The Inverse Function Theorem

We give the classical contraction mapping proof, in four stages: normalization, local injectivity and a contraction estimate, surjectivity onto an open set via the contraction principle, and differentiability of the inverse.

Stage 0: Reduction by an invertible linear change. Let $A = DF(a)$. Since $A$ is invertible, define $\widetilde F(x) = A^{-1}F(x)$. Then $\widetilde F$ is $C^1$ (composition of $C^1$ with linear), $D\widetilde F(x) = A^{-1}DF(x)$ by the The Chain Rule in Several Variables, and $D\widetilde F(a) = A^{-1}A = I$, the identity matrix. If we prove the theorem for $\widetilde F$ — producing a local $C^1$ inverse $\widetilde G$ near $\widetilde F(a)$ — then $G(y) := \widetilde G(A^{-1}y)$ is a local $C^1$ inverse for $F$ near $F(a)$, since $F = A\widetilde F$. So without loss of generality we assume $DF(a) = I$.

Stage 1: Setting up the contraction. Since $DF$ is continuous at $a$ (this is where $C^1$, not mere differentiability, is used) and $DF(a) = I$, choose $r>0$ so small that the closed ball $\overline{B}(a,2r) \subset U$ and

where $\|\cdot\|$ denotes the operator norm on matrices. (Such $r$ exists because $x \mapsto \|DF(x)-I\|$ is continuous and equals $0$ at $x=a$.) For $y \in \mathbb{R}^n$ define the auxiliary map

Note that $x$ is a fixed point of $\varphi_y$ if and only if $F(x) = y$. We show $\varphi_y$ is a contraction on $\overline{B}(a,2r)$ (for suitable $y$), with Lipschitz constant $\tfrac12$. The derivative of $\varphi_y$ is $D\varphi_y(x) = I - DF(x)$, so $\|D\varphi_y(x)\| \le \tfrac12$ for all $x \in \overline{B}(a,2r)$. By the mean value inequality (applied componentwise, or directly to vector-valued $C^1$ maps: for $x_1,x_2$ in a convex set, $\|\varphi_y(x_1)-\varphi_y(x_2)\| \le \sup_{t}\|D\varphi_y(x_1+t(x_2-x_1))\| \cdot \|x_1-x_2\|$, a standard consequence of the fundamental theorem of calculus applied to $t\mapsto \varphi_y(x_1+t(x_2-x_1))$), we get for all $x_1,x_2 \in \overline{B}(a,2r)$:

In particular, taking $y = F(x_2)$ (so $\varphi_y(x_2) = x_2$... we do not yet know that, so instead directly apply $(\ast)$ with $x_2 = a$, $y$ arbitrary):

Injectivity of $F$ on $\overline{B}(a,2r)$. If $F(x_1) = F(x_2)$ for $x_1,x_2\in \overline B(a,2r)$, then $\varphi_y(x_1) - \varphi_y(x_2) = (x_1-x_2) - (F(x_1)-F(x_2)) = x_1 - x_2$ for $y=F(x_1)$. Combined with $(\ast)$: $\|x_1-x_2\| \le \tfrac12\|x_1-x_2\|$, forcing $x_1=x_2$. So $F$ is injective on $\overline{B}(a,2r)$.

Stage 2: Existence of the inverse on a genuinely open set (surjectivity via contraction). Let $W_0 := B\big(F(a), r/2\big)$, an open ball. We claim: for every $y \in W_0$, there is a unique $x \in B(a,2r)$ with $F(x) = y$, and moreover $x \in B(a,r)$. Fix $y \in W_0$. We show $\varphi_y$ maps the closed ball $\overline{B}(a,r)$ into itself and is a contraction there, hence has a unique fixed point by the contraction mapping principle (a standard consequence of completeness of $\mathbb{R}^n$, i.e. of Compactness in R^n together with the Cauchy criterion). For $x \in \overline{B}(a,r)$:

By $(\ast)$, $\|\varphi_y(x)-\varphi_y(a)\| \le \tfrac12\|x-a\| \le \tfrac{r}{2}$. Also $\varphi_y(a) - a = y - F(a)$, so $\|\varphi_y(a)-a\| = \|y - F(a)\| < r/2$ since $y \in W_0$. Hence

so $\varphi_y(x) \in B(a,r) \subset \overline{B}(a,r)$. Thus $\varphi_y : \overline{B}(a,r) \to \overline{B}(a,r)$ maps this complete metric space into itself, and by $(\ast)$ it is a contraction with constant $\tfrac12 < 1$. By the Banach contraction mapping principle, $\varphi_y$ has a unique fixed point $x = x(y) \in \overline{B}(a,r)$, i.e. a unique $x\in \overline B(a,r)$ with $F(x) = y$; the argument above (bounding $\|\varphi_y(x)-a\|<r$ strictly) shows this fixed point actually lies in the open ball $B(a,r)$. Combined with injectivity on the larger ball $\overline B(a,2r)$ from Stage 1, this $x$ is the unique preimage of $y$ in all of $\overline B(a,2r)$. Define $V := F^{-1}(W_0) \cap B(a,r)$ and $W := W_0$. By the above, $F: V \to W$ is a bijection, with inverse $G := F^{-1}|_W$. We should check $V$ is open: since $F$ is continuous, and we can show below that $G$ is continuous (in fact Lipschitz), $V = G(W)$ is the continuous image of an open set under a map that is, as we'll see, an open map — alternatively, directly: for any $x_0 \in V$, $F(x_0) \in W_0$ open, and by continuity of $F$ and the estimate below there's a ball around $x_0$ mapping into $W_0 \cap$ appropriate region; we verify this rigorously via the Lipschitz estimate on $G$ next, which shows $G$ is continuous, hence $V=G(W)$ is open as the preimage of the open set... to avoid circularity we argue directly:

$V$ is open. Let $x_0\in V$, so $F(x_0)=y_0\in W_0$. Since $W_0$ open, pick $\rho>0$ with $B(y_0,\rho)\subset W_0$. For any $y\in B(y_0,\rho)$, the fixed-point argument above (applied verbatim with $y$ in place of the earlier generic point, since $y \in W_0$) produces $x=x(y)\in B(a,r)$ with $F(x)=y$; we must check $x$ is close to $x_0$. From $(\ast)$ with $x_1 = x(y), x_2 = x_0$:

using $\varphi_y(x_0) - \varphi_{y_0}(x_0) = (y - y_0)$ directly from the definition of $\varphi_y$:

so $\|x(y)-x_0\| \le 2\|y-y_0\|$. As $y$ ranges over $B(y_0,\rho)$, $x(y)$ ranges over a subset of $B(x_0, 2\rho)$; choosing $\rho$ small enough that $B(x_0,2\rho)\subset V\cap B(a,r)$ (possible since $B(a,r)$ is open and $x_0 \in B(a,r)$), this shows $F^{-1}(B(y_0,\rho)) \cap B(a,r) \subset B(x_0,2\rho)$, and this whole preimage lies in $V$. Since $x_0$ was arbitrary and every point of $F(V)=W_0$ has such a neighborhood pulling back into $V$, we conclude $V$ is open. $\;\square$ (Stage 2)

Also, the same inequality shows $G$ is Lipschitz continuous: for $y_1, y_2 \in W$,

(Same computation as above, with $y_0 \to y_1$, $x_0\to G(y_1)$.)

Stage 3: Differentiability of $G$. Fix $y \in W$ and let $x = G(y) \in V$. Since $DF(x)$ is invertible — indeed $\|DF(x)-I\|\le 1/2$ on $\overline B(a,2r)$ forces $DF(x)$ invertible, because if $DF(x)v=0$ for unit vector $v$ then $\|v - DF(x)v\| = \|v\| =1 \le \|I - DF(x)\|\cdot\|v\| \le 1/2$, a contradiction — let $T = [DF(x)]^{-1}$. We claim $DG(y) = T$. Let $k \in \mathbb{R}^n$ be small and nonzero with $y+k \in W$. Let $h := G(y+k) - G(y)$, so $x+h = G(y+k) \in V$, and $F(x+h) = y+k$, $F(x) = y$. By $(\ast\ast)$, $\|h\| \le 2\|k\|$, so $h \to 0$ as $k \to 0$; in particular $h \neq 0$ whenever $k\neq 0$ (by injectivity of $F$ on $V$), and more importantly $\|h\|/\|k\|$ stays bounded. By differentiability of $F$ at $x$:

Apply $T = [DF(x)]^{-1}$ to both sides:

Thus

As $k \to 0$, $h \to 0$ (shown above), so $\varepsilon(h) \to 0$, and the right side $\to 0$. This is exactly the statement that $G$ is differentiable at $y$ with $DG(y) = T = [DF(x)]^{-1} = [DF(G(y))]^{-1}$.

Stage 4: Continuity of $DG$. We have, for all $y \in W$,

The map $y \mapsto G(y)$ is continuous (Lipschitz, Stage 2); the map $x \mapsto DF(x)$ is continuous ($F$ is $C^1$); and matrix inversion $M \mapsto M^{-1}$ is continuous on the open set of invertible matrices (each entry of $M^{-1}$ is, by Cramer's rule, a ratio of polynomials in the entries of $M$ divided by $\det M \neq 0$, hence continuous). Composing three continuous maps, $y \mapsto DG(y)$ is continuous on $W$. Hence $G \in C^1(W)$. This proves all three parts of the theorem, with $V,W$ as constructed. $\blacksquare$