Unordered sums in Hilbert spaces

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 $\le$ on $J$ that is reflexive ($\alpha \le \alpha$), antisymmetric (if both $\alpha \le \beta$ and $\beta \le \alpha$ then $\alpha =\beta$), and transitive (if both $\alpha \le \beta$ and $\beta \le \gamma$ then $\alpha \le \gamma$). A directed set is a partially ordered set $(J,\le )$ such that if $\alpha ,\beta \in J$ then there is some $\gamma \in J$ such that $\alpha \le \gamma$ and $\beta \le \gamma$. 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 $\alpha \le \beta$ 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 $f(x)$.

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

$$S(F)=\sum _{i\in F}f(i)\in X,F\in ℱ.$$

$S$ 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, $f\mathrm{:}I\mathrm{\to }X$ is a function, $x\mathrm{\in }X$, and $I_{\mathrm{0}}$ is a subset of $I$ such that if $i\mathrm{\in }I\mathrm{\setminus }{I}_{\mathrm{0}}$ then $f\mathit{}\mathrm{(}i\mathrm{)}\mathrm{=}\mathrm{0}$, then ${\mathrm{\sum }}_{i\mathrm{\in }I}f\mathit{}\mathrm{(}i\mathrm{)}$ converges to $x$ if and only if ${\mathrm{\sum }}_{i\mathrm{\in }{I}_{\mathrm{0}}}f\mathit{}\mathrm{(}i\mathrm{)}$ converges to $x$.

Proof.

Let $ℱ$ be the set of all finite subsets of $I$, let ${ℱ}_0$ be the set of all finite subsets of $I_0$, define $S:ℱ\to X$ by $S(F)={\sum }_{i\in F}f(i)$, and let $S_0$ be the restriction of $S$ to ${ℱ}_0$. Suppose that ${\sum }_{i\in I}f(i)$ converges to $x$, and let $ϵ>0$. There is some $F_{ϵ}\in ℱ$ such that if $F_{ϵ}\subseteq F\in ℱ$ then . Let $G_{ϵ}=F_{ϵ}\cap I_0$. If $G_{ϵ}\subseteq G\in {ℱ}_0$, then

$$S_0(G)-x=\sum _{i\in G}f(i)-x=\sum _{i\in F}f(i)-x=S(F)-x,$$

giving $\parallel S_0(G)-x\parallel =\parallel S(F)-x\parallel$. Hence $G_{ϵ}\subseteq G\in {ℱ}_0$ implies that , 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 $ϵ>0$. There is some $G_{ϵ}\in {ℱ}_0$ such that if $G_{ϵ}\subseteq G\in {ℱ}_0$ then . If $G_{ϵ}\subseteq F\in ℱ$, 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 $G_{ϵ}\subseteq F\in ℱ$ implies that . 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, $f\mathrm{:}I\mathrm{\to }X$ is a function, and ${\mathrm{\sum }}_{i\mathrm{\in }I}f\mathit{}\mathrm{(}i\mathrm{)}$ converges, then $\mathrm{\{}i\mathrm{\in }I\mathrm{:}f\mathit{}\mathrm{(}i\mathrm{)}\mathrm{\ne }\mathrm{0}\mathrm{\}}$ is countable.

Proof.

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

If $G\in ℱ$ and $G\cap F_n=\mathrm{\varnothing }$, then

Let $J={\bigcup }_{n\in \mathbb{N}}{F}_n$. If $i\in I\setminus J$, then for each $n\in \mathbb{N}$, we have $\{i\}\cap F_n=\mathrm{\varnothing }$, whence . That is, if $i\in I\setminus J$ then for each $n\in \mathbb{N}$ we have , which implies that if $i\in I\setminus J$ then $f(i)=0$. Therefore $\{i\in I:f(i)\ne 0\}\subseteq J$, and as $J$ is countable, the set $\{i\in I:f(i)\ne 0\}$ 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, $x_n\mathrm{\in }X$, and ${\mathrm{\sum }}_{n\mathrm{\in }\mathrm{N}}{x}_n$ converges to $x$, then ${\mathrm{\sum }}_{n\mathrm{=}\mathrm{1}}^N{x}_n\mathrm{\to }x$ as $N\mathrm{\to }\mathrm{\infty }$.

Proof.

Let $ϵ>0$, let $ℱ$ be the set of all finite subsets of $\mathbb{N}$, and let $S:ℱ\to X$ be $S(F)={\sum }_{n\in F}{x}_n$. The net $S$ converges to $x$, so there is some $F_{ϵ}\in ℱ$ such that if $F_{ϵ}\subseteq F$ then . Let $N_{ϵ}=\max F_{ϵ}$. If $N\ge N_{ϵ}$, then for $F=\{1,\mathrm{\dots },N\}$ we have $F_{ϵ}\subseteq F$ and so

showing that ${\sum }_{n=1}^N{x}_n\to x$ as $N\to \mathrm{\infty }$. ∎

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 $\sigma :\mathbb{N}\to \mathbb{N}$, ${\sum }_{n=1}^{\mathrm{\infty }}{x}_{\sigma (n)}={\sum }_{n\in \mathbb{N}}{x}_n$. If $x_n$ is a sequence in a normed space and for every bijection $\sigma :\mathbb{N}\to \mathbb{N}$ the series ${\sum }_{n=1}^{\mathrm{\infty }}{x}_{\sigma (n)}$ 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, $x_n\mathrm{\in }X$, and , then ${\mathrm{\sum }}_{n\mathrm{\in }\mathrm{N}}{x}_n$ converges.

Proof.

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

suppose that if then . Define

$$v_k=\sum _{n=1}^{K(k)}{x}_n.$$

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

$$\parallel v_k-v_j\parallel =\parallel \sum _{n=1}^{K(k)}{x}_n-\sum _{n=1}^{K(j)}{x}_n\parallel =\parallel \sum _{n=K(j)+1}^{K(k)}{x}_n\parallel \le \sum _{n=K(j)+1}^{K(k)}\parallel x_n\parallel \le \sum _{n=K(j)+1}^{\mathrm{\infty }}\parallel x_n\parallel ,$$

hence if $k>j\ge N$, then . This shows that $v_k$ is a Cauchy sequence, and hence $v_k$ converges to some $x\in X$.

Let $ℱ$ be the set of all finite subsets of $\mathbb{N}$ and define $S:ℱ\to X$ by $S(F)={\sum }_{n\in F}{x}_n$. Let $ϵ>0$, and as $v_k\to x$ there is some $N_1$ such that if $k\ge N_1$ then . Let $N_2>\frac{1}{ϵ}$, put $N=\max \{N_1,N_2\}$, and put $F_{ϵ}=\{1,\mathrm{\dots },K(N)\}$. If $F_{ϵ}\subseteq F\in ℱ$, then

$\parallel S(F)-x\parallel$ $=$ $\parallel {\displaystyle \sum _{n\in F}}{x}_n-x\parallel$

$\le$ $\parallel {\displaystyle \sum _{n\in F}}{x}_n-{\displaystyle \sum _{n\in F_{ϵ}}}{x}_n\parallel +\parallel {\displaystyle \sum _{n\in F_{ϵ}}}{x}_n-x\parallel$

$=$ $\parallel {\displaystyle \sum _{n\in F\setminus F_{ϵ}}}{x}_n\parallel +\parallel v_N-x\parallel$

${\displaystyle \sum _{n\in F\setminus F_{ϵ}}}\parallel x_n\parallel +ϵ$

$\le$ ${\displaystyle \sum _{n=K(N)+1}^{\mathrm{\infty }}}\parallel x_n\parallel +ϵ$

${\displaystyle \frac{1}{N}}+ϵ$

$2ϵ.$

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 $X=ℂ$. The other direction follows from the Riemann rearrangement theorem.

Theorem 5.

If ${\alpha }_n\mathrm{\in }\mathrm{C}$, then ${\mathrm{\sum }}_{n\mathrm{\in }\mathrm{N}}{\alpha }_n$ converges if and only if .

Let $X$ be a normed space and $z:J\to X$ a net. We say that $z$ is Cauchy if for every $ϵ>0$ there is some $\alpha \in J$ such that $\alpha \le \beta$ and $\alpha \le \gamma$ together imply that .

Theorem 6.

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

Proof.

Let ${\alpha }_1\in J$ such that if ${\alpha }_1\le \alpha$ then , and for $n>1$ let ${\alpha }_n\in J$ be such that if ${\alpha }_n\le \alpha$ then and such that ${\alpha }_{n-1}\le {\alpha }_n$. Define $x_n=z({\alpha }_n)$. For $ϵ>0$, let $N>\frac{1}{ϵ}$. If $n\ge m\ge N$, then, as ${\alpha }_n\ge {\alpha }_m$,

showing that $x_n$ is a Cauchy sequence in $X$. Hence there is some $x\in X$ such that $x_n\to x$.

Let $ϵ>0$, let $N_1>\frac{1}{ϵ}$, let $N_2$ be such that if $n\ge N_2$ then , and set $N=\max \{N_1,N_2\}$. If ${\alpha }_N\le \alpha$, then, by construction of the sequence ${\alpha }_n$,

$\parallel z(\alpha )-x\parallel$ $\le$ $\parallel z(\alpha )-z({\alpha }_N)\parallel +\parallel z({\alpha }_N)-x\parallel$

$=$ $\parallel z(\alpha )-z({\alpha }_N)\parallel +\parallel x_N-x\parallel$

${\displaystyle \frac{1}{N}}+ϵ$

$2ϵ,$

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

Theorem 7.

If $H$ is an infinite dimensional Hilbert space and $\mathrm{\{}{e}_n\mathrm{:}n\mathrm{\in }\mathrm{N}\mathrm{\}}$ is an orthonormal set in $H$, then ${\mathrm{\sum }}_{n\mathrm{\in }\mathrm{N}}\frac{\mathrm{1}}{n}\mathit{}{e}_n$ converges.

Proof.

Let $ℱ$ be the set of finite subsets of $\mathbb{N}$ and let $S(F)={\sum }_{n\in F}\frac{1}{n}{e}_n$, $F\in ℱ$. Define $v_N={\sum }_{n=1}^N\frac{1}{n}{e}_n$. If $N_1>N_2\ge N$, then, as $e_n$ are orthonormal,

so $v_N$ is a Cauchy sequence in $H$ and hence converges to some $h\in H$. For $ϵ>0$, let $N_1>\frac{1}{ϵ}$, let , put $N=\max \{N_1,N_2\}$, and put $F_{ϵ}=\{1,\mathrm{\dots },N\}$. If $F_{ϵ}\subseteq F\in ℱ$, then, using that $e_n$ are orthonormal and $0\le {(a-b)}^2=a^2-2ab+b^2$,

${\parallel S(F)-h\parallel }^2$ $\le$ ${\left(\parallel S(F)-S(F_{ϵ})\parallel +\parallel S(F_{ϵ})-h\parallel \right)}^2$

$\le$ $2{\parallel S(F)-S(F_{ϵ})\parallel }^2+2{\parallel S(F_{ϵ})-h\parallel }^2$

$=$ $2{\parallel {\displaystyle \sum _{n\in F\setminus F_{ϵ}}}{\displaystyle \frac{1}{n}}{e}_n\parallel }^2+2{\parallel v_N-h\parallel }^2$

$=$ $2{\displaystyle \sum _{n\in F\setminus F_{ϵ}}}{\displaystyle \frac{1}{n^2}}+2{\parallel v_N-h\parallel }^2$

$4ϵ.$

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 ${\sum }_{n=1}^{\mathrm{\infty }}\parallel \frac{1}{n}{e}_n\parallel ={\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n}=\mathrm{\infty }$. 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 ${\sum }_{n\in \mathbb{N}}\parallel x_n\parallel =\mathrm{\infty }$.

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 $i\ne j$ implies that $S_i⟂S_j$, 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

$$\underset{n\in \mathbb{N}}{\oplus }{M}_n=\underset{n\in \mathbb{N}}{\bigvee }{M}_n,$$

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

$$\parallel h-v_h\parallel =\underset{v\in M}{inf}\parallel h-v\parallel ,$$

and $h-v_h\in M^{⟂}$. This gives

$$H=M\oplus M^{⟂}.$$

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

$$P(h_1+h_2)=h_1,h_1\in M,h_2\in M^{⟂}.$$

It is straightforward to check that $P$ is linear, $\parallel P\parallel \le 1$ ($\parallel P\parallel =1$ if and only if $M$ is nonzero), $P^2=P$, and $\ker P=M^{⟂}$ and $P(H)=M$. 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 inequality states that if $\{e_n:n\in \mathbb{N}\}$ is an orthonormal set in a Hilbert space $H$ and $h\in H$, then

$$\sum _{n=1}^{\mathrm{\infty }}{|⟨h,e_n⟩|}^2\le {\parallel h\parallel }^2.$$ (1)

Theorem 8.

If $H$ is a Hilbert space, $\mathrm{E}$ is an orthonormal set in $H$, and $h\mathrm{\in }H$, then there are only countably many $e\mathrm{\in }\mathrm{E}$ such that $\mathrm{⟨}h\mathrm{,}e\mathrm{⟩}\mathrm{\ne }\mathrm{0}$.

Proof.

Let

$${ℰ}_n=\{e\in ℰ:|⟨h,e⟩|\ge \frac{1}{n}\}.$$

If ${ℰ}_n$ were infinite, let $\{e_j:j\in \mathbb{N}\}$ be a subset of it, and this gives us a contradiction by (1). Therefore each ${ℰ}_n$ is finite. But if $⟨h,e⟩\ne 0$ then there is some $n$ such that $|⟨h,e⟩|\ge \frac{1}{n}$, so

$$ℰ=\bigcup _{n=1}^{\mathrm{\infty }}{ℰ}_n.$$

Therefore $ℰ$ 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, $\mathrm{E}$ is an orthonormal set in $H$, and $h\mathrm{\in }H$, then

$$\sum _{e\in ℰ}{|⟨h,e⟩|}^2\le {\parallel h\parallel }^2.$$

Proof.

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

$$\sum _{n=1}^{\mathrm{\infty }}{|⟨h,e_n⟩|}^2\le {\parallel h\parallel }^2.$$

Theorem 4 states that if $X$ is a Banach space, $x_n\in X,n\in \mathbb{N}$, and , then the unordered sum ${\sum }_{n\in \mathbb{N}}{x}_n$ converges. Thus, with $X=ℂ$ and $x_n={|⟨h,e_n⟩|}^2$, the unordered sum ${\sum }_{n\in \mathbb{N}}{|⟨h,e_n⟩|}^2$ converges, say to $S$. Because ${\sum }_{n\in \mathbb{N}}{|⟨h,e_n⟩|}^2$ converges to $S$, by Theorem 3 the series ${\sum }_{n=1}^{\mathrm{\infty }}{|⟨h,e_n⟩|}^2$ converges to $S$. But we already know that this series is $\le {\parallel h\parallel }^2$, so

$$\sum _{n\in \mathbb{N}}{|⟨h,e_n⟩|}^2\le {\parallel h\parallel }^2.$$

By Theorem 1, the unordered sum ${\sum }_{e\in ℰ}{|⟨h,e⟩|}^2$ converges if and only if the unordered sum ${\sum }_{n\in \mathbb{N}}{|⟨h,e_n⟩|}^2$ converges, and if they converge they have the same value. Therefore, the unordered sum ${\sum }_{e\in ℰ}{|⟨h,e⟩|}^2$ indeed converges, and it is $\le {\parallel h\parallel }^2$. ∎

4 Convergence of unordered sums in the strong operator topology

Let $H$ be a Hilbert space and let $ℬ(H)$ be the set of bounded linear maps $H\to H$. It is straightforward to check that $ℬ(H)$ is a normed space with the operator norm $\parallel T\parallel ={sup}_{\parallel h\parallel \le 1}\parallel Th\parallel$. (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 $ℬ(H)$ can be characterized in the following way: a net $f:I\to ℬ(H)$ converges to $T\in ℬ(H)$ in the strong operator topology if for all $h\in H$ the net $f(i)h$ converges to $Th$ in $H$.

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

$$S(F)=\sum _{i\in I}f(i)\in ℬ(H).$$

$S$ is a net in $ℬ(H)$, and if the net converges to $T\in ℬ(H)$ 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

$$f\otimes g(h)=⟨h,g⟩f.$$

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

$$\parallel f\otimes g(h)\parallel =\parallel ⟨h,g⟩f\parallel =|⟨h,g⟩|\parallel f\parallel \le \parallel h\parallel \parallel g\parallel \parallel f\parallel ,$$

so $\parallel f\otimes g\parallel \le \parallel f\parallel \parallel g\parallel$, giving $f\otimes g\in ℬ(H)$. Additionally,

$$⟨f\otimes g(h_1),h_2⟩=⟨⟨h_1,g⟩f,h_2⟩=⟨h_1,g⟩⟨f,h_2⟩=⟨h_1,⟨h_2,f⟩g⟩=⟨h_1,g\otimes f(h_2)⟩,$$

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

Theorem 10.

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

Proof.

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

$$\sum _{e\in ℰ}{|⟨h,e⟩|}^2=\sum _{n\in \mathbb{N}}{|⟨h,e_n⟩|}^2=\sum _{n=1}^{\mathrm{\infty }}{|⟨h,e_n⟩|}^2\le {\parallel h\parallel }^2.$$ (2)

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

$$S(F)=\sum _{n\in F}⟨h,e_n⟩e_n\in H.$$

If $ϵ>0$, then by (2) there is some $N$ such that . If $F_{ϵ}=\{1,\mathrm{\dots },N\}$ and $F,G\in ℱ$ both contain $F_{ϵ}$, then, because the $e_n$ are orthonormal,

${\parallel S(F)-S(G)\parallel }^2$ $=$ ${\parallel {\displaystyle \sum _{n\in F}}⟨h,e_n⟩e_n-{\displaystyle \sum _{n\in G}}⟨h,e_n⟩e_n\parallel }^2$

$=$ ${\displaystyle \sum _{n\in (F\cup G)\setminus (F\cap G)}}{\parallel ⟨h,e_n⟩e_n\parallel }^2$

$=$ ${\displaystyle \sum _{n\in (F\cup G)\setminus (F\cap G)}}{|⟨h,e_n⟩|}^2$

$\le$ ${\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{|⟨h,e_n⟩|}^2$

${ϵ}^2.$

Therefore, if $F,G\in ℱ$ both contain $F_{ϵ}$ then . This means that $S$ is a Cauchy net, and hence, by Theorem 6, has a limit $v\in H$. That is, the unordered sum ${\sum }_{n\in \mathbb{N}}⟨h,e_n⟩e_n$ converges to $v$.

As the unordered sum ${\sum }_{n\in \mathbb{N}}⟨h,e_n⟩e_n$ converges to $v$ we have

$$\underset{N\to \mathrm{\infty }}{lim}\sum _{n=1}^N⟨h,e_n⟩e_n=v.$$

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

$$\underset{N\to \mathrm{\infty }}{lim}\sum _{n=1}^N⟨h,e_n⟩⟨e_n,e_m⟩=⟨v,e_m⟩,$$

which is

$$⟨h,e_m⟩=⟨v,e_m⟩.$$

Let $Q$ be the orthogonal projection onto ${\bigvee }_{n\in \mathbb{N}}\{e_n\}$. On the one hand, because $⟨h,e⟩=0$ for $e\notin \{e_n:n\in \mathbb{N}\}$, we check that $Ph=Qh$. On the other hand, we check that $Qh=v$. Therefore, $v=Ph$, i.e.

$$\sum _{e\in ℰ}e\otimes e(h)=\sum _{e\in ℰ}⟨h,e⟩e=\sum _{n\in \mathbb{N}}⟨h,e_n⟩e_n=Ph,$$

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

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