inverse.tex

\chapter{ Proofs of the Inverse and the Implicit Functions Theorems} \label{sec:proof-inverse-implicit}
% 
%   In this section , we  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).

\section{Banach's Fixed Point Theorem}\label{banachs-fixed-point-theorem}

\begin{df}
 Let $M$  be a  subset  of $\bbR^n$.  A function \(f: M \to M\) is  a
\negrito{strict contraction} if  there exists a constant
\(0 \le \lambda < 1\) such that

\[\forall x, y \in M: \dist(f(x), f(y)) \le \lambda \dist(x, y).\]
\end{df}




\begin{thm}[Banach's Fixed Point]
 Let $M$  be a closed subset\footnote{Tthe Banach Fixed Point can be proved for a  complete metric space  \((M,d)\)} of $\bbR^n$ and let \(f: M \to M\) be a
\negrito{strict contraction}.  Then \(f\)
has a unique \negrito{fixed point}, which means that there is a unique
\(z \in M\) such that \(f(z) = z\). Furthermore, if we start with a
completely arbitrary point \(y \in M,\) then the sequence

\[y, f(y), f(f(y)), f(f(f(y))), \ldots\] converges to \(z\).
\end{thm}

\begin{proof}
We start  proving the uniqueness of the fixed point. 

Assume  that \(x, y\) are fixed points. Then
\[\dist(x, y) = \dist(f(x), f(y)) \le \lambda \dist(x, y) \Rightarrow (1 - \lambda) \dist(x, y) = 0.\]
Since \(0 \le \lambda < 1,\) we have 
\(\dist(x, y) = 0\) and so $ x = y$. So we have the  the uniqueness of the fixed point. 


Now we prove the existence of the fixed point.

Let $z_n$  be the sequence recursively defined  by  \(z_0 := y\) and \(z_{n+1} = f(z_n)\). This sequence 
\((z_n)_{n \in \mathbb N}\) is nothing else but the sequence
\(y, f(y), f(f(y)), f(f(f(y))), \ldots\).

Let \(n \ge 0\). We claim that

\[\dist(z_{n+1}, z_n) \le \lambda^n \dist(z_1, z_0).\] Indeed, this follows by
induction on \(n\). The case \(n=0\) is trivial, and if the claim is
true for \(n,\) then
\(\dist(z_{n+2}, z_{n+1}) = \dist(f(z_{n+1}), f(z_n)) \le \lambda \dist(z_{n+1}, z_n) \le \lambda \cdot \lambda^n \dist(z_1, z_0)\).

Hence, by the triangle inequality,

\begin{align}
\dist(z_{n + m}, z_n) & \le \sum_{j=n+1}^{n+m} \dist(z_j, z_{j-1}) \\
& \le \sum_{j=n+1}^{n+m} \lambda^{j-1} \dist(z_1, z_0) \\
& \le \sum_{j=n+1}^\infty \lambda^{j-1} \dist(z_1, z_0) \\
& = \dist(z_1, z_0) \lambda^n \frac{1}{1 - \lambda}
\end{align}

We note that the latter expression goes to zero as \(n \to \infty\)
and so  $z_n$ is a Cauchy sequence. 
Since $M$ is a  closed subset of $\bbR^n$, the sequence converges to a limit \(z\). This limit  is the
fixed point:

\[x = \lim_{n \to \infty} z_n = \lim_{n \to \infty} f(z_{n-1}) = f(\lim_{n \to \infty} z_{n-1}) = f(z).\]

\end{proof}



\begin{lemma}
Let \(g: \overline{B_r(0)} \to \overline{B_r(0)}\) be  Lipschitz continuous function  with Lipschitz
constant less or equal \(1/2\) such that \(g(0) = 0\). Then the function

\[f: \overline{B_r(0)} \to \mathbb R^n, f(x) := g(x) + x\] is injective
and \(B_{r/2}(0) \subseteq f(B_r(0))\).
\end{lemma}


\begin{proof}
First, we note that for \(y \in B_{r/2}(0)\) the function

\[h: \overline{B_r(0)} \to \mathbb R^n, h(z) := y - g(z)\] is a strict
contraction. This fact follows from

\[\|y - g(z) - (y - g(z'))\| = \|g(z') - g(z)\| \le \frac{1}{2}\|z - z'\|.\]
Furthermore, it maps the closed ball  \(\overline{B_r(0)}\) to itself, since for
\(z \in \overline{B_r(0)}\)

\[\|y - g(z)\| \le \|y\| + \|g(z - 0)\| \le \frac{r}{2} + \frac{1}{2}\|z\| \le r.\]
Hence, the Banach Fixed Point Theorem is applicable to the function \(h\). Now  the fact that the point \(x\)
is a fixed point of the function  \(h\) is equivalent to

\[f(x) = y,\] and thus \(B_{r/2}(0) \subseteq f(B_r(0))\) follows from
the existence of fixed points. Furthermore, if \(f(x) = f(x'),\) then

\[\frac{1}{2} \|x - x'\| \ge \|g(x) - g(x')\| = \|f(x) - x - (f(x') - x')\| = \|x - x'\|\]
and so \(x = x'\). So  the function is injective.
\end{proof}


\section{The Inverse Function Theorem}\label{the-inverse-function-theorem}

\begin{thm}
Let \(\vector{f}: \mathbb R^n \to \mathbb R^n\) be a continuously differentiable function  in a neighborhood \(\vector{x}_0 \in \mathbb R^n\)
such that \(\vector{f}'(\vector{x}_0)\) is invertible. Then there exists an open set
\(U \subseteq \mathbb R^n\) with \(\vector{x}_0 \in U\) such that \(\vector{f}|_U\) is a
bijective function with an inverse \(\vector{f}^{-1} : \vector{f}(U) \to U\) which is
differentiable at \(\vector{x}_0\) and satisfies

\[(\vector{f}^{-1})'(\vector{x}_0) = (\vector{f}'(\vector{x}_0))^{-1}.\]
\end{thm}


\begin{proof}
We first reduce the analysis  to the case \(\vector{f}(\vector{x}_0) = 0,\) \(\vector{x}_0 = 0\) and
\(\vector{f}'(\vector{x}_0) = \text{I}\). 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
\(\vector{x}_0\)). We set

\[\tilde h(x) := h'(\vector{x}_0)^{-1}(h(\vector{x}_0 - x) - h(\vector{x}_0))\] 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 approximation by
\(h'(\vector{x}_0)^{-1}\) 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

\[h^{-1}(y) :=  (\tilde h^{-1} (h'(\vector{x}_0)^{-1}(y - h(\vector{x}_0))) - \vector{x}_0),\] 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 \(\vector{f}\) be a function such that \(\vector{f}(0) = 0,\) \(\vector{f}\) is invertible
at \(0\) and \(\vector{f}'(0) = \text{Id}\). We define

\[g(x) := \vector{f}(x) - x.\] The differential of this function is zero. Since the function \(g\) is also
continuously differentiable at a small neighborhood 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
\(g(0) = \vector{f}(0) - 0 = 0,\) the general mean-value theorem and Cauchy's
inequality imply that for \(k \in \{1, \ldots, n\}\) and
\(x \in B_\delta(0),\)

\[|g_k(x)| = |\langle x, \frac{\partial g}{\partial x_j}(t_k x) \rangle| \le \|x\| n \frac{1}{2n^2}\]
for suitable \(t_k \in [0, 1]\). Hence,

\[\|g(x)\| \le |g_1(x)| + \cdots + |g_n(x)| \le \frac{1}{2} \|x\|\]
and thus, we note that the lemma is
applicable, and \(\vector{f}\) is a bijection on \(\overline{B_\delta(0)},\)
whose image is contained within the open set
\(\overline{B_{\delta/2}(0)}\). So we may pick
\(U := \vector{f}^{-1}(B_{\delta/2}(0)),\) 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 \(\vector{f}^{-1}\) at \(0\). Now
we even prove the slightly stronger claim that the differential of
\(\vector{f}^{-1}\) at \(\vector{x}_0\) 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 \(\vector{f}\):

\[\frac{1}{2}\|x\| \le \|\vector{f}(x)\| \le \frac{3}{2} \|x\|.\] The second
bound follows from

\[\|\vector{f}(x)\| \le \|\vector{f}(x) - x\| + \|x\| = \|g(x)\| + \|x\| \le \frac{3}{2} \|x\|,\]
and the first bound follows from

\[\|\vector{f}(x)\| \ge |\|\vector{f}(x) - x\| - \|x\|| = \left| \|g(x)\| - \|x\| \right| \ge \frac{1}{2} \|x\|.\]

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

\begin{align}
\lim_{\mathbf h \to 0} \frac{\|\vector{f}^{-1}(\mathbf h) - \vector{f}^{-1}(0) - \operatorname{Id} (\mathbf h - 0)\|}{\|\mathbf h\|} & = \lim_{\mathbf h \to 0} \frac{\|\vector{f}^{-1}(\vector{f}(\mathbf h)) - \vector{f}(\mathbf h)\|}{\|\vector{f}(\mathbf h)\|} \\
& = \lim_{\mathbf h \to 0} \frac{\|\mathbf h - f(\mathbf h)\|}{\|f(\mathbf h)\|},
\end{align}
where the last expression converges to zero due to the
differentiability of \(\vector{f}\) at \(0\) with differential the identity, and
the sandwhich criterion applied to the expressions

\[\frac{\|\mathbf h - \vector{f}(\mathbf h)\|}{\frac{3}{2} \|\mathbf h\|}\] and

\[\frac{\|\mathbf h - f(\mathbf h)\|}{\frac{1}{2} \|\mathbf h\|}.\]
\end{proof}

\section{The Implicit Function Theorem}\label{the-implicit-function-theorem}

\begin{thm}
Let \(\vector{f}: \mathbb R^n \to \mathbb R\) be a continuously differentiable
function, and consider the set

\[S := \{(x_1, \ldots, x_n) \in \mathbb R^n | \vector{f}(x_1, \ldots, x_n) = 0\}.\]
If we are given some \(y \in S\) such that \(\partial_n \vector{f}(y) \neq 0,\)
then we find \(U \subseteq \mathbb R^{n-1}\) open with
\((y_1, \ldots, y_{n-1}) \in U\) and \(g: U \to S\) such that

\[y = g(y_1, \ldots, y_{n-1})\] and
\(\{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\} \subseteq S,\)
% where
% \(\{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\}\)


Furthermore, \(g\) is a differentiable function.
\end{thm}


\begin{proof}
 
We define a new function

\[F: \mathbb R^n \to \mathbb R^n, F(x_1, \ldots, x_n) := (x_1, \ldots, x_{n-1}, f(x_1, \ldots, x_n)).\]
The differential of this function looks like this:

\[F'(x) = \begin{pmatrix}
1 & 0 & \cdots & & 0 \\
0 & 1 & & & \vdots \\
\vdots & & \ddots & & \\
0 & \cdots & 0 & 1 & 0 \\
\partial_1 f(x) & & \cdots & & \partial_n f(x)
\end{pmatrix}\] Since we assumed that \(\partial_n f(y) \neq 0,\)
\(F'(y)\) is invertible, and hence the Inverse Function Theorem implies
the existence of a small open neighbourhood
\(\tilde V \subseteq \mathbb R^n\) containing \(y\) such that
\emph{restricted to that neighbourhood} \(F\) is  invertible, with
a differentiable inverse \(F^{-1},\) which is itself defined on an open
set \(\tilde U\) containing \(F(y)\). Now set first

\[U := \{(x_1, \ldots, x_{n-1}) | (x_1, \ldots, x_{n-1}, 0) \in \tilde U \},\]
) and then

\[g: U \to \mathbb R, g(x_1, \ldots, x_{n-1}) := \pi_n(F^{-1}(x_1, \ldots, x_{n-1}, 0)),\]
the \(n\)-th component of \(F^{-1}(x_1, \ldots, x_{n-1}, 0)\). We claim
that \(g\) has the desired properties.

Indeed, we first note that
\(F^{-1}(x_1, \ldots, x_{n-1}, 0) = (x_1, \ldots, x_{n-1}, g(x_1, \ldots, x_{n-1})),\)
since applying \(F\) leaves the first \(n-1\) components unchanged, and
thus we get the identity by observing \(F(F^{-1}(x)) = x\). Let thus
\((z_1, \ldots, z_{n-1}) \in U\). Then

\begin{align}
f(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) & = (\pi_n \circ F)(F^{-1}(z_1, \ldots, z_{n-1}, 0)) \\
& = \pi_n((F \circ F^{-1})(z_1, \ldots, z_{n-1}, 0)) = 0.
\end{align} Furthermore, the set



\[\{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\} = S \cap \tilde V.\]
For \(\subseteq,\) we first note that the set on the left hand side is
in \(S,\) since all points in it are mapped to zero by \(f\). Further,

\[F(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) = (z_1, \ldots, z_{n-1}, 0) \in \tilde U\]
and hence \(\subseteq\) is completed when applying \(F^{-1}\). For the
other direction, let a point \((x_1, \ldots, x_n)\) in
\(S \cap \tilde V\) be given, apply \(F\) to get

\[F((x_1, \ldots, x_n)) = (x_1, \ldots, x_{n-1}, 0) \in \tilde U\] and
hence \((x_1, \ldots, x_{n-1}) \in U\); further

\[(x_1, \ldots, x_{n-1}, g(x_1, \ldots, x_{n-1})) = (x_1, \ldots, x_n)\]
by applying \(F\) to both sides of the equation.

Now \(g\) is automatically differentiable since $g$ is  the component of a
differentiable function.
 
\end{proof}