# Differentiation done correctly: 4. Inverse and implicit functions

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Now we’re going to prove the inverse function and implicit function theorems for Banach spaces.

Theorem 32 (Contraction principle). Let $$(X,d)$$ be a complete metric space and let $$\varphi:X\to X$$ be a map satisfying $$d(\varphi(x),\varphi(y)) \le cd(x,y)$$ for all $$x,y\in X$$ and some constant $$c < 1$$. Then there is exactly one $$x\in X$$ for which $$\varphi(x)=x$$.

Proof. Choose any $$x_0\in X$$ and define $$x_{n+1}=\varphi(x_n)$$. For all $$n\ge 1$$ we have $$d(x_{n+1},x_n)=d(\varphi(x_n),\varphi(x_{n-1}))\le cd(x_n,x_{n-1}),$$ so $$d(x_{n+1},x_n)\le c^n d(x_1,x_0)$$ by induction. For all $$m > n$$, \begin{align}
d(x_n,x_m) &\le d(x_n,x_{n+1})+\cdots+d(x_{m-1},x_m) \\
&\le (c^n+\cdots+c^{m-1})d(x_1,x_0) \\
&\le c^n(1-c)^{-1}d(x_1,x_0),
\end{align} which shows that $$\{x_n\}$$ is a Cauchy sequence. Since $$X$$ is complete, $$x_n\to x$$ for some $$x\in X$$. Furthermore, $$x=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\varphi(x_n)=\varphi(x)$$ since $$\varphi$$ is continuous. Uniqueness is obvious. $$\square$$

Theorem 33 (Inverse function theorem). Let $$A\subseteq E$$ be an open set and let $$f:A\to F$$ be of class $$C^p$$ (with $$p\ge 1$$). Suppose that $$f'(p)$$ is invertible for some $$p\in A$$. Then there is a neighborhood $$U\subseteq A$$ of $$p$$ such that $$f(U)$$ is open and $$f|_U:U\to f(U)$$ is a $$C^p$$ diffeomorphism.

Proof. Let $$\iota:E\to E$$ be the identity map. By replacing $$f$$ with $$f'(p)^{-1}\circ f$$, we may assume that $$E=F$$ and $$f'(p)=\iota$$. Since $$f’$$ is continuous at $$p$$, there exists an open ball $$U\subseteq A$$ around $$p$$ such that $$|f'(x)-\iota| < \frac{1}{2}$$ for all $$x\in U$$. For $$y\in f(U)$$, define the map $$\varphi_y(x)=x-f(x)+y$$. Note that $$x$$ is a fixed point of $$\varphi_y$$ if and only if $$f(x)=y$$. For $$y\in f(U)$$ we have $$|\varphi'_y(x)|=|f'(x)-\iota|<\frac{1}{2}$$ for all $$x\in U$$, so by Corollary 16 we have $$|\varphi_y(x_1)-\varphi_y(x_2)| \le \frac{1}{2}|x_1-x_2|\tag{*}$$ for all $$x_1,x_2\in U$$. Using the uniqueness argument in Theorem 32, we conclude that $$f|_U:U\to f(U)$$ is a bijection.

Now let $$b\in f(U)$$ so that $$b=f(a)$$ for some $$a\in U$$. Let $$B$$ be an open ball with radius $$r$$ around $$a$$ such that $$\overline{B}\subseteq U$$, and let $$B’$$ be an open ball of radius $$r/2$$ around $$b$$. We want to show that $$B’\subseteq f(U)$$, thus proving that $$f(U)$$ is open. Let $$y\in B’$$. If $$x\in\overline{B}$$ then \begin{align}
|\varphi_y(x)-a| &\le |\varphi_y(x)-\varphi_y(a)|+|\varphi_y(a)-a| \\
&< \frac{1}{2}|x-a|+|y-b| \\ &< r, \end{align} so $$\varphi_y(x)\in B$$. This together with (*) shows that $$\varphi_y|_{\overline{B}}:\overline{B}\to\overline{B}$$ is a contraction mapping, and since $$\overline{B}$$ is complete we can apply Theorem 32 to obtain a fixed point $$x\in\overline{B}$$ of $$\varphi_y|_{\overline{B}}$$, which implies that $$f(x)=y$$ and $$y\in f(U)$$. For the last part of the proof, we denote $$f|_U$$ by $$f$$ and $$(f|_U)^{-1}$$ by $$f^{-1}$$ for convenience. Let $$y\in f(U)$$ and $$y+k\in f(U)$$ with $$k\ne 0$$; there exist $$x\in U$$ and $$x+h\in U$$ with $$y=f(x)$$ and $$y+k=f(x+h)$$, noting that $$h\ne 0$$. In fact we have \begin{align} |h-k| &= |h-f(x+h)+f(x)| \\ &= |\varphi_y(x+h)-\varphi_y(x)| \\ &\le \frac{1}{2}|h| \end{align} from (*), so $$|h|\le 2|k|$$. Then $$h\to 0$$ as $$k\to 0$$ and \begin{align} \frac{|f^{-1}(y+k)-f^{-1}(y)-f'(x)^{-1}k|}{|k|} &= \frac{|f'(x)^{-1}(f(x+h)-f(x))-h|}{|k|} \\ &\le |f'(x)^{-1}|\frac{|f(x+h)-f(x)-f'(x)h|}{|k|} \\ &\le 2|f'(x)^{-1}|\frac{|f(x+h)-f(x)-f'(x)h|}{|h|} \\ &\to 0 \end{align} as $$h\to 0$$. (Note that $$f'(x)$$ is invertible since $$|f'(x)-\iota|<\frac{1}{2}$$.) This proves that $$(f^{-1})'(y)=f'(x)^{-1}=f'(f^{-1}(y))^{-1},\tag{**}$$ so $$f^{-1}$$ is continuous and differentiable on $$f(U)$$. Furthermore, (**) shows that $$(f^{-1})'$$ is of class $$C^p$$ since the maps $$f^{-1}$$, $$f'$$ and $$\lambda\mapsto\lambda^{-1}$$ (operator inversion) are all of class $$C^p$$. $$\square$$ Theorem 34 (Implicit function theorem). Let $$A\subseteq E$$ and $$B\subseteq F$$ be open sets and let $$f:A\times B\to G$$ be of class $$C^p$$ (with $$p\ge 1$$). Suppose $$(a,b)\in A\times B$$ such that $$f(a,b)=0$$ and $$D_2 f(a,b):F\to G$$ is invertible. Then there exists a neighborhood $$U$$ of $$a$$ and a $$C^p$$ map $$g:U\to B$$ with the following properties:

1. $$g(a)=b$$.
2. $$f(x,g(x))=0$$ for all $$x\in A$$.
3. $$g'(a)=-[D_2 f(a,b)]^{-1}\circ D_1 f(a,b)$$.

Proof. Let $$\iota:E\to E$$ be the identity map. Define \begin{align}
\widetilde{f}:A\times B &\to E\times G \\
(x,y) &\mapsto (x,f(x,y))
\end{align} and compute $$\widetilde{f}'(a,b) = \begin{bmatrix} \iota & 0 \\ D_1 f(a,b) & D_2 f(a,b) \end{bmatrix}.$$ Then $$\widetilde{f}'(a,b)$$ is invertible, with $$\widetilde{f}'(a,b)^{-1} = \begin{bmatrix} \iota & 0 \\ -[D_2 f(a,b)]^{-1}\circ D_1 f(a,b) & [D_2 f(a,b)]^{-1} \end{bmatrix}.\tag{*}$$ By the inverse function theorem, there exist neighborhoods $$V\subseteq A\times B$$ of $$(a,b)$$ and $$W\subseteq E\times G$$ of $$(a,0)$$ such that $$\widetilde{f}|_V:V\to W$$ is a $$C^p$$ diffeomorphism. Let $$U=\{x\in E:(x,0)\in W\}$$; it is clear that $$U$$ is a neighborhood of $$a$$. Define $$g:U\to B$$ by $$g=\pi\circ(\widetilde{f}|_V)^{-1}\circ i$$ where $$\pi:A\times B\to B$$ is the canonical projection and $$i:A\to A\times B$$ is given by $$i(x)=(x,0)$$. To complete the proof, we check the three required properties. Firstly, $$g(a)=\pi((\widetilde{f}|_V)^{-1}(a,0))=\pi(a,b)=b$$ since $$\widetilde{f}(a,b)=(a,0)$$. If $$x\in U$$ then $$(x,0)\in W$$, so $$(x,f(x,y))=\widetilde{f}(x,y)=(x,0)$$ for a unique $$y\in B$$ and $$f(x,g(x))=f(x,\pi((\widetilde{f}|_V)^{-1}(x,0)))=f(x,\pi(x,y))=f(x,y)=0.$$ Lastly, $$g'(b)$$ is simply the bottom left entry of (*). $$\square$$

In the next and final post, we will look at some applications of Taylor’s theorem and the implicit function theorem to finding minima and maxima of maps from Banach spaces to $$\mathbb{R}$$.

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