Proof of Chain Rule for Differentiable Maps Between Euclidean Spaces

Throughout, $\lVert\,\cdot\,\rVert$ denotes the Euclidean norm, used on $\mathbb{R}^n$, $\mathbb{R}^m$ and $\mathbb{R}^p$ alike; sums, differences and scalar multiples of points are those of Sum of Points of $\mathbb{R}^n$, Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$ and Scalar Multiple of a Point of $\mathbb{R}^n$, and each of $\mathbb{R}^n$, $\mathbb{R}^m$, $\mathbb{R}^p$ is regarded as a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, with origin $0_{\mathbb{R}^n}$ and so on as in The Origin of $\mathbb{R}^n$. The order arithmetic of the ordered field $\mathbb{R}$ of real numbers is that of Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field; in particular we use repeatedly that non-strict inequalities may be added, that is, that $x\le y$ and $x'\le y'$ imply $x+x'\le y+y'$, which follows from claims 2 and 3 of Elementary Arithmetic in an Ordered Field, and that a non-strict inequality may be multiplied by a nonnegative factor, which is claim 5 of that lemma. Write $2=1+1$, so that $0<2$ and $\varepsilon\cdot 2^{-1}+\varepsilon\cdot 2^{-1}=\varepsilon$ for every real $\varepsilon$ with $0<\varepsilon$, by claim 8 of Elementary Order Arithmetic in an Ordered Field.

Setup. Put $b=f(a)$; then $b\in V$. By claim 3 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product there are real numbers $C_A$ and $C_B$ with $0\le C_A$ and $0\le C_B$ such that

$$\lVert A\,z\rVert\le C_A\lVert z\rVert\ \ \text{for every }z\in\mathbb{R}^n,\qquad \lVert B\,u\rVert\le C_B\lVert u\rVert\ \ \text{for every }u\in\mathbb{R}^m .$$

Put $K=C_A+1$ and $L=C_B+1$. Since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, claim 3 of that lemma gives $0<K$ and $0<L$; and since $0\le 1$ by claim 1 of Elementary Arithmetic in an Ordered Field, claim 3 of that lemma gives $C_A\le K$ and $C_B\le L$. By claim 7 of Elementary Order Arithmetic in an Ordered Field the inverses $K^{-1}$ and $L^{-1}$ exist and are positive.

Let $\varepsilon$ be a real number with $0<\varepsilon$. By claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<\varepsilon\cdot 2^{-1}$, and then by claim 5 of that lemma

$$\eta=\varepsilon\cdot 2^{-1}\cdot L^{-1}\quad\text{and}\quad \varepsilon_2=\varepsilon\cdot 2^{-1}\cdot K^{-1}$$

are positive. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is a real number $\varepsilon_1$ with $\varepsilon_1\le 1$, $\varepsilon_1\le\eta$, and $\varepsilon_1$ equal to $1$ or to $\eta$; since both $1$ and $\eta$ are positive, $0<\varepsilon_1$.

Applying Differentiability at a Point for Maps Between Euclidean Spaces to $g$ at $b$ with the tolerance $\varepsilon_2$, there is a real $\delta_2$ with $0<\delta_2$ such that every $u\in\mathbb{R}^m$ with $0<\lVert u\rVert<\delta_2$ satisfies $b+u\in V$ and

$$\lVert g(b+u)-g(b)-B\,u\rVert\le\varepsilon_2\lVert u\rVert . \qquad (\ast)$$

Applying the same definition to $f$ at $a$ with the tolerance $\varepsilon_1$, there is a real $\delta_1$ with $0<\delta_1$ such that every $h\in\mathbb{R}^n$ with $0<\lVert h\rVert<\delta_1$ satisfies $a+h\in U$ and

$$\lVert f(a+h)-f(a)-A\,h\rVert\le\varepsilon_1\lVert h\rVert . \qquad (\ast\ast)$$

Since $0<K^{-1}$, claim 5 of Elementary Order Arithmetic in an Ordered Field gives $0<\delta_2K^{-1}$, so by claim 9 of that lemma there is a real $\delta$ with $\delta\le\delta_1$, $\delta\le\delta_2K^{-1}$, and $\delta$ equal to $\delta_1$ or to $\delta_2K^{-1}$; in either case $0<\delta$.

The estimate. Fix $h\in\mathbb{R}^n$ with $0<\lVert h\rVert<\delta$, and put

$$u=f(a+h)-f(a)\in\mathbb{R}^m .$$

By claim 2 of Elementary Order Arithmetic in an Ordered Field and $\delta\le\delta_1$ we have $\lVert h\rVert<\delta_1$, so $a+h\in U$ and $(\ast\ast)$ holds; note also that $0\le\lVert h\rVert$ by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. In the notation just introduced, $(\ast\ast)$ reads $\lVert u-A\,h\rVert\le\varepsilon_1\lVert h\rVert$.

Step 1: $\lVert u\rVert\le K\lVert h\rVert$ and $\lVert u\rVert<\delta_2$. By claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space we have $u=(u-A\,h)+A\,h$, so the triangle inequality, claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, gives

$$\lVert u\rVert\le\lVert u-A\,h\rVert+\lVert A\,h\rVert\le\varepsilon_1\lVert h\rVert+C_A\lVert h\rVert=(\varepsilon_1+C_A)\lVert h\rVert ,$$

adding the two non-strict inequalities and using distributivity. From $\varepsilon_1\le 1$ and claim 3 of Elementary Arithmetic in an Ordered Field we get $\varepsilon_1+C_A\le 1+C_A=K$, and multiplying by the nonnegative factor $\lVert h\rVert$ gives $(\varepsilon_1+C_A)\lVert h\rVert\le K\lVert h\rVert$. Hence $\lVert u\rVert\le K\lVert h\rVert$.

Furthermore $\lVert h\rVert<\delta_2K^{-1}$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, so multiplying by the positive factor $K$ (claim 10 of that lemma) gives $K\lVert h\rVert<K\delta_2K^{-1}=\delta_2$. Combining with the previous inequality by claim 2 of that lemma yields $\lVert u\rVert<\delta_2$.

Step 2: the decomposition. Since $a+h\in U$ we have $f(a+h)\in V$, and by claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space

$$b+u=f(a)+\bigl(f(a+h)-f(a)\bigr)=f(a+h) ,$$

so $g(b+u)=g\bigl(f(a+h)\bigr)=(g\circ f)(a+h)$ and $g(b)=(g\circ f)(a)$. By claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product we have $(BA)h=B(A\,h)$, and by claim 1 of that lemma $B\,u-B(A\,h)=B(u-A\,h)$. Working in the vector space $\mathbb{R}^p$ of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space,

$$(g\circ f)(a+h)-(g\circ f)(a)-(BA)h=\bigl(g(b+u)-g(b)-B\,u\bigr)+B(u-A\,h) ,$$

since the two sides differ only by the regrouping of the term $B\,u$ with its additive inverse. The triangle inequality therefore gives

$$\bigl\lVert (g\circ f)(a+h)-(g\circ f)(a)-(BA)h\bigr\rVert\le\lVert g(b+u)-g(b)-B\,u\rVert+\lVert B(u-A\,h)\rVert . \qquad (\dagger)$$

Step 3: the second term of $(\dagger)$. By the choice of $C_B$ and then $(\ast\ast)$ multiplied by the nonnegative factor $C_B$,

$$\lVert B(u-A\,h)\rVert\le C_B\lVert u-A\,h\rVert\le C_B\varepsilon_1\lVert h\rVert .$$

Since $0\le C_B$ and $\varepsilon_1\le\eta$, multiplying by $C_B$ gives $C_B\varepsilon_1\le C_B\eta$; since $0\le\eta$ and $C_B\le L$, multiplying by $\eta$ gives $C_B\eta\le L\eta=L\cdot\varepsilon\cdot2^{-1}\cdot L^{-1}=\varepsilon\cdot 2^{-1}$. Hence $C_B\varepsilon_1\le\varepsilon\cdot2^{-1}$, and multiplying by the nonnegative factor $\lVert h\rVert$ gives

$$\lVert B(u-A\,h)\rVert\le\varepsilon\cdot2^{-1}\,\lVert h\rVert .$$

Step 4: the first term of $(\dagger)$. Suppose first that $u\ne 0_{\mathbb{R}^m}$. Then $\lVert u\rVert\ne 0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and $0\le\lVert u\rVert$ by claim 1 of that lemma, so $0<\lVert u\rVert$. Together with $\lVert u\rVert<\delta_2$ from step 1, the estimate $(\ast)$ applies and gives $\lVert g(b+u)-g(b)-B\,u\rVert\le\varepsilon_2\lVert u\rVert$. Multiplying $\lVert u\rVert\le K\lVert h\rVert$ by the nonnegative factor $\varepsilon_2$ gives $\varepsilon_2\lVert u\rVert\le\varepsilon_2K\lVert h\rVert=\varepsilon\cdot2^{-1}\cdot K^{-1}\cdot K\,\lVert h\rVert=\varepsilon\cdot2^{-1}\lVert h\rVert$, whence

$$\lVert g(b+u)-g(b)-B\,u\rVert\le\varepsilon\cdot2^{-1}\,\lVert h\rVert .$$

Suppose instead that $u=0_{\mathbb{R}^m}$. Then $b+u=b$, so $g(b+u)-g(b)=0_{\mathbb{R}^p}$ by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field applied coordinatewise, and $B\,u=B\,0_{\mathbb{R}^m}=0_{\mathbb{R}^p}$ by claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product; hence $g(b+u)-g(b)-B\,u=0_{\mathbb{R}^p}$ and its norm is $0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. Since $0\le\varepsilon\cdot2^{-1}$ and $0\le\lVert h\rVert$, claim 5 of Elementary Arithmetic in an Ordered Field gives $0\le\varepsilon\cdot2^{-1}\lVert h\rVert$, so the displayed bound holds in this case too.

Step 5: conclusion. Adding the bounds of steps 3 and 4 in $(\dagger)$ and using distributivity together with $\varepsilon\cdot2^{-1}+\varepsilon\cdot2^{-1}=\varepsilon$,

$$\bigl\lVert (g\circ f)(a+h)-(g\circ f)(a)-(BA)h\bigr\rVert\le\varepsilon\cdot2^{-1}\lVert h\rVert+\varepsilon\cdot2^{-1}\lVert h\rVert=\varepsilon\,\lVert h\rVert .$$

Thus for every real $\varepsilon$ with $0<\varepsilon$ there is a real $\delta$ with $0<\delta$ such that every $h\in\mathbb{R}^n$ with $0<\lVert h\rVert<\delta$ satisfies $a+h\in U$ and the displayed inequality. Since $BA$ is a real matrix with $p$ rows and $n$ columns, Differentiability at a Point for Maps Between Euclidean Spaces says exactly that $g\circ f$ is differentiable at $a$ with derivative matrix $BA$.

The Jacobian form. Claim 2 of A Derivative Matrix is the Jacobian Matrix, and is Unique, applied to $f$ at $a$, to $g$ at $b=f(a)$, and to $g\circ f$ at $a$, gives that $Df(a)$, $Dg(f(a))$ and $D(g\circ f)(a)$ are defined and equal $A$, $B$ and $BA$ respectively. Hence $D(g\circ f)(a)=Dg(f(a))\,Df(a)$.