Proof of Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation

Each result cited is universally quantified over the data in its own statement.

By Pointwise Operations on Real-Valued Maps §operations, the sum and the product on $\mathbb{R}^{I}$ are the values of the binary operations of Pointwise Operations on Real-Valued Maps Exist and Are Unique §binary, the scalar multiple and the negative are the values of the maps of Pointwise Operations on Real-Valued Maps Exist and Are Unique §scalar-multiple and Pointwise Operations on Real-Valued Maps Exist and Are Unique §negative, and the constant maps $0,1\in\mathbb{R}^{I}$ are those of Pointwise Operations on Real-Valued Maps §constants. The elements of $\mathbb{R}^{I}$ are maps from $I$ to $\mathbb{R}$, by The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set §members, and all have domain $I$ by Functions, Values of a Function, and Functions from One Class to Another §map; so two of them are equal as soon as their values agree at every $p\in I$, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §equality. We refer to this as (E).

$\mathbb{R}$, with $+$, $\cdot$, $0$, $1$ and $\le$, is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, hence a field and so a commutative ring, by Ordered Fields §ordered-field and Fields §field. Each identity of values below holds in $\mathbb{R}$ by the axioms of Commutative Rings §ring or by the rule of Rules of Arithmetic and Order in an Ordered Field cited with it, the values of the operations on $\mathbb{R}^{I}$ being those of Pointwise Operations on Real-Valued Maps §operations and Pointwise Operations on Real-Valued Maps §constants; by (E) it yields the stated equality in $\mathbb{R}^{I}$. Let $f,g,h\in\mathbb{R}^{I}$, $s,t\in\mathbb{R}$ and $p\in I$.

Clause ring. We have

$$((f+g)+h)(p)=(f(p)+g(p))+h(p)=f(p)+(g(p)+h(p))=(f+(g+h))(p),\qquad (f+g)(p)=f(p)+g(p)=g(p)+f(p)=(g+f)(p),$$ $$(f+0)(p)=f(p)+0=f(p),\qquad ((fg)h)(p)=(f(p)g(p))h(p)=f(p)(g(p)h(p))=(f(gh))(p),\qquad (fg)(p)=f(p)g(p)=g(p)f(p)=(gf)(p),$$ $$(f1)(p)=f(p)\cdot1=f(p),\qquad (f(g+h))(p)=f(p)\bigl(g(p)+h(p)\bigr)=f(p)g(p)+f(p)h(p)=(fg+fh)(p),$$

and $(f+(-f))(p)=f(p)+(-f(p))=0=0(p)$, the middle equality by Negatives, Differences, Reciprocals and Quotients §negative in $\mathbb{R}$. By (E), $(f+g)+h=f+(g+h)$, $f+g=g+f$, $f+0=f$, $(fg)h=f(gh)$, $fg=gf$, $f1=f$, $f(g+h)=fg+fh$ and $f+(-f)=0$. Since the sum and the product are binary operations on the set $\mathbb{R}^{I}$ and $0,1\in\mathbb{R}^{I}$, these are all the requirements of Commutative Rings §ring, $-f$ serving as the element $w$ with $f+w=0$. So $\mathbb{R}^{I}$ is a commutative ring. Its negative of $f$ is the unique $w\in\mathbb{R}^{I}$ with $f+w=0$, by Negatives, Differences, Reciprocals and Quotients §negative and Additive and Multiplicative Inverses Are Unique §negative; as $-f$ is such a $w$, it is $-f$. In particular the difference $f-g=f+(-g)$ of Pointwise Operations on Real-Valued Maps §operations is also the difference in this ring.

Clause values. $(f-g)(p)=(f+(-g))(p)=f(p)+(-g)(p)=f(p)+(-g(p))=f(p)-g(p)$, the last step by Negatives, Differences, Reciprocals and Quotients §negative in $\mathbb{R}$.

Clause scalar-laws. Here $1$ is the unit of $\mathbb{R}$ used as a scalar. The reading as the product with the constant map $1$ gives the same element, by the clause constant-multiple with $t=1$ and the clause ring. We have

$$(s(tf))(p)=s\,(t\,f(p))=(st)\,f(p)=((st)f)(p),\qquad (1f)(p)=1\cdot f(p)=f(p),$$ $$((s+t)f)(p)=(s+t)f(p)=s\,f(p)+t\,f(p)=(sf+tf)(p),\qquad (t(f+g))(p)=t\bigl(f(p)+g(p)\bigr)=t\,f(p)+t\,g(p)=(tf+tg)(p),$$ $$(t(fg))(p)=t\bigl(f(p)g(p)\bigr)=\bigl(t\,f(p)\bigr)g(p)=((tf)g)(p),$$

by associativity, commutativity, the unit law and distributivity in $\mathbb{R}$; (E) gives the five laws.

Clause constant-multiple. Let $h$ be the constant map with value $t$. Then $(hf)(p)=h(p)f(p)=t\,f(p)=(tf)(p)$, so $tf=hf$ by (E). Further $(0f)(p)=0\cdot f(p)=0=0(p)$ by Rules of Arithmetic and Order in an Ordered Field §zero; with $-1\in\mathbb{R}$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §negative, $((-1)f)(p)=(-1)f(p)=-(1\cdot f(p))=-f(p)=(-f)(p)$ by Rules of Arithmetic and Order in an Ordered Field §signs and, for $-(1\cdot f(p))=-f(p)$, by the unit law and commutativity of Commutative Rings §ring in $\mathbb{R}$; and $(t0)(p)=t\cdot0=0\cdot t=0=0(p)$ by commutativity and Rules of Arithmetic and Order in an Ordered Field §zero. By (E), $0f=0$, $(-1)f=-f$ and $t0=0$.

Clause zero-product. If $t=0$, then $tf=0f=0$, and if $f=0$, then $tf=t0=0$, both by the clause constant-multiple. Conversely, let $tf=0$ and $t\neq0$. For every $q\in I$, $t\,f(q)=(tf)(q)=0(q)=0$, so $f(q)=0$ by Rules of Arithmetic and Order in an Ordered Field §zero, as $t\neq0$; thus $f(q)=0(q)$ for every $q\in I$, and $f=0$ by (E).

Clause evaluation. Fix $p\in I$. For $u\in\mathbb{R}^{I}$, the value $u(p)$ lies in $\mathbb{R}$, and the expression $u(p)$ is built from the defined set symbol of Functions, Values of a Function, and Functions from One Class to Another §value, used properly, with no quantifier over a class variable, so it is predicative by Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols §closure (see also Predicative Substitution of Values of Functions and Binary Operations, and Predicative Formulas for Negatives, Differences, Absolute Values, Distances, Reciprocals and Quotients in an Ordered Field §value-substitution). Hence, by Sets and Maps: Ordinary Notation §maps and Maps and Relations Given by Formulas §map, there is a map $\varepsilon:\mathbb{R}^{I}\to\mathbb{R}$ with $\varepsilon(u)=u(p)$ for every $u\in\mathbb{R}^{I}$. Likewise, for a finite set $A$ and $F:A\to\mathbb{R}^{I}$, each $F(a)$ with $a\in A$ is a map from $I$ to $\mathbb{R}$, so $F(a)(p)\in\mathbb{R}$, and the expression $F(a)(p)$, two values of maps, is predicative by the same clause; so the map $a\mapsto F(a)(p)$ from $A$ to $\mathbb{R}$ exists by Sets and Maps: Ordinary Notation §maps, and it is the map $a\mapsto\varepsilon(F(a))$, these having the same values, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §equality. Now $\varepsilon(0)=0(p)=0$, and $\varepsilon(u+v)=(u+v)(p)=u(p)+v(p)=\varepsilon(u)+\varepsilon(v)$ for $u,v\in\mathbb{R}^{I}$. The sum of $\mathbb{R}^{I}$ is associative and commutative with neutral element $0$ by the clause ring (on both sides by commutativity), and so is the sum of $\mathbb{R}$, these being the facts by which finite sums are defined in a commutative ring in Commutative Rings, Fields and Ordered Fields: Standard Notation §rings. Therefore Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §homomorphism, applied with $X=\mathbb{R}^{I}$, $Y=\mathbb{R}$, both operations $+$, $e=e'=0$, $\varphi=\varepsilon$ and $f=F$, gives

$$\Bigl(\sum_{a\in A}F(a)\Bigr)(p)=\varepsilon\Bigl(\sum_{a\in A}F(a)\Bigr)=\sum_{a\in A}\varepsilon(F(a))=\sum_{a\in A}F(a)(p).$$

This includes the case $A=\emptyset$, in which both sides are $0$ by Sums and Products over a Finite Set and over an Interval §empty.