Proof of The Implicit Function Theorem

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); the metric spaces $(\mathbb{R}^{n},d_{E})$ and $(\mathbb{R},d_{\mathbb{R}})$ through $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (i) and The Absolute Value Metric on the Real Numbers (i); the indices of $[n+m]$ split into the first $n$ and the last $m$ as in Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (i); and, distances being nonnegative, $s<t$ holds for distances $s,t$ exactly when $s^{2}<t^{2}$.

The idea is to apply the inverse function theorem to the map that keeps $x$ and replaces $y$ by $F(x,y)$. Define $\Phi:U\to\mathbb{R}^{n+m}$ by

$$\Phi(x,y)=\bigl(x,F(x,y)\bigr).$$

The first $n$ component maps of $\Phi$ are the coordinate functions $(x,y)\mapsto x_{j}$, $j\in[n]$, which are smooth on $U$ by Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (iii) and hence of class $C^{k}$ by Basic Properties of Cᵏ Functions (ii); its last $m$ component maps are $F_{1},\dots,F_{m}$, which are of class $C^{k}$ by hypothesis. So $\Phi$ is of class $C^{k}$ on $U$ (Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii)). Its Jacobian matrix is formed from the partial derivatives of its components (Jacobian Matrix of a Function Between Euclidean Spaces (i)); the coordinate function $(x,y)\mapsto x_{j}$ has gradient the $j$-th standard basis vector (Rules for Derivatives of Functions on Open Subsets of ℝⁿ (iii)), and the rows of $F$ contribute $D_{x}F$ and $D_{y}F$. In the block form of matrices (with $n$ and $m$ in place of $m$ and $n$) this reads

$$D\Phi(x,y)=\begin{pmatrix}I_{n}&0_{n\times m}\\D_{x}F(x,y)&D_{y}F(x,y)\end{pmatrix}\qquad\bigl((x,y)\in U\bigr).$$

We next show that, at every $(x,y)\in U$, the matrix $D\Phi(x,y)$ is invertible exactly when $D_{y}F(x,y)$ is. Write $P=D_{x}F(x,y)$ and $Q=D_{y}F(x,y)$. If $Q$ is invertible, multiplying the blocks (an entrywise check) gives

$$\begin{pmatrix}I_{n}&0_{n\times m}\\P&Q\end{pmatrix}\begin{pmatrix}I_{n}&0_{n\times m}\\-Q^{-1}P&Q^{-1}\end{pmatrix}=I_{n+m},$$

and the same product in the other order is $I_{n+m}$ as well; so $D\Phi(x,y)$ is invertible. Conversely, let $D\Phi(x,y)$ be invertible, and let $v\in\mathbb{R}^{m}$ satisfy $Qv=0$. Then $D\Phi(x,y)\,\iota(0,v)=\iota(0,Qv)=0$, so $\iota(0,v)=0$ by the equivalence of Criteria for Invertibility of a Square Real Matrix (vi) and Criteria for Invertibility of a Square Real Matrix (i) for the invertible matrix $D\Phi(x,y)$, and hence $v=0$. Thus $Qv=0$ only for $v=0$, and the square matrix $Q$ is therefore invertible, by the same criterion for $Q$.

Since $k\ge1$, $\Phi$ is of class $C^{1}$ on $U$ (Basic Properties of Cᵏ Functions (ii)), and $D\Phi(a,b)$ is invertible because $D_{y}F(a,b)$ is. By The Inverse Function Theorem (i), applied to $\Phi$ with $n+m$ in place of $n$, there is an open set $W$ with $(a,b)\in W\subseteq U$ such that $V=\Phi(W)$ is open, $\Phi$ is injective on $W$, $D\Phi(x,y)$ is invertible for every $(x,y)\in W$, and the inverse function $\Psi:V\to W$ of $\Phi|_{W}$ is of class $C^{1}$ on $V$. As $\Phi$ is of class $C^{k}$ on $U$, The Inverse Function Theorem (ii) shows that $\Psi$ is of class $C^{k}$ on $V$.

Now we construct $A$, $B$ and $g$. Since $W$ is open, there is $r>0$ with $B((a,b),r)\subseteq W$. Let $A_{0}=B(a,r/2)\subseteq\mathbb{R}^{n}$, and let $B\subseteq\mathbb{R}^{m}$ be the open ball of radius $r/2$ about $b$; both are open by Open Ball in a Metric Space is Open (i). By Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (vi), for all $x,x'\in\mathbb{R}^{n}$ and $y,y'\in\mathbb{R}^{m}$,

$$d_{E}\bigl((x,y),(x',y')\bigr)^{2}=d_{E}(x,x')^{2}+d_{E}(y,y')^{2}.$$

Taking $(x',y')=(a,b)$, a point of $A_{0}\times B$ has squared distance less than $r^{2}/4+r^{2}/4<r^{2}$ from $(a,b)$, and hence distance less than $r$; so $A_{0}\times B\subseteq B((a,b),r)\subseteq W$. The same identity shows that $A_{0}\times B$ is open. Let $(x,y)\in A_{0}\times B$, and choose $\varepsilon>0$ such that $B(x,\varepsilon)\subseteq A_{0}$ and the open ball of radius $\varepsilon$ about $y$ lies in $B$. Every point of $\mathbb{R}^{n+m}$ is of the form $(x',y')$, as $\iota$ is a bijection; if it lies at distance less than $\varepsilon$ from $(x,y)$, then $d_{E}(x,x')<\varepsilon$ and $d_{E}(y,y')<\varepsilon$, so it lies in $A_{0}\times B$.

Next we define $A$ and show that it is open. Let

$$A=\bigl\{x\in A_{0}:\ (x,0)\in V\ \text{and}\ \Psi(x,0)\in A_{0}\times B\bigr\},$$

and let $\ell:\mathbb{R}^{n}\to\mathbb{R}^{n+m}$ be the map $\ell(x)=(x,0)$. The first $n$ component maps of $\ell$ are the coordinate functions $x\mapsto x_{j}$, $j\in[n]$, which are smooth by Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (iii), and its last $m$ component maps are the zero function, which is smooth by Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (i). All of them are therefore of class $C^{k}$ (Basic Properties of Cᵏ Functions (ii)), so $\ell$ is of class $C^{k}$ on $\mathbb{R}^{n}$ (Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii); $\mathbb{R}^{n}$ is open in itself by Euclidean Space Is Open in Itself (i)), and in particular continuous (Basic Properties of Cᵏ Functions (iv)); the same argument applies on every open subset of $\mathbb{R}^{n}$. As $V$ is open, $\ell^{-1}(V)$ is open by Continuity of a Map Between Metric Spaces via Preimages of Open Sets (i), and so

$$A_{1}=A_{0}\cap\ell^{-1}(V)=\bigl\{x\in A_{0}:\ (x,0)\in V\bigr\}$$

is open, as an intersection of two open sets (The Open Subsets of a Metric Space Form a Topology (i)). On the open set $A_{1}$, the map $\ell$ is of class $C^{k}$ and takes values in $V$, and $\Psi$ is of class $C^{k}$ on $V$; so $\Psi\circ\ell:x\mapsto\Psi(x,0)$ is of class $C^{k}$ on $A_{1}$ by Composites of Cᵏ Functions Are of Class Cᵏ (ii) (with $n+m$ in place of both $m$ and $p$), hence continuous on $A_{1}$ (Basic Properties of Cᵏ Functions (iv)), and

A=\bigl\{x\in A_{1}:\ \Psi\bigl(\ell(x)\bigr)\in A_{0}\times B\bigr\}. $$ By [Continuity on a Subset Is Continuity on the Metric Subspace (ii)](/theorems/144cd2ad-8743-4508-8a60-f15659ae9191?v=a1de772c-5c77-4e42-9f4f-d77bb4521688#clause-on-subset), $\Psi\circ\ell$ is therefore continuous as a map from the metric space $(A_{1},d_{A_{1}})$, where $d_{A_{1}}$ is the restriction of $d_{E}$, to $(\mathbb{R}^{n+m},d_{E})$. Since $A_{0}\times B$ is open, its preimage $A$ under this map is open in $(A_{1},d_{A_{1}})$ by [Continuity of a Map Between Metric Spaces via Preimages of Open Sets (i)](/theorems/2ebe68ef-94ea-467f-b8cb-16bce8b8dd8a?v=e78bca54-4f20-403e-99ac-b51201e4e163#clause-preimage-open). By [Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (v)](/theorems/4dbd668d-2478-4092-8e35-0abc764e2e4a?v=f7f86610-100a-45ef-b7f4-03629253d546#clause-subspace-open), applied to $(\mathbb{R}^{n},d_{E})$ and its subset $A_{1}$, $A=O\cap A_{1}$ for some $O$ open in $(\mathbb{R}^{n},d_{E})$, and so $A$ is open, as an intersection of two open sets ([The Open Subsets of a Metric Space Form a Topology (i)](/theorems/ebeca980-0321-48ab-b205-de4800ab87d6?v=e3277819-1a1a-4845-941d-d661d3de60da#clause-topology)). Moreover $a\in A$: indeed $(a,b)\in W$ and $\Phi(a,b)=(a,0)$, so $(a,0)\in V$ and $\Psi(a,0)=(a,b)\in A_{0}\times B$. Now we define $g$. For $x\in A$ write $\Psi(x,0)=(x',y')$ with $x'\in A_{0}$ and $y'\in B$. Then $\Phi(x',y')=(x,0)$, that is, $x'=x$ and $F(x,y')=0$. We put $g(x)=y'$, the vector of the last $m$ components of $\Psi(x,0)$; thus $g:A\to B$ and

\Psi(x,0)=\bigl(x,g(x)\bigr),\qquad F\bigl(x,g(x)\bigr)=0\qquad(x\in A).

Finally $A\times B\subseteq A_{0}\times B\subseteq W\subseteq U$. We prove [(i)](/theorems/826f61c3-34a4-4f85-babc-1de039c02325?v=d42026c5-18e4-49f7-a1a1-5ceec13f67fd#clause-implicit). Let $x\in A$ and $y\in B$. If $y=g(x)$, then $F(x,y)=0$ by the display above. Conversely, let $F(x,y)=0$. Both $(x,y)$ and $(x,g(x))=\Psi(x,0)$ lie in $W$, and

\Phi(x,y)=(x,0)=\Phi\bigl(x,g(x)\bigr),

so $y=g(x)$, as $\Phi$ is injective on $W$. Since $F(a,b)=0$ and $b\in B$, the equivalence at $x=a$ gives $g(a)=b$. For [(ii)](/theorems/826f61c3-34a4-4f85-babc-1de039c02325?v=d42026c5-18e4-49f7-a1a1-5ceec13f67fd#clause-unique), let $h:A\to B$ satisfy $F(x,h(x))=0$ for every $x\in A$. For each $x\in A$, the equivalence just proved, applied with $y=h(x)\in B$, gives $h(x)=g(x)$; so $h=g$. It remains to prove [(iii)](/theorems/826f61c3-34a4-4f85-babc-1de039c02325?v=d42026c5-18e4-49f7-a1a1-5ceec13f67fd#clause-regularity). For $i\in[m]$ the component map $g_{i}$ is $\Psi_{n+i}\circ\ell:x\mapsto\Psi_{n+i}(x,0)$ on $A$. On the open set $A$, the map $\ell$ is of class $C^{k}$, as shown above, and takes values in $V$ by the definition of $A$; and $\Psi_{n+i}$ is of class $C^{k}$ on $V$. By [Composites of Cᵏ Functions Are of Class Cᵏ (ii)](/theorems/e747fa8a-f8fe-4930-81ce-c6431002b59a?v=4eb27564-5897-4955-9803-460ff1dca0ae#clause-composition) every $g_{i}$ is of class $C^{k}$ on $A$, so $g$ is of class $C^{k}$ on $A$. Now let $x\in A$. Then $(x,g(x))=\Psi(x,0)\in W$, so $D\Phi(x,g(x))$ is invertible, and hence so is $D_{y}F(x,g(x))$, by the equivalence proved above. To compute $Dg(x)$, let $\gamma:A\to\mathbb{R}^{n+m}$, $\gamma(x)=(x,g(x))$; it takes values in $A\times B\subseteq U$, and $F\circ\gamma$ vanishes on $A$, so that, as a constant function, it has derivative matrix $0_{m\times n}$ at $x$ ([Rules for Derivatives of Functions on Open Subsets of ℝⁿ (i)](/theorems/1543e075-0648-43b6-a96f-de4b9e31dc4a?v=0080898d-4b3c-40a1-acf6-c04e740549e5#clause-constant)). Since $F$ and $g$ are of class $C^{1}$, they are differentiable with derivative matrices $DF$ and $Dg$ ([Basic Properties of Cᵏ Functions (v)](/theorems/caa72c92-4555-4959-920b-4777511fb791?v=8ec0d350-a30f-402f-b5c2-c9b31345a60d#clause-c1-differentiable)). So is $\gamma$: its first $n$ component maps are the coordinate functions $x\mapsto x_{j}$, $j\in[n]$, smooth on $A$ by [Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (iii)](/theorems/beb1dfb9-d732-4283-bcc1-450e1cc3fad7?v=48753fbe-ab3f-42ef-9a29-3ff5dc406afc#clause-coordinates) and hence of class $C^{1}$ ([Basic Properties of Cᵏ Functions (ii)](/theorems/caa72c92-4555-4959-920b-4777511fb791?v=8ec0d350-a30f-402f-b5c2-c9b31345a60d#clause-hierarchy)), and its last $m$ component maps are $g_{1},\dots,g_{m}$; thus $\gamma$ is of class $C^{1}$ on $A$ ([Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii)](/theorems/bee1cc38-102c-42f6-b8d4-717b11ed035c?v=327097f3-cf97-462b-82b4-f793a60f1edc#clause-vector)) and differentiable there by [Basic Properties of Cᵏ Functions (v)](/theorems/caa72c92-4555-4959-920b-4777511fb791?v=8ec0d350-a30f-402f-b5c2-c9b31345a60d#clause-c1-differentiable). The coordinate function $x\mapsto x_{j}$ has gradient $e_{j}$ ([Rules for Derivatives of Functions on Open Subsets of ℝⁿ (iii)](/theorems/1543e075-0648-43b6-a96f-de4b9e31dc4a?v=0080898d-4b3c-40a1-acf6-c04e740549e5#clause-coordinates)), so the first $n$ rows of the Jacobian matrix $D\gamma(x)$ ([Jacobian Matrix of a Function Between Euclidean Spaces (i)](/theorems/c9747fd7-a455-485e-ae47-c9e320444d22?v=bf909ac6-ed54-4a27-92f4-2deca356d619#clause-jacobian)) form $I_{n}$ and its last $m$ rows form $Dg(x)$. By the chain rule [Chain Rule for Differentiable Functions Between Euclidean Spaces (i)](/theorems/fa4192c2-f4d6-4fbe-bac6-abc720b65f65?v=ba4541d2-a607-45c9-9b84-946be2ee14f1#clause-chain), applied to $\gamma$ and $F$, both differentiable as just shown, $D(F\circ\gamma)(x)=DF(\gamma(x))\,D\gamma(x)$; equivalently, entrywise,

\partial_{j}(F\circ\gamma){i}(x)=\sum{l\in[n+m]}\partial_{l}F_{i}\bigl(\gamma(x)\bigr),\partial_{j}\gamma_{l}(x)\qquad(i\in[m],\ j\in[n])

by [Composites of Cᵏ Functions Are of Class Cᵏ (i)](/theorems/e747fa8a-f8fe-4930-81ce-c6431002b59a?v=4eb27564-5897-4955-9803-460ff1dca0ae#clause-partials), as $\gamma$ and $F$ are of class $C^{1}$, which is the block product below. Hence

0_{m\times n}=DF\bigl(x,g(x)\bigr),D\gamma(x)=D_{x}F\bigl(x,g(x)\bigr)+D_{y}F\bigl(x,g(x)\bigr),Dg(x),

and multiplying on the left by $D_{y}F(x,g(x))^{-1}$ yields

Dg(x)=-D_{y}F\bigl(x,g(x)\bigr)^{-1}D_{x}F\bigl(x,g(x)\bigr).

$$$$