Proof of Finite Sums of Cᵏ Functions Are of Class Cᵏ

Let $k$, $N$ and $f_{1},\dots,f_{N}$ be as in the statement. For $M\in\mathbb{N}$ with $M\le N$ write $g_{M}:U\to\mathbb{R}$ for the function $x\mapsto\sum_{i=1}^{M}f_{i}(x)$; thus $g_{N}$ is the function of the statement. By the recursion for finite sums, applied for each $x\in U$ to the map $i\mapsto f_{i}(x)$ (restricted to $[M+1]$ as in Iterated Operations: Finite Sums and Finite Products (ii)), we have $g_{1}=f_{1}$ and, whenever $M+1\le N$,

$$g_{M+1}(x)=g_{M}(x)+f_{M+1}(x)\qquad(x\in U),$$

that is, $g_{M+1}=g_{M}+f_{M+1}$ as a pointwise sum in the sense of Functions of Class Cᵏ: Notation (v).

Proof of (i). Suppose that every $f_{i}$ is of class $C^{k}$ on $U$. We show by induction on $M$, as in The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (viii), that $g_{M}$ is of class $C^{k}$ on $U$ for every $M\in\mathbb{N}$ with $M\le N$ (the induction condition is expressed by a predicative formula by Iterated Operations Are Given by a Predicative Formula (i)). For $M=1$, $g_{1}=f_{1}$ is of class $C^{k}$. Let $M\in\mathbb{N}$ satisfy the claim, and suppose $M+1\le N$. Then $M\le N$, so $g_{M}$ is of class $C^{k}$ on $U$, and so is $f_{M+1}$; hence $g_{M+1}=g_{M}+f_{M+1}$ is of class $C^{k}$ on $U$ by Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (iv). This completes the induction, and $M=N$ gives the claim.

Proof of (ii). Suppose that every $f_{i}$ is smooth on $U$, and let $j\in\mathbb{N}_{0}$. By Smooth Functions on an Open Subset of ℝⁿ (i), every $f_{i}$ is of class $C^{j}$ on $U$, so $g_{N}$ is of class $C^{j}$ on $U$ by (i) with $j$ in place of $k$. Since $j$ was arbitrary, $g_{N}$ is smooth on $U$ by Smooth Functions on an Open Subset of ℝⁿ (i).