Proof of Composites of Cᵏ Functions Are of Class Cᵏ

Throughout, the arithmetic and order of $\mathbb{R}$ are used through The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ix), Rules of Arithmetic and Order in an Ordered Field (iii) and The Quotient 1/x, Order Rules for Sums, and Nesting of Closed Intervals in ℝ (i), norms and distances in $\mathbb{R}^{n}$ through Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi) and Standard Basis Vectors of ℝ^q: Coordinates, Orthonormality and Unit Length (iii), and $(\mathbb{R}^{n},d_{E})$ and $(\mathbb{R},d_{\mathbb{R}})$ are metric spaces by $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (i) and The Absolute Value Metric on the Real Numbers (i); membership in $\mathcal{C}^{0}$ and $\mathcal{C}^{k+1}$ of $U$ and of $V$ is read through Functions of Class Cᵏ on an Open Subset of ℝⁿ (i), $(g\circ f)(x)=g(f(x))$, and the condition used in the induction below is predicative.

We first prove (i), beginning with a real-valued $g:V\to\mathbb{R}$ of class $C^{1}$ on $V$. Fix $x\in U$. By Basic Properties of Cᵏ Functions (v), $f$ is differentiable at $x$ with derivative matrix $Df(x)$, and $g$, as a function on the open set $V\subseteq\mathbb{R}^{m}$, is differentiable at $f(x)$ with derivative vector $\nabla g(f(x))$. By Chain Rule for Differentiable Functions Between Euclidean Spaces (ii), $g\circ f$ is differentiable at $x$ with derivative vector

$$\nabla(g\circ f)(x)=Df(x)^{\top}\,\nabla g\bigl(f(x)\bigr).$$

By The Derivative Is the Jacobian Matrix, and Is Unique (i), $\partial_{i}(g\circ f)(x)$ exists and is the $i$-th component of this vector. Expanding the transpose (Transpose of a Real Matrix (i)), the matrix-vector product (Matrix-Vector Product (i)), the gradient (Gradient of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (i)) and the Jacobian entries $(Df(x))_{li}=\partial_{i}f_{l}(x)$ (Jacobian Matrix of a Function Between Euclidean Spaces (i)), we get

$$\partial_{i}(g\circ f)(x)=\sum_{l\in[m]}\bigl(Df(x)\bigr)_{li}\,\partial_{l}g\bigl(f(x)\bigr)=\sum_{l\in[m]}\partial_{l}g\bigl(f(x)\bigr)\,\partial_{i}f_{l}(x).$$

Now let $g:V\to\mathbb{R}^{p}$ be of class $C^{1}$ on $V$. Every component $g_{j}$ is of class $C^{1}$ on $V$ (Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii)), and $(g\circ f)_{j}=g_{j}\circ f$; the real-valued case applied to $g_{j}$ gives the formula of (i).

Next we prove (ii) for real-valued $g$. We show by induction on $k$ (The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (viii)) that for every $f:U\to\mathbb{R}^{m}$ of class $C^{k}$ on $U$ with $f(x)\in V$ for every $x\in U$, and every $g\in\mathcal{C}^{k}(V)$, we have $g\circ f\in\mathcal{C}^{k}(U)$. For $k=0$, $f$ is continuous on $U$ by Basic Properties of Cᵏ Functions (iv), $g$ is continuous on $V$, and so $g\circ f$ is continuous on $U$ by Continuity of Identity Maps, Compositions and Reciprocals (ii). For the inductive step, let $f$ and $g$ be of class $C^{k+1}$. By Basic Properties of Cᵏ Functions (ii), both are of class $C^{1}$ and of class $C^{k}$, so the real-valued case of (i) shows that every $\partial_{i}(g\circ f)$ exists on $U$, with

$$\partial_{i}(g\circ f)=\sum_{l\in[m]}\bigl(\partial_{l}g\circ f\bigr)\,\partial_{i}f_{l}.$$

Here $\partial_{l}g\in\mathcal{C}^{k}(V)$ and $\partial_{i}f_{l}\in\mathcal{C}^{k}(U)$ by Basic Properties of Cᵏ Functions (iii), and $\partial_{l}g\circ f\in\mathcal{C}^{k}(U)$ by the induction hypothesis, applied to $f$ and $\partial_{l}g$. Each summand therefore lies in $\mathcal{C}^{k}(U)$ by Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (v), and so does their sum over $l\in[m]$, by Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (iv) and induction on the number of summands. Hence $g\circ f\in\mathcal{C}^{k+1}(U)$.

For vector-valued $g:V\to\mathbb{R}^{p}$ of class $C^{k}$, the component maps of $g\circ f$ are the $g_{j}\circ f$, each of class $C^{k}$ by the real-valued case, so $g\circ f$ is of class $C^{k}$ by Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii); this completes (ii). Applying (ii) for every $k\in\mathbb{N}_{0}$ gives (iii), since smoothness means being of class $C^{k}$ for every $k$ (Smooth Functions on an Open Subset of ℝⁿ (i), Smooth Functions on an Open Subset of ℝⁿ (ii)).