In this chapter, we want to prove the inverse function theorem (which asserts that if a function has invertible differential at a point, then it is locally invertible itself) and the implicit function theorem (which asserts that certain sets are the graphs of functions).
Theorem:
Let
be a complete metric space, and let
be a strict contraction; that is, there exists a constant
such that
.
Then
has a unique fixed point, which means that there is a unique
such that
. Furthermore, if we start with a completely arbitrary point
, then the sequence
![{\displaystyle y,f(y),f(f(y)),f(f(f(y))),\ldots }](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9CyjsPo2e5aDCPaDs2oDnBaDFAoqe2ygo5agdAotnDaqnBz2w0ntCO)
converges to
.
Proof:
First, we prove uniqueness of the fixed point. Assume
are both fixed points. Then
.
Since
, this implies
.
Now we prove existence and simultaneously the claim about the convergence of the sequence
. For notation, we thus set
and if
is already defined, we set
. Then the sequence
is nothing else but the sequence
.
Let
. We claim that
.
Indeed, this follows by induction on
. The case
is trivial, and if the claim is true for
, then
.
Hence, by the triangle inequality,
.
The latter expression goes to zero as
and hence we are dealing with a Cauchy sequence. As we are in a complete metric space, it converges to a limit
. This limit further is a fixed point, as the continuity of
(
is Lipschitz continuous with constant
) implies
.![{\displaystyle \Box }](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8QaDBEzNJAajBBoArFnDK1aDFAoNGNaqnBztK4zjFEntsPnDeOzjlE)
A corollary to this important result is the following lemma, which shall be the main ingredient for the proof of the inverse function theorem:
Lemma:
Let
(
denoting the closed ball of radius
) be a function which is Lipschitz continuous with Lipschitz constant less or equal
such that
. Then the function
![{\displaystyle f:{\overline {B_{r}(0)}}\to \mathbb {R} ^{n},f(x):=g(x)+x}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO80a2dEajoOotCPoNi5yjrFzAo3aAsNzNK4ygePztBDo2w5zDm5zAsQ)
is injective and
.
Proof:
First, we note that for
the function
![{\displaystyle h:{\overline {B_{r}(0)}}\to \mathbb {R} ^{n},h(z):=y-g(z)}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Qa2w1ajsPzthCytdBztzFntJCyqaNygiQoArEnqhCnjKNztJAntw1)
is a strict contraction; this is due to
.
Furthermore, it maps
to itself, since for
.
Hence, the Banach fixed-point theorem is applicable to
. Now
being a fixed point of
is equivalent to
,
and thus
follows from the existence of fixed points. Furthermore, if
, then
![{\displaystyle {\frac {1}{2}}\|x-x'\|\geq \|g(x)-g(x')\|=\|f(x)-x-(f(x')-x')\|=\|x-x'\|}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO84o2wQoNG5oDlBoAvCotC5aAiPz2a4zgnEyjo2nAnBzqrFzAdBaDBE)
and hence
. Thus injectivity.
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
![{\displaystyle {\tilde {h}}(x):=h'(x_{0})^{-1}(h(x_{0}-x)-h(x_{0}))}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9FyjwQzgrCytdEnqo5a2e5zDCQyqzFyjs5zjs3o2hEnDi0aDa0a2zD)
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
![{\displaystyle {\frac {\partial g}{\partial x_{j}}}(x)<{\frac {1}{2n^{2}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Aoqi1z2wNngzDoNC2zqaOyts5aqoPnDJAzNK5nqo1yjJEaNlCotG2)
for all
and
. Since further
, the general mean-value theorem and Cauchy's inequality imply that for
and
,
![{\displaystyle |g_{k}(x)|=|\langle x,{\frac {\partial g}{\partial x_{j}}}(t_{k}x)\rangle |\leq \|x\|n{\frac {1}{2n^{2}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO80ztwPnDwQaqw4agwOaNi4nDs0njrCz2dDyjs1zAzBate2zNm2zqs5)
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
):
![{\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}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9FoAs4oDJDyti2oAo0zAs2aAzBytCQatJFoDmNa2w0aqi0oDGPoAvE)
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
![{\displaystyle {\frac {\|\mathbf {h} -f(\mathbf {h} )\|}{{\frac {3}{2}}\|\mathbf {h} \|}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO84nqvFotJFz2sPnqnEoNK2oDo1aArAnDi5yja4nqw3ygs5ntGOnge0)
and
.![{\displaystyle \Box }](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8QaDBEzNJAajBBoArFnDK1aDFAoNGNaqnBztK4zjFEntsPnDeOzjlE)
Theorem:
Let
be a continuously differentiable function, and consider the set
.
If we are given some
such that
, then we find
open with
and
such that
and
,
where
is open with respect to the subspace topology of
.
Furthermore,
is a differentiable function.
Proof:
We define a new function
.
The differential of this function looks like this:
![{\displaystyle F'(x)={\begin{pmatrix}1&0&\cdots &&0\\0&1&&&\vdots \\\vdots &&\ddots &&\\0&\cdots &0&1&0\\\partial _{1}f(x)&&\cdots &&\partial _{n}f(x)\end{pmatrix}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Oags0aNs2ztoNnAzCoNhDyja0ztvFotrFaDiNnqnCyqe3ztaOntiN)
Since we assumed that
,
is invertible, and hence the inverse function theorem implies the existence of a small open neighbourhood
containing
such that restricted to that neighbourhood
is itself invertible, with a differentiable inverse
, which is itself defined on an open set
containing
. Now set first
,
which is open with respect to the subspace topology of
, and then
,
the
-th component of
. We claim that
has the desired properties.
Indeed, we first note that
, since applying
leaves the first
components unchanged, and thus we get the identity by observing
. Let thus
. Then
.
Furthermore, the set
![{\displaystyle \{(z_{1},\ldots ,z_{n-1},g(z_{1},\ldots ,z_{n-1}))|(z_{1},\ldots ,z_{n-1})\in U\}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Pnjw3yqoNagrBzNe0njJAyjnEzgw5yjiNoDK3atJDatlFyjC3o2vE)
is open with respect to the subspace topology on
. Indeed, we show
.
For
, we first note that the set on the left hand side is in
, since all points in it are mapped to zero by
. Further,
![{\displaystyle F(z_{1},\ldots ,z_{n-1},g(z_{1},\ldots ,z_{n-1}))=(z_{1},\ldots ,z_{n-1},0)\in {\tilde {U}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO83njBAajdBatwQaDi5zNaOygi4ajrEygoNaNnDnqvEo2w3nDGOyje0)
and hence
is completed when applying
. For the other direction, let a point
in
be given, apply
to get
![{\displaystyle F((x_{1},\ldots ,x_{n}))=(x_{1},\ldots ,x_{n-1},0)\in {\tilde {U}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8PoDo0zga2njo5aqe0oDa0ytaNaDnFztnAyts3aqzAyjK4yjdBoNsO)
and hence
; further
![{\displaystyle (x_{1},\ldots ,x_{n-1},g(x_{1},\ldots ,x_{n-1}))=(x_{1},\ldots ,x_{n})}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8QaDiNzqsOaDm1ztCNoNoOa2eNoDK2zja0ote4atGNz2i0yjK4zjoN)
by applying
to both sides of the equation.
Now
is automatically differentiable as the component of a differentiable function.
Informally, the above theorem states that given a set
, one can choose the first
coordinates as a "base" for a function, whose graph is precisely a local bit of that set.