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

Theorem (inverse function theorem):

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

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

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

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

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

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

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

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

${\displaystyle g(x):=f(x)-x}$.

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

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

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

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

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

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

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

Thus, the most important part of the theorem is already done. All that is left to do is to prove differentiability of ${\displaystyle f^{-1}}$ at ${\displaystyle 0}$. Now we even prove the slightly stronger claim that the differential of ${\displaystyle f^{-1}}$ at ${\displaystyle x_{0}}$ 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 ${\displaystyle g}$ implies the following bounds on ${\displaystyle f}$:

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

The second bound follows from

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

and the first bound follows from

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

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

{\displaystyle {\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 ${\displaystyle f}$ at ${\displaystyle 0}$ with differential the identity, and the sandwhich criterion applied to the expressions

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

and

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