TheoremBase

The partial-sum functions are the iterates of the pointwise sum on [0,∞]^X; evaluation at a point is a homomorphism, so their values are the partial sums of the values.

Proof

Write P=[0,∞]XP=[0,\infty]^{X}, a set whose elements are exactly the functions X→[0,∞]X\to[0,\infty] (The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (ii), The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (i)), and let [n][n], for n∈Nn\in\mathbb{N}, be the segment {1,…,n}\{1,\dots,n\}, a subset of N\mathbb{N} (Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i)).

The pointwise sum and evaluation. For g,h∈Pg,h\in P, the sum g+hg+h of Sums, Multiples and Products of Functions into [0,∞] (i) lies in PP, so there is a function ⊕:P×P→P\oplus:P\times P\to P with g⊕h=g+hg\oplus h=g+h for all g,h∈Pg,h\in P (Maps and Relations Given by Formulas (ii)); it is a binary operation on PP. For x∈Xx\in X, the function εx:P→[0,∞]\varepsilon_{x}:P\to[0,\infty], g↦g(x)g\mapsto g(x), exists by Maps and Relations Given by Formulas (i), and by Sums, Multiples and Products of Functions into [0,∞] (i)

εx(g⊕h)=(g+h)(x)=g(x)+h(x)=εx(g)+εx(h)(g,h∈P).\varepsilon_{x}(g\oplus h)=(g+h)(x)=g(x)+h(x)=\varepsilon_{x}(g)+\varepsilon_{x}(h)\qquad(g,h\in P).

The partial-sum functions. As (fk)(f_{k}) is a sequence in PP, Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set (i) gives a function H:N→PH:\mathbb{N}\to P with

H(n)=⊕k=1nfk(n∈N),H(n)=\mathop{\oplus}\limits_{k=1}^{n}f_{k}\qquad(n\in\mathbb{N}),

the iterated operation of ⊕\oplus being read as in Iterated Operations: Finite Sums and Finite Products (ii). Thus HH is a sequence (Hn)(H_{n}), Hn=H(n)H_{n}=H(n), of functions X→[0,∞]X\to[0,\infty].

Their values. Let n∈Nn\in\mathbb{N}, x∈Xx\in X, and let uu be a sequence in [0,∞][0,\infty] with uk=fk(x)u_{k}=f_{k}(x) for every k∈Nk\in\mathbb{N}. Let a:[n]→Pa:[n]\to P and b:[n]→[0,∞]b:[n]\to[0,\infty] be the restrictions of (fk)(f_{k}) and of uu to [n][n], so that ak=fka_{k}=f_{k} and bk=ukb_{k}=u_{k} for k∈[n]k\in[n] (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ix)). By the display above and Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (ii), applied to εx\varepsilon_{x},

Hn(x)=εx(⊕k=1nak)=∑k=1nεx(ak).H_{n}(x)=\varepsilon_{x}\Bigl(\mathop{\oplus}\limits_{k=1}^{n}a_{k}\Bigr)=\sum_{k=1}^{n}\varepsilon_{x}(a_{k}).

The function [n]→[0,∞][n]\to[0,\infty], k↦εx(ak)=fk(x)=ukk\mapsto\varepsilon_{x}(a_{k})=f_{k}(x)=u_{k}, has the domain and values of bb, so it is bb (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (i)). Hence

Hn(x)=∑k=1nbk=∑k=1nuk,H_{n}(x)=\sum_{k=1}^{n}b_{k}=\sum_{k=1}^{n}u_{k},

the nn-th partial sum of uu (Iterated Operations: Finite Sums and Finite Products (ii), Sum of a Sequence in [0,∞] (i)). This is (i).

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