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

**Theorem (inverse function theorem)**:

Let be a function which is continuously differentiable in a neighbourhood such that is invertible. Then there exists an open set with such that is a bijective function with an inverse which is differentiable at and satisfies

- .

**Proof:** We first reduce to the case , and . Indeed, suppose for all those functions the theorem holds, and let now be an arbitrary function satisfying the requirements of the theorem (where the differentiability is given at ). We set

and obtain that is differentiable at with differential and ; the first property follows since we multiply both the function and the linear-affine approximation by and only shift the function, and the second one is seen from inserting . Hence, we obtain an inverse of with it's differential at , and if we now set

- ,

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

Thus let be a function such that , is invertible at and . We define

- .

The differential of this function is zero (since taking the differential is linear and the differential of the function is the identity). Since the function is also continuously differentiable at a small neighbourhood of , we find such that

for all and . Since further , the general mean-value theorem and Cauchy's inequality imply that for and ,

for suitable . Hence,

- (triangle inequality),

and thus, we obtain that our preparatory lemma is applicable, and is a bijection on , whose image is contained within the open set ; thus we may pick , which is open due to the continuity of .

Thus, the most important part of the theorem is already done. All that is left to do is to prove differentiability of at . Now we even prove the slightly stronger claim that the differential of at 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 implies the following bounds on :

- .

The second bound follows from

- ,

and the first bound follows from

- .

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

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

and

- .