Square-summable families on an arbitrary index set and the space ${\mathrm{\ell }}^2(I)$

Throughout, $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ and families are indexed by an arbitrary set $I$, with no enumeration or countability assumed. Finite real lists use Finite sums and finite products, by recursion, and sums over finite subsets use A finite sum in a commutative monoid indexed by an arbitrary finite set in the additive monoid of the scalar field. Disjoint splitting is Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule, and scalar modulus estimates use Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive; all suprema and infima of real sets below are taken in the complete ordered field $\mathbb{R}$ (Complete ordered field (least-upper-bound property)) or in $[0,+\mathrm{\infty }]\subseteq \bar{\mathbb{R}}$ (The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined, Every subset of $\bar{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\bar{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$).

Sums of nonnegative families. Let $(c_i{)}_{i\in I}$ be a family of nonnegative reals and let $\mathrm{Fin}(I)$ be the set of finite subsets of $I$, ordered by inclusion. Define

${\sum }_{i\in I}{c}_i:=\sup \{{\sum }_{i\in F}{c}_i:F\in \mathrm{Fin}(I)\}\in [0,+\mathrm{\infty }]\mathrm{.}$

The set of finite subsums is nonempty, since $\mathrm{\varnothing }$ contributes the empty sum $0$, so the supremum exists in $[0,+\mathrm{\infty }]$; it is a real number exactly when the finite subsums are bounded above in $\mathbb{R}$, and $+\mathrm{\infty }$ otherwise. For finite $I$ the finite subsum at $I$ is the largest of all finite subsums, because all terms are nonnegative, so the definition agrees with the finite sum; in particular ${\sum }_{i\in I}{c}_i=0$ when every $c_i$ is $0$, and $I=\mathrm{\varnothing }$ gives the empty sum $0$.

Splitting identity and small tails. Fix a finite $F\subseteq I$. Every finite $G\subseteq I$ splits as the disjoint union $(G\cap F)\cup (G\setminus F)$, so ${\sum }_{i\in G}{c}_i={\sum }_{i\in G\cap F}{c}_i+{\sum }_{i\in G\setminus F}{c}_i\le {\sum }_{i\in F}{c}_i+{\sum }_{i\in I\setminus F}{c}_i$; conversely the finite $G$ with $F\subseteq G$ satisfy ${\sum }_{i\in G}{c}_i={\sum }_{i\in F}{c}_i+{\sum }_{i\in G\setminus F}{c}_i$. Taking suprema, with the constant ${\sum }_{i\in F}{c}_i$ passing through the supremum,

${\sum }_{i\in I}{c}_i={\sum }_{i\in F}{c}_i+{\sum }_{i\in I\setminus F}{c}_i\mathrm{.}(1)$

Consequently, if $S:={\sum }_{i\in I}{c}_i$ is finite, then for every real $\epsilon >0$ there is a finite $F\subseteq I$ with ${\sum }_{i\in I\setminus F}{c}_i<\epsilon$: the finite subsums form a nonempty bounded-above set with supremum $S$, so by the epsilon characterisation of the supremum (Epsilon characterisation of the supremum) some finite $F$ has $S-\epsilon <{\sum }_{i\in F}{c}_i$, and (1) gives ${\sum }_{i\in I\setminus F}{c}_i=S-{\sum }_{i\in F}{c}_i<\epsilon$. Such an $F$ is called a tail-control set for $\epsilon$.

Sums of scalar families. Now let $(a_i{)}_{i\in I}$ be a family in $\mathbb{F}$ and put $s_F:={\sum }_{i\in F}{a}_i$ for finite $F\subseteq I$. The set $\mathrm{Fin}(I)$ with inclusion is a directed preorder: it is nonempty and $F\cup G$ is a common upper bound of $F$ and $G$ (Directed preorders and nets). Hence $(s_F{)}_{F\in \mathrm{Fin}(I)}$ is a net in $\mathbb{F}$, the finite-subset net of the family, and the family is summable when this net converges (Convergence and cluster points of a net in a topological space). The scalar metric is the usual real metric or The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane. These metric spaces are Hausdorff by Distinct points of a metric space have disjoint balls around them. A net in $\mathbb{F}$ has at most one limit (A topological space is Hausdorff if and only if every net has at most one limit), so for a summable family the limit is unique and we write ${\sum }_{i\in I}{a}_i:={\lim }_F{s}_F$ for it. The family is absolutely summable when ${\sum }_{i\in I}\mathrm{\mid }{a}_i\mathrm{\mid }<+\mathrm{\infty }$ in the sense above.

Every absolutely summable family is summable, in ZF. Assume $S:={\sum }_{i\in I}\mathrm{\mid }{a}_i\mathrm{\mid }$ is finite and put $S_F:={\sum }_{i\in F}\mathrm{\mid }{a}_i\mathrm{\mid }$ for finite $F$. For finite $F,G\subseteq I$ the triangle inequality for finite sums and the fact that a finite subsum is at most the whole nonnegative sum give

$\mathrm{\mid }{s}_F-s_G\mathrm{\mid }\le {\sum }_{i\in F\mathrm{△}G}\mathrm{\mid }{a}_i\mathrm{\mid }\le S-S_{F\cap G}\mathrm{.}(2)$

Now fix a real $\epsilon >0$, let $F_0$ be a tail-control set for $\epsilon$, and let $F,G\supseteq F_0$ be finite. Then $F\cap G\supseteq F_0$, so $S-S_{F\cap G}\le S-S_{F_0}<\epsilon$ and (2) gives $\mathrm{\mid }{s}_F-s_G\mathrm{\mid }<\epsilon$.

First suppose $\mathbb{F}=\mathbb{R}$. For each finite $F$ define $A_F:=\inf \{s_G:G\supseteq F\}$ and $B_F:=\sup \{s_G:G\supseteq F\}$ over finite $G$. Both are real numbers, because $\mathrm{\mid }{s}_G\mathrm{\mid }\le S_G\le S$, so the two sets are nonempty and bounded, and $A_F\le B_F$. If $H\supseteq F$ then $A_H\ge A_F$ and $B_H\le B_F$. Let $L:={\sup }_F{A}_F$ and $U:={\inf }_F{B}_F$. Since $A_F\le s_{F\cup H}\le B_H$ for all finite $F,H$, we get $L\le U$. Moreover, the preceding estimate holds for every pair $F,G\supseteq F_0$; taking the supremum over $F$ and the infimum over $G$ gives $B_{F_0}-A_{F_0}\le \epsilon$. Hence $0\le U-L\le B_{F_0}-A_{F_0}\le \epsilon$ for every $\epsilon >0$, so $L=U=:s$. If $F\supseteq F_0$, then both $s_F$ and $s$ lie in $[A_{F_0},B_{F_0}]$, and therefore $\mathrm{\mid }{s}_F-s\mathrm{\mid }\le \epsilon$.

If $\mathbb{F}=\mathbb{C}$, apply the real argument just proved to the families $(\mathrm{Re}{a}_i)$ and $(\mathrm{Im}{a}_i)$. They are absolutely summable because $\mathrm{\mid }\mathrm{Re}{a}_i\mathrm{\mid },\mathrm{\mid }\mathrm{Im}{a}_i\mathrm{\mid }\le \mathrm{\mid }{a}_i\mathrm{\mid }$. Their finite-subset nets converge to real numbers $r$ and $t$, respectively, so $s_F\to r+it$ in $\mathbb{C}$. Thus every absolutely summable real or complex family is summable. No choice principle is used: the construction uses only two-sided suprema in $\mathbb{R}$.

Linearity and absolute value. If $(a_i)$ and $(b_i)$ are absolutely summable and $\lambda \in \mathbb{F}$, then so are $(a_i+b_i)$ and $(\lambda a_i)$, and

${\sum }_{i\in I}(a_i+b_i)={\sum }_{i\in I}{a}_i+{\sum }_{i\in I}{b}_i,{\sum }_{i\in I}\lambda a_i=\lambda {\sum }_{i\in I}{a}_i,\mid {\sum }_{i\in I}{a}_i\mid \le {\sum }_{i\in I}\mathrm{\mid }{a}_i\mathrm{\mid },$

because the corresponding identities hold for every finite subsum, both sides are limits of the corresponding finite-subset nets, and addition, scalar multiplication and the modulus are continuous. The splitting identity (1) likewise passes to absolutely summable scalar families: ${\sum }_{i\in I}{a}_i={\sum }_{i\in F}{a}_i+{\sum }_{i\in I\setminus F}{a}_i$ for every finite $F$, because finite subsums over sets containing $F$ converge to the left-hand side and equal the finite sum over $F$ plus the finite subsum of the tail, whose net converges to the tail sum.

The finite-dimensional Cauchy-Schwarz inequality. Let $(u_k{)}_{0\le k<m}$ and $(v_k{)}_{0\le k<m}$ be scalar lists, for $m\in \mathbb{N}$ and let $t\in \mathbb{R}$. Every term of ${\sum }_{k<m}(\mathrm{\mid }{u}_k\mathrm{\mid }-t\mathrm{\mid }{v}_k\mathrm{\mid }{)}^2$ is nonnegative, so for all real $t$

$0\le {\sum }_{k<m}\mathrm{\mid }{u}_k{\mathrm{\mid }}^2-2t{\sum }_{k<m}\mathrm{\mid }{u}_k\mathrm{\mid }\mathrm{\mid }{v}_k\mathrm{\mid }+t^2{\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2\mathrm{.}$

If ${\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2>0$, substituting $t=\left({\sum }_{k<m}\mathrm{\mid }{u}_k\mathrm{\mid }\mathrm{\mid }{v}_k\mathrm{\mid }\right)\mathrm{/}\left({\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2\right)$ gives $({\sum }_{k<m}\mathrm{\mid }{u}_k\mathrm{\mid }\mathrm{\mid }{v}_k\mathrm{\mid }{)}^2\le ({\sum }_{k<m}\mathrm{\mid }{u}_k{\mathrm{\mid }}^2\left)\right({\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2)$; if ${\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2=0$ then every $v_k=0$ and both sides are $0$. In either case ${\sum }_{k<m}\mathrm{\mid }{u}_k\mathrm{\mid }\mathrm{\mid }{v}_k\mathrm{\mid }\le \left({\sum }_{k<m}\mathrm{\mid }{u}_k{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\right({\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ after taking square roots, and applying the modulus inequality for finite sums to $u_k\overline{v_k}$ also gives

$\mid {\sum }_{k<m}{u}_k\overline{v_k}\mid \le \left({\sum }_{k<m}\mathrm{\mid }{u}_k{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\right({\sum }_{k<m}\mathrm{\mid }{v}_k{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\mathrm{.}(3)$

The space ${\mathrm{\ell }}^2(I)$. For a family $a=(a_i{)}_{i\in I}$ in $\mathbb{F}$ define $Q(a):={\sum }_{i\in I}\mathrm{\mid }{a}_i{\mathrm{\mid }}^2$. If $Q(a)$ is finite, set $\mathrm{\parallel }a{\mathrm{\parallel }}_2:=\sqrt{Q(a)}$, using the nonnegative real square root of Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$; if $Q(a)=+\mathrm{\infty }$, set $\mathrm{\parallel }a{\mathrm{\parallel }}_2:=+\mathrm{\infty }$. This is a case definition, not exponentiation of an extended real. Let

${\mathrm{\ell }}^2(I,\mathbb{F}):=\{a=(a_i{)}_{i\in I}:\mathrm{\parallel }a{\mathrm{\parallel }}_2<+\mathrm{\infty }\}\mathrm{.}$

The set ${\mathrm{\ell }}^2(I,\mathbb{F})$ is a vector space over $\mathbb{F}$: it contains the zero family, is closed under scalar multiplication because $\mathrm{\parallel }\lambda a{\mathrm{\parallel }}_2=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }a{\mathrm{\parallel }}_2$, and is closed under addition because (3) applied to finite subsums gives $\mathrm{\parallel }a+b{\mathrm{\parallel }}_2\le \mathrm{\parallel }a{\mathrm{\parallel }}_2+\mathrm{\parallel }b{\mathrm{\parallel }}_2<+\mathrm{\infty }$. The same inequality is the triangle inequality for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$, which is moreover nonnegative, vanishes only for the zero family (an arbitrary sum of nonnegative terms with supremum $0$ has every term $0$) and satisfies $\mathrm{\parallel }\lambda a{\mathrm{\parallel }}_2=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }a{\mathrm{\parallel }}_2$; thus $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ is a norm on ${\mathrm{\ell }}^2(I,\mathbb{F})$. Finally the pairing $⟨a,b⟩:={\sum }_{i\in I}{a}_i\overline{b_i}$ is well defined on ${\mathrm{\ell }}^2(I,\mathbb{F})\times {\mathrm{\ell }}^2(I,\mathbb{F})$, because $\mathrm{\mid }{a}_i\overline{b_i}\mathrm{\mid }=\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }{b}_i\mathrm{\mid }$ has finite nonnegative sum by (3); it is linear in the first variable, conjugate symmetric, positive definite, and satisfies $⟨a,a⟩=\mathrm{\parallel }a{\mathrm{\parallel }}_2^2$, all by the corresponding finite identities and the linearity of the sum. Consequently ${\mathrm{\ell }}^2(I,\mathbb{F})$ is an inner-product space, and $e_i\in {\mathrm{\ell }}^2(I,\mathbb{F})$ denotes the family that is $1$ at $i$ and $0$ elsewhere. Nothing here asserts that ${\mathrm{\ell }}^2(I,\mathbb{F})$ is complete; that follows later, from an orthonormal basis of a Hilbert space.