Differentiation and Integration in Several Real Variables/The inverse and implicit function theorems

Theorem (inverse function theorem):

Let $f:\mathbb {R} ^{n}\to \mathbb {R} ^{n}$ be a function which is continuously differentiable in a neighbourhood $x_{0}\in \mathbb {R} ^{n}$ such that $f'(x_{0})$ is invertible. Then there exists an open set $U\subseteq \mathbb {R} ^{n}$ with $x_{0}\in U$ such that $f|_{U}$ is a bijective function with an inverse $f^{-1}:f(U)\to U$ which is differentiable at $x_{0}$ and satisfies

$(f^{-1})'(x_{0})=(f'(x_{0}))^{-1}$.

Proof: We first reduce to the case $f(x_{0})=0$, $x_{0}=0$ and $f'(x_{0})={\text{Id}}$. Indeed, suppose for all those functions the theorem holds, and let now $h$ be an arbitrary function satisfying the requirements of the theorem (where the differentiability is given at $x_{0}$). We set

${\tilde {h}}(x):=h'(x_{0})^{-1}(h(x_{0}-x)-h(x_{0}))$

and obtain that ${\tilde {h}}$ is differentiable at $0$ with differential ${\text{Id}}$ and ${\tilde {h}}(0)=0$; the first property follows since we multiply both the function and the linear-affine approximation by $h'(x_{0})^{-1}$ and only shift the function, and the second one is seen from inserting $x=0$. Hence, we obtain an inverse of ${\tilde {h}}$ with it's differential at ${\tilde {h}}(0)=0$, and if we now set

$h^{-1}(y):=({\tilde {h}}^{-1}(h'(x_{0})^{-1}(y-h(x_{0})))-x_{0})$,

it can be seen that $h^{-1}$ is an inverse of $h$ with all the required properties (which is a bit of a tedious exercise, but involves nothing more than the definitions).

Thus let $f$ be a function such that $f(0)=0$, $f$ is invertible at $0$ and $f'(0)={\text{Id}}$. We define

$g(x):=f(x)-x$.

The differential of this function is zero (since taking the differential is linear and the differential of the function $x\mapsto x$ is the identity). Since the function $g$ is also continuously differentiable at a small neighbourhood of $0$, we find $\delta >0$ such that

${\frac {\partial g}{\partial x_{j}}}(x)<{\frac {1}{2n^{2}}}$

for all $j\in \{1,\ldots ,n\}$ and $x\in B_{\delta }(0)$. Since further $g(0)=f(0)-0=0$, the general mean-value theorem and Cauchy's inequality imply that for $k\in \{1,\ldots ,n\}$ and $x\in B_{\delta }(0)$,

$|g_{k}(x)|=|\langle x,{\frac {\partial g}{\partial x_{j}}}(t_{k}x)\rangle |\leq \|x\|n{\frac {1}{2n^{2}}}$

for suitable $t_{k}\in [0,1]$. Hence,

$\|g(x)\|\leq |g_{1}(x)|+\cdots +|g_{n}(x)|\leq {\frac {1}{2}}\|x\|$ (triangle inequality),

and thus, we obtain that our preparatory lemma is applicable, and $f$ is a bijection on ${\overline {B_{\delta }(0)}}$, whose image is contained within the open set ${\overline {B_{\delta /2}(0)}}$; thus we may pick $U:=f^{-1}(B_{\delta /2}(0))$, which is open due to the continuity of $f$.

Thus, the most important part of the theorem is already done. All that is left to do is to prove differentiability of $f^{-1}$ at $0$. Now we even prove the slightly stronger claim that the differential of $f^{-1}$ at $x_{0}$ is given by the identity, although this would also follow from the chain rule once differentiability is proven.

Note now that the contraction identity for $g$ implies the following bounds on $f$:

${\frac {1}{2}}\|x\|\leq \|f(x)\|\leq {\frac {3}{2}}\|x\|$.

The second bound follows from

$\|f(x)\|\leq \|f(x)-x\|+\|x\|=\|g(x)\|+\|x\|\leq {\frac {3}{2}}\|x\|$,

and the first bound follows from

$\|f(x)\|\geq |\|f(x)-x\|-\|x\||=\left|\|g(x)\|-\|x\|\right|\geq {\frac {1}{2}}\|x\|$.

Now for the differentiability at $0$. We have, by substitution of limits (as $f$ is continuous and $f(0)=0$):

${\begin{aligned}\lim _{\mathbf {h} \to 0}{\frac {\|f^{-1}(\mathbf {h} )-f^{-1}(0)-\operatorname {Id} (\mathbf {h} -0)\|}{\|\mathbf {h} \|}}&=\lim _{\mathbf {h} \to 0}{\frac {\|f^{-1}(f(\mathbf {h} ))-f(\mathbf {h} )\|}{\|f(\mathbf {h} )\|}}\\&=\lim _{\mathbf {h} \to 0}{\frac {\|\mathbf {h} -f(\mathbf {h} )\|}{\|f(\mathbf {h} )\|}},\end{aligned}}$

where the last expression converges to zero due to the differentiability of $f$ at $0$ with differential the identity, and the sandwhich criterion applied to the expressions

${\frac {\|\mathbf {h} -f(\mathbf {h} )\|}{{\frac {3}{2}}\|\mathbf {h} \|}}$

and

${\frac {\|\mathbf {h} -f(\mathbf {h} )\|}{{\frac {1}{2}}\|\mathbf {h} \|}}$. $\Box$