_{\epsilon})% -h\right\|^{2}$ $\displaystyle=$ $\displaystyle 2\left\|\sum_{n\in F\setminus F_{\epsilon}}\frac{1}{n}e_{n}% \right\|^{2}+2\left\|v_{N}-h\right\|^{2}$ $\displaystyle=$ $\displaystyle 2\sum_{n\in F\setminus F_{\epsilon}}\frac{1}{n^{2}}+2\left\|v_{N% }-h\right\|^{2}$ $\displaystyle<$ $\displaystyle 4\epsilon.$

This shows that the net $S$ converges to $h$, that is, that $\sum_{n\in\mathbb{N}}\frac{1}{n}e_{n}$ converges to $h$. ∎

We have proved that if $H$ is an infinite dimensional Hilbert space and $\{e_{n}:n\in\mathbb{N}\}$ is an orthonormal set in $H$, then $\sum_{n\in\mathbb{N}}\frac{1}{n}e_{n}$ converges, although $\sum_{n=1}^{\infty}\left\|\frac{1}{n}e_{n}\right\|=\sum_{n=1}^{\infty}\frac{1}% {n}=\infty$. This shows that the converse of Theorem 4 is false. In fact, the Dvoretsky-Rogers theorem states that if $X$ is an infinite dimensional Banach space then there is some countable subset $\{x_{n}:n\in\mathbb{N}\}$ of $X$ such that $\sum_{n\in\mathbb{N}}x_{n}$ converges but $\sum_{n\in\mathbb{N}}\left\|x_{n}\right\|=\infty$.55 5 Joseph Diestel, Sequences and Series in Banach Spaces, p. 59, chapter VI.

## 3 Orthogonal projections

If $S_{i},i\in I$, are subsets of a Hilbert space $H$, we define $\bigvee_{i\in I}S_{i}$ to be the closure of the span of $\bigcup_{i\in I}S_{i}$. If $i\neq j$ implies that $S_{i}\perp S_{j}$, we say that the sets $S_{i}$ are mutually orthogonal. To say that $\{e_{i}:i\in I\}$ is an orthonormal basis for $H$ is to say that $\{e_{i}:i\in I\}$ is an orthonormal set and that $H=\bigvee_{i\in I}\{e_{i}\}$.

If $M_{n},n\in\mathbb{N}$, are mutually orthogonal closed subspaces of $M$, we denote

 $\bigoplus_{n\in\mathbb{N}}M_{n}=\bigvee_{n\in\mathbb{N}}M_{n},$

which we call an orthogonal direct sum.

If $H$ is a Hilbert space and $M$ is a closed subspace of $H$, then for every $h\in H$ there is a unique $v_{h}\in M$ such that

 $\left\|h-v_{h}\right\|=\inf_{v\in M}\left\|h-v\right\|,$

and $h-v_{h}\in M^{\perp}$.66 6 John B. Conway, A Course in Functional Analysis, second ed., p. 9, Theorem 2.6. This gives

 $H=M\oplus M^{\perp}.$

The orthogonal projection of $H$ onto $M$ is the map $P:H\to H$ defined by

 $P(h_{1}+h_{2})=h_{1},\qquad h_{1}\in M,h_{2}\in M^{\perp}.$

It is straightforward to check that $P$ is linear, $\left\|P\right\|\leq 1$ ($\left\|P\right\|=1$ if and only if $M$ is nonzero), $P^{2}=P$, and $\ker P=M^{\perp}$ and $P(H)=M$.77 7 John B. Conway, A Course in Functional Analysis, second ed., p. 10, Theorem 2.7. Rather than specifying a closed subspace of $H$ and talking about the orthogonal projection onto $M$, we can talk about an orthogonal projection in $H$, which is the orthogonal projection onto its image.

Bessel’s inequality88 8 John B. Conway, A Course in Functional Analysis, second ed., p. 15, Theorem 4.8. states that if $\{e_{n}:n\in\mathbb{N}\}$ is an orthonormal set in a Hilbert space $H$ and $h\in H$, then

 $\sum_{n=1}^{\infty}|\langle h,e_{n}\rangle|^{2}\leq\left\|h\right\|^{2}.$ (1)
###### Theorem 8.

If $H$ is a Hilbert space, $\mathscr{E}$ is an orthonormal set in $H$, and $h\in H$, then there are only countably many $e\in\mathscr{E}$ such that $\langle h,e\rangle\neq 0$.

###### Proof.

Let

 $\mathscr{E}_{n}=\left\{e\in\mathscr{E}:|\langle h,e\rangle|\geq\frac{1}{n}% \right\}.$

If $\mathscr{E}_{n}$ were infinite, let $\{e_{j}:j\in\mathbb{N}\}$ be a subset of it, and this gives us a contradiction by (1). Therefore each $\mathscr{E}_{n}$ is finite. But if $\langle h,e\rangle\neq 0$ then there is some $n$ such that $|\langle h,e\rangle|\geq\frac{1}{n}$, so

 $\mathscr{E}=\bigcup_{n=1}^{\infty}\mathscr{E}_{n}.$

Therefore $\mathscr{E}$ is countable. ∎

Bessel’s inequality makes sense for an orthonormal set of any cardinality in a Hilbert space, rather than just for a countable orthonormal set.

###### Theorem 9 (Bessel’s inequality).

If $H$ is a Hilbert space, $\mathscr{E}$ is an orthonormal set in $H$, and $h\in H$, then

 $\sum_{e\in\mathscr{E}}|\langle h,e\rangle|^{2}\leq\left\|h\right\|^{2}.$
###### Proof.

By Theorem 8, there are only countably many $e\in\mathscr{E}$ such that $\langle h,e\rangle\neq 0$; let them be $\{e_{n}:n\in\mathbb{N}\}$. $\{e_{n}:n\in\mathbb{N}\}$ is an orthonormal set, so by (1) we have

 $\sum_{n=1}^{\infty}|\langle h,e_{n}\rangle|^{2}\leq\left\|h\right\|^{2}.$

Theorem 4 states that if $X$ is a Banach space, $x_{n}\in X,n\in\mathbb{N}$, and $\sum_{n=1}^{\infty}\left\|x_{n}\right\|<\infty$, then the unordered sum $\sum_{n\in\mathbb{N}}x_{n}$ converges. Thus, with $X=\mathbb{C}$ and $x_{n}=|\langle h,e_{n}\rangle|^{2}$, the unordered sum $\sum_{n\in\mathbb{N}}|\langle h,e_{n}\rangle|^{2}$ converges, say to $S$. Because $\sum_{n\in\mathbb{N}}|\langle h,e_{n}\rangle|^{2}$ converges to $S$, by Theorem 3 the series $\sum_{n=1}^{\infty}|\langle h,e_{n}\rangle|^{2}$ converges to $S$. But we already know that this series is $\leq\left\|h\right\|^{2}$, so

 $\sum_{n\in\mathbb{N}}|\langle h,e_{n}\rangle|^{2}\leq\left\|h\right\|^{2}.$

By Theorem 1, the unordered sum $\sum_{e\in\mathscr{E}}|\langle h,e\rangle|^{2}$ converges if and only if the unordered sum $\sum_{n\in\mathbb{N}}|\langle h,e_{n}\rangle|^{2}$ converges, and if they converge they have the same value. Therefore, the unordered sum $\sum_{e\in\mathscr{E}}|\langle h,e\rangle|^{2}$ indeed converges, and it is $\leq\left\|h\right\|^{2}$. ∎

## 4 Convergence of unordered sums in the strong operator topology

Let $H$ be a Hilbert space and let $\mathscr{B}(H)$ be the set of bounded linear maps $H\to H$. It is straightforward to check that $\mathscr{B}(H)$ is a normed space with the operator norm $\left\|T\right\|=\sup_{\left\|h\right\|\leq 1}\left\|Th\right\|$. (In fact it is a Banach space, actually a Banach algebra, actually a $C^{*}$-algebra; each of these statements implies the previous one.) The strong operator topology on $\mathscr{B}(H)$ can be characterized in the following way: a net $f:I\to\mathscr{B}(H)$ converges to $T\in\mathscr{B}(H)$ in the strong operator topology if for all $h\in H$ the net $f(i)h$ converges to $Th$ in $H$.99 9 For the strong operator topology see John B. Conway, A Course in Functional Analysis, second ed., p. 256.

If $I$ is a set, $\mathscr{F}$ is the set of all finite subsets of $I$, and $f:I\to\mathscr{B}(H)$ is a function, define $S:\mathscr{F}\to\mathscr{B}(H)$ by

 $S(F)=\sum_{i\in I}f(i)\in\mathscr{B}(H).$

$S$ is a net in $\mathscr{B}(H)$, and if the net converges to $T\in\mathscr{B}(H)$ in the strong operator topology we say that the unordered sum $\sum_{i\in I}f(i)$ converges strongly to $T$. To say that the net $S$ converges to $T$ in the strong operator topology is to say that if $h\in H$ then $\sum_{i\in I}f(i)h$ converges to $Th$ in $H$.

If $f,g\in H$, we define $f\otimes g:H\to H$ by

 $f\otimes g(h)=\langle h,g\rangle f.$

It is apparent that $f\otimes g$ is linear, and

 $\left\|f\otimes g(h)\right\|=\left\|\langle h,g\rangle f\right\|=|\langle h,g% \rangle|\left\|f\right\|\leq\left\|h\right\|\left\|g\right\|\left\|f\right\|,$

so $\left\|f\otimes g\right\|\leq\left\|f\right\|\left\|g\right\|$, giving $f\otimes g\in\mathscr{B}(H)$. Additionally,

 $\langle f\otimes g(h_{1}),h_{2}\rangle=\langle\langle h_{1},g\rangle f,h_{2}% \rangle=\langle h_{1},g\rangle\langle f,h_{2}\rangle=\langle h_{1},\langle h_{% 2},f\rangle g\rangle=\langle h_{1},g\otimes f(h_{2})\rangle,$

showing that $(f\otimes g)^{*}=g\otimes f$.

###### Theorem 10.

If $H$ is a Hilbert space, $\mathscr{E}$ is an orthonormal set in $H$, and $P$ is the orthogonal projection onto $\bigvee\mathscr{E}$, then $\sum_{e\in\mathscr{E}}e\otimes e$ converges strongly to $P$.

###### Proof.

Let $h\in H$. By Theorem 8 there are only countably many $e\in\mathscr{E}$ such that $\langle h,e\rangle\neq 0$, and we denote these by $\{e_{n}:n\in\mathbb{N}\}$. By Bessel’s inequality,

 $\sum_{e\in\mathscr{E}}|\langle h,e\rangle|^{2}=\sum_{n\in\mathbb{N}}|\langle h% ,e_{n}\rangle|^{2}=\sum_{n=1}^{\infty}|\langle h,e_{n}\rangle|^{2}\leq\left\|h% \right\|^{2}.$ (2)

Let $\mathscr{F}$ be the set of all finite subsets of $\mathbb{N}$ and for $F\in\mathscr{F}$ let

 $S(F)=\sum_{n\in F}\langle h,e_{n}\rangle e_{n}\in H.$

If $\epsilon>0$, then by (2) there is some $N$ such that $\sum_{n=N+1}^{\infty}|\langle h,e_{n}\rangle|^{2}<\epsilon^{2}$. If $F_{\epsilon}=\{1,\ldots,N\}$ and $F,G\in\mathscr{F}$ both contain $F_{\epsilon}$, then, because the $e_{n}$ are orthonormal,

 $\displaystyle\left\|S(F)-S(G)\right\|^{2}$ $\displaystyle=$ $\displaystyle\left\|\sum_{n\in F}\langle h,e_{n}\rangle e_{n}-\sum_{n\in G}% \langle h,e_{n}\rangle e_{n}\right\|^{2}$ $\displaystyle=$ $\displaystyle\sum_{n\in(F\cup G)\setminus(F\cap G)}\left\|\langle h,e_{n}% \rangle e_{n}\right\|^{2}$ $\displaystyle=$ $\displaystyle\sum_{n\in(F\cup G)\setminus(F\cap G)}|\langle h,e_{n}\rangle|^{2}$ $\displaystyle\leq$ $\displaystyle\sum_{n=N+1}^{\infty}|\langle h,e_{n}\rangle|^{2}$ $\displaystyle<$ $\displaystyle\epsilon^{2}.$

Therefore, if $F,G\in\mathscr{F}$ both contain $F_{\epsilon}$ then $\left\|S(F)-S(G)\right\|<\epsilon$. This means that $S$ is a Cauchy net, and hence, by Theorem 6, has a limit $v\in H$. That is, the unordered sum $\sum_{n\in\mathbb{N}}\langle h,e_{n}\rangle e_{n}$ converges to $v$.

As the unordered sum $\sum_{n\in\mathbb{N}}\langle h,e_{n}\rangle e_{n}$ converges to $v$ we have

 $\lim_{N\to\infty}\sum_{n=1}^{N}\langle h,e_{n}\rangle e_{n}=v.$

If $m\in\mathbb{N}$ then it follows that

 $\lim_{N\to\infty}\sum_{n=1}^{N}\langle h,e_{n}\rangle\langle e_{n},e_{m}% \rangle=\langle v,e_{m}\rangle,$

which is

 $\langle h,e_{m}\rangle=\langle v,e_{m}\rangle.$

Let $Q$ be the orthogonal projection onto $\bigvee_{n\in\mathbb{N}}\{e_{n}\}$. On the one hand, because $\langle h,e\rangle=0$ for $e\not\in\{e_{n}:n\in\mathbb{N}\}$, we check that $Ph=Qh$. On the other hand, we check that $Qh=v$. Therefore, $v=Ph$, i.e.

 $\sum_{e\in\mathscr{E}}e\otimes e(h)=\sum_{e\in\mathscr{E}}\langle h,e\rangle e% =\sum_{n\in\mathbb{N}}\langle h,e_{n}\rangle e_{n}=Ph,$

showing that the unordered sum $\sum_{e\in\mathscr{E}}e\otimes e$ converges strongly to $P$. ∎

In particular, if $\mathscr{E}$ is an orthonormal basis for $H$, then $\sum_{e\in\mathscr{E}}e\otimes e$ converges strongly to $\mathrm{id}_{H}$.