Proof of The Inverse 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); $(\mathbb{R}^{n},d_{E})$ 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); and continuity on a set is that of Continuous Map Between Metric Spaces (ii).

We begin with (i). Let $A=Df(a)$, and let $\lambda>0$ be the real number with $2\lambda\lVert A^{-1}\rVert=1$; it exists because $A^{-1}\neq0$ (as $AA^{-1}=I_{n}\neq0$), so that $\lVert A^{-1}\rVert>0$. Since $f$ is of class $C^{1}$, the Jacobian function $x\mapsto Df(x)$ is continuous on $U$ (Basic Properties of Cᵏ Functions (v)). As $U$ is open, there is therefore a radius $r>0$ such that the open ball $B(a,r)$ lies in $U$ (Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (x)) and

$$\lVert Df(x)-A\rVert<\lambda\qquad\text{for every }x\in B(a,r).$$

Put $W=B(a,r)$. It is an open set containing $a$, and it is convex: for $x,x'\in W$ and $t\in[0,1]$, by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v),

$$\lVert(1-t)x+tx'-a\rVert\le(1-t)\lVert x-a\rVert+t\lVert x'-a\rVert<r .$$

For $x\in W$ we have $\lVert A^{-1}\rVert\,\lVert Df(x)-A\rVert<\lambda\lVert A^{-1}\rVert=\tfrac12<1$, where the matrix distance is $d_{n}(A,Df(x))=\lVert A-Df(x)\rVert=\lVert Df(x)-A\rVert$ by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (v). So the perturbation bound Perturbations of an Invertible Matrix: the Invertible Matrices Form an Open Set and Inversion Is Continuous (i), applied to $A$ and the companion matrix $B=Df(x)$, shows that $Df(x)$ is invertible for every $x\in W$.

The heart of the argument is a family of contractions. For $y\in\mathbb{R}^{n}$ define $\varphi_{y}:W\to\mathbb{R}^{n}$ by

$$\varphi_{y}(x)=x+A^{-1}\bigl(y-f(x)\bigr).$$

Since $A^{-1}$ is invertible, $\varphi_{y}(x)=x$ holds exactly when $f(x)=y$: the fixed points of $\varphi_{y}$ are the solutions in $W$ of the equation $f(x)=y$. The restriction $f|_{W}$ is differentiable at every point of $W$, with $D(f|_{W})=Df$ there (Basic Properties of Cᵏ Functions (v), Restriction of a Cᵏ Function to an Open Subset (i)). For $x,x+h\in W$, the algebraic rules Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (ii), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v) give

$$\varphi_{y}(x+h)-\varphi_{y}(x)-A^{-1}\bigl(A-Df(x)\bigr)h=-A^{-1}\bigl(f(x+h)-f(x)-Df(x)h\bigr),$$

whose norm is at most $\lVert A^{-1}\rVert$ times the remainder of $f$ at $x$ (Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii)). Hence $\varphi_{y}$ is differentiable on $W$ with $D\varphi_{y}(x)=A^{-1}\bigl(A-Df(x)\bigr)$ (The Derivative Is the Jacobian Matrix, and Is Unique (ii)), and by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (vii) and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (v),

$$\lVert D\varphi_{y}(x)\rVert\le\lVert A^{-1}\rVert\,\lVert Df(x)-A\rVert<\tfrac12\qquad(x\in W).$$

The mean value inequality The Mean Value Theorem and Inequality on Convex Open Sets (ii) on the convex open set $W$, with bound $M=\tfrac12$, now gives

$$\lVert\varphi_{y}(x_{1})-\varphi_{y}(x_{2})\rVert\le\tfrac12\lVert x_{1}-x_{2}\rVert\qquad(x_{1},x_{2}\in W).\tag{$*$}$$

In particular $f$ is injective on $W$: if $f(x_{1})=f(x_{2})=y$, then $x_{1}$ and $x_{2}$ are both fixed points of $\varphi_{y}$, and $(*)$ gives $\lVert x_{1}-x_{2}\rVert\le\tfrac12\lVert x_{1}-x_{2}\rVert$, so $x_{1}=x_{2}$.

Next we show that $V=f(W)$ is open. Let $y_{0}\in V$, say $y_{0}=f(x_{0})$ with $x_{0}\in W$. Since $W$ is open, there is $\rho>0$ such that the closed ball $K=\bar{B}_{d_{E}}(x_{0},\rho)$ lies in $W$ (take a radius smaller than the one given by Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (x)). Let $\lVert y-y_{0}\rVert<\lambda\rho$. Since $\varphi_{y}(x_{0})-x_{0}=A^{-1}(y-y_{0})$, for every $x\in K$ the estimate $(*)$ and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii) give

$$\lVert\varphi_{y}(x)-x_{0}\rVert\le\lVert\varphi_{y}(x)-\varphi_{y}(x_{0})\rVert+\lVert A^{-1}(y-y_{0})\rVert<\tfrac{\rho}{2}+\tfrac{\rho}{2}=\rho .$$

So $\varphi_{y}$ maps $K$ into $K$, and by $(*)$ it is a contraction of $K$ with constant $\tfrac12$ for the restricted metric, which is a metric on $K$ by Subsets of a Metric Space: the Restricted Metric, Sequences, Limits in Closed Subsets and Completeness of Closed Subsets (i). The ball $K$ is compact by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere (ii), hence closed by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere (i); since $(\mathbb{R}^{n},d_{E})$ is complete ($\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (iv)) and a closed subset of a complete metric space is complete (Subsets of a Metric Space: the Restricted Metric, Sequences, Limits in Closed Subsets and Completeness of Closed Subsets (iv)), $K$ is complete, and it is nonempty, as $x_{0}\in K$. The contraction mapping theorem Contraction Mapping Theorem on a Nonempty Complete Metric Space (i) therefore provides a fixed point $x\in K$ of $\varphi_{y}$, that is, a point $x\in W$ with $f(x)=y$. Hence $B(y_{0},\lambda\rho)\subseteq V$, and $V$ is open.

Let $g:V\to W$ be the inverse function of $f|_{W}$. We show that $g$ is Lipschitz. For $y,y+v\in V$ put $x=g(y)$ and $h=g(y+v)-x$, so that $f(x)=y$ and $f(x+h)=y+v$. Then $\varphi_{y}(x+h)-\varphi_{y}(x)=h-A^{-1}v$ (Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v)), and $(*)$ gives $\lVert h-A^{-1}v\rVert\le\tfrac12\lVert h\rVert$. Hence $\lVert h\rVert\le\lVert h-A^{-1}v\rVert+\lVert A^{-1}v\rVert\le\tfrac12\lVert h\rVert+\lVert A^{-1}v\rVert$, and by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii)

$$\lVert h\rVert\le2\lVert A^{-1}v\rVert\le\frac{\lVert v\rVert}{\lambda} .\tag{$**$}$$

In particular $g$ is continuous on $V$: given $\varepsilon>0$, the choice $\delta=\lambda\varepsilon$ works at every point.

Now we show that $g$ is differentiable. Fix $y\in V$, and let $x=g(y)$ and $T=Df(x)^{-1}$, which exists because $x\in W$. With $v$ and $h$ as before, $g(y+v)-g(y)=h$ and $v=f(x+h)-f(x)$, so by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v)

$$g(y+v)-g(y)-Tv=-T\bigl(f(x+h)-f(x)-Df(x)h\bigr).$$

We choose in this order. Given $\varepsilon>0$, the differentiability of $f$ at $x$ (Basic Properties of Cᵏ Functions (v)) provides $\eta>0$ such that $\lVert f(x+h)-f(x)-Df(x)h\rVert\le\varepsilon\lambda\lVert h\rVert/(\lVert T\rVert+1)$ whenever $\lVert h\rVert<\eta$ and $x+h\in U$. Then choose $\delta>0$ with $\delta\le\lambda\eta$ and $B(y,\delta)\subseteq V$, possible as $V$ is open (Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (x)). If $\lVert v\rVert<\delta$, then $y+v\in V$, and $(**)$ gives $\lVert h\rVert<\eta$; hence, by Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii) and $(**)$ again,

$$\lVert g(y+v)-g(y)-Tv\rVert\le\lVert T\rVert\,\frac{\varepsilon\lambda\lVert h\rVert}{\lVert T\rVert+1}\le\varepsilon\lVert v\rVert .$$

So $g$ is differentiable at $y$ with derivative matrix $T$, and $Dg(y)=T$ by The Derivative Is the Jacobian Matrix, and Is Unique (ii); that is,

$$Dg\bigl(f(x)\bigr)=Df(x)^{-1}\qquad\text{for every }x\in W .$$

It remains to see that $g$ is of class $C^{1}$. The function $y\mapsto Dg(y)=Df(g(y))^{-1}$ is the composite of $g:V\to W$, continuous by $(**)$, of the Jacobian function $x\mapsto Df(x)$ on $W$, continuous by Basic Properties of Cᵏ Functions (v) and with values in $\mathrm{GL}_{n}(\mathbb{R})$ as shown above, and of the inversion map $B\mapsto B^{-1}$ on $\mathrm{GL}_{n}(\mathbb{R})$, continuous by Perturbations of an Invertible Matrix: the Invertible Matrices Form an Open Set and Inversion Is Continuous (iv). It is therefore continuous on $V$, by Continuity of Identity Maps, Compositions and Reciprocals (ii) applied twice. Its entries are the partial derivatives $\partial_{i}g_{l}$, which exist on $V$ by The Derivative Is the Jacobian Matrix, and Is Unique (ii), and they are continuous on $V$ by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (ii) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v). So every partial derivative of every component $g_{l}$ exists on $V$ and is of class $C^{0}$ there (Basic Properties of Cᵏ Functions (i)); hence each $g_{l}$ lies in $\mathcal{C}^{1}(V)$ by Functions of Class Cᵏ on an Open Subset of ℝⁿ (i), and $g$ is of class $C^{1}$ on $V$ by Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii). This proves (i).

We turn to (ii). Let $k$, $W$ and $g$ be as there, and let $V=f(W)$. For $x\in W$, $f|_{W}$ is differentiable at $x$ with derivative $Df(x)$ (Basic Properties of Cᵏ Functions (ii), Basic Properties of Cᵏ Functions (v), Restriction of a Cᵏ Function to an Open Subset (i)), and $g$ is differentiable at $f(x)$ because it is of class $C^{1}$ on $V$ (Basic Properties of Cᵏ Functions (v)). The chain rule Chain Rule for Differentiable Functions Between Euclidean Spaces (i), applied to $g\circ f|_{W}=\mathrm{id}_{W}$, whose derivative is $I_{n}$ at every point (Rules for Derivatives of Functions on Open Subsets of ℝⁿ (ii) with $B=I_{n}$ and $c=0$, as $I_{n}x=x$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v)), gives $Dg(f(x))\,Df(x)=I_{n}$. So $Df(x)$ is invertible (a square matrix with a left inverse is invertible) and

$$Dg\bigl(f(x)\bigr)=Df(x)^{-1}\qquad(x\in W).$$

We first show, by induction on $j\in\mathbb{N}_{0}$, the following claim about the inverse of a matrix-valued function.

Claim. If $V'\subseteq\mathbb{R}^{n}$ is open and $B:V'\to\mathcal{M}_{n}(\mathbb{R})$ has all entries of class $C^{j}$ on $V'$ and $B(y)$ invertible for every $y\in V'$, then all entries of $y\mapsto B(y)^{-1}$ are of class $C^{j}$ on $V'$.

For $j=0$ the entries of $B$ are continuous, so $B$ is continuous on $V'$ (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (ii), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)); then $y\mapsto B(y)^{-1}$, the composite of $B$ with inversion, is continuous by Perturbations of an Invertible Matrix: the Invertible Matrices Form an Open Set and Inversion Is Continuous (iv) and Continuity of Identity Maps, Compositions and Reciprocals (ii), and so are its entries. For the step from $j$ to $j+1$, suppose the entries of $B$ are of class $C^{j+1}$, and let $y\in V'$ and $i\in[n]$. On a small interval around $0$ (Lines, Scalar Multiples, Basis Directions, Small Balls and Matrix Rows in ℝⁿ (xi), The Line t ↦ x+th and Values of a Function along a Line in ℝⁿ, as Functions (ii)), the matrix function $u\mapsto B(y+ue_{i})$ is differentiable at $0$, with derivative the matrix $\partial_{i}B(y)$ of the partial derivatives $\partial_{i}B_{pq}(y)$ (Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ (i), Derivatives along Lines in ℝⁿ: the Line t ↦ x+th, and Partial Derivatives as Derivatives along Coordinate Axes (ii)). By the derivative of the inverse Continuity and Derivative of the Inverse of a Matrix-Valued Map (iii), read entrywise through Derivatives along Lines in ℝⁿ: the Line t ↦ x+th, and Partial Derivatives as Derivatives along Coordinate Axes (iii), the partial derivatives of the entries of $B^{-1}$ exist and

$$\partial_{i}\bigl(B^{-1}\bigr)(y)=-B(y)^{-1}\,\partial_{i}B(y)\,B(y)^{-1}.$$

Each entry of the right-hand side is the negative of a sum, over finitely many index pairs, of products of an entry of $B^{-1}$, an entry of $\partial_{i}B$ and an entry of $B^{-1}$. The entries of $B^{-1}$ are of class $C^{j}$ by the induction hypothesis, which applies because the entries of $B$ are of class $C^{j}$ (Basic Properties of Cᵏ Functions (ii)); the entries $\partial_{i}B_{pq}$ are of class $C^{j}$ by Basic Properties of Cᵏ Functions (iii). Products, sums of finitely many terms and negatives of functions of class $C^{j}$ are of class $C^{j}$ (Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (v), Finite Sums of Cᵏ Functions Are of Class Cᵏ (i), Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions (iv)); the sum over index pairs is handled by applying Finite Sums of Cᵏ Functions Are of Class Cᵏ (i) twice, first to the sum over one index and then to the sum over the other. So every partial derivative of every entry of $B^{-1}$ is of class $C^{j}$, and the entries of $B^{-1}$ are of class $C^{j+1}$ by Functions of Class Cᵏ on an Open Subset of ℝⁿ (i). This proves the claim.

Now we show by induction that $g$ is of class $C^{j}$ on $V$ for every $j\in[k]$; the case $j=1$ is the hypothesis. Suppose $g$ is of class $C^{j}$ with $j<k$, and let $B=Df\circ g$ on $V$, whose entries are the functions $\partial_{i}f_{l}\circ g$. Each $\partial_{i}f_{l}$ is of class $C^{k-1}$ on $U$ (Basic Properties of Cᵏ Functions (iii)), hence on $W$ (Restriction of a Cᵏ Function to an Open Subset (ii)), hence of class $C^{j}$ there, as $j\le k-1$ (Basic Properties of Cᵏ Functions (ii)); so $\partial_{i}f_{l}\circ g$ is of class $C^{j}$ on $V$ by Composites of Cᵏ Functions Are of Class Cᵏ (ii). By the claim, applied to $B$ on $V$, the entries $\partial_{i}g_{l}$ of $Dg=B^{-1}$ are of class $C^{j}$, so $g$ is of class $C^{j+1}$ on $V$ (Functions of Class Cᵏ on an Open Subset of ℝⁿ (i), Functions of Class Cᵏ on an Open Subset of ℝⁿ (iii)). Hence $g$ is of class $C^{k}$ on $V$.

As $f|_{W}$ is of class $C^{k}$ on $W$ (Restriction of a Cᵏ Function to an Open Subset (ii)) and is a bijection of $W$ onto $V$ whose inverse $g$ is of class $C^{k}$ on $V$, $f|_{W}$ is a $C^{k}$ diffeomorphism of $W$ onto $V$ (Cᵏ Diffeomorphisms Between Open Subsets of ℝⁿ (i)). Finally, since $Df(a)$ is invertible, (i) provides such a set $W$ with $a\in W$; applying the above to it, $f$ is a local $C^{k}$ diffeomorphism at $a$ (Local Cᵏ Diffeomorphisms (i)).