View source for The Inverse Function Theorem and the Implicit Function Theorem

In this section, 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).

== The inverse function theorem ==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
'''Theorem''':

:Let <math>f: \mathbb R^n \to \mathbb R^n</math> be a function which is continuously differentiable in a neighbourhood <math>x_0 \in \mathbb R^n</math> such that <math>f'(x_0)</math> is invertible. Then there exists an open set <math>U \subseteq \mathbb R^n</math> with <math>x_0 \in U</math> such that <math>f|_U</math> is a bijective function with an inverse <math>f^{-1} : f(U) \to U</math> which is differentiable at <math>x_0</math> and satisfies
::<math>(f^{-1})'(f(x_0)) = (f'(x_0))^{-1}</math>.
</blockquote>

'''Proof''':

We first reduce to the case <math>f(x_0) = 0</math>, <math>x_0 = 0</math> and <math>f'(x_0) = \text{Id}</math>. Indeed, suppose for all those functions the theorem holds, and let now <math>h</math> be an arbitrary function satisfying the requirements of the theorem (where the differentiability is given at <math>x_0</math>). We set
:<math>\tilde h(x) := h'(x_0)^{-1}(h(x_0 - x) - h(x_0))</math>
and obtain that <math>\tilde h</math> is differentiable at <math>0</math> with differential <math>\text{Id}</math> and <math>\tilde h(0) = 0</math>; the first property follows since we multiply both the function and the linear-affine approximation by <math>h'(x_0)^{-1}</math> and only shift the function, and the second one is seen from inserting <math>x = 0</math>. Hence, we obtain an inverse of <math>\tilde h</math> with it's differential at <math>\tilde h(0) = 0</math>, and if we now set
:<math>h^{-1}(y) :=  (\tilde h^{-1} (h'(x_0)^{-1}(y - h(x_0))) - x_0)</math>,
it can be seen that <math>h^{-1}</math> is an inverse of <math>h</math> with all the required properties (which is a bit of a tedious exercise, but involves nothing more than the definitions).

Thus let <math>f</math> be a function such that <math>f(0) = 0</math>, <math>f</math> is invertible at <math>0</math> and <math>f'(0) = \text{Id}</math>. We define
:<math>g(x) := f(x) - x</math>.
The differential of this function is zero (since taking the differential is linear and the differential of the function <math>x \mapsto x</math> is the identity). Since the function <math>g</math> is also continuously differentiable at a small neighbourhood of <math>0</math>, we find <math>\delta > 0</math> such that
:<math>\frac{\partial g}{\partial x_j}(x) < \frac{1}{2n^2}</math>
for all <math>j \in \{1, \ldots, n\}</math> and <math>x \in B_\delta(0)</math>. Since further <math>g(0) = f(0) - 0 = 0</math>, the general mean-value theorem and Cauchy's inequality imply that for <math>k \in \{1, \ldots, n\}</math> and <math>x \in B_\delta(0)</math>,
:<math>|g_k(x)| = |\langle x, \frac{\partial g}{\partial x_j}(t_k x) \rangle| \le \|x\| n \frac{1}{2n^2}</math>
for suitable <math>t_k \in [0, 1]</math>. Hence,
:<math>\|g(x)\| \le |g_1(x)| + \cdots + |g_n(x)| \le \frac{1}{2} \|x\|</math> (triangle inequality),
and thus, we obtain that our preparatory lemma is applicable, and <math>f</math> is a bijection on <math>\overline{B_\delta(0)}</math>, whose image is contained within the open set <math>\overline{B_{\delta/2}(0)}</math>; thus we may pick <math>U := f^{-1}(B_{\delta/2}(0))</math>, which is open due to the continuity of <math>f</math>.

Thus, the most important part of the theorem is already done. All that is left to do is to prove differentiability of <math>f^{-1}</math> at <math>0</math>. Now we even prove the slightly stronger claim that the differential of <math>f^{-1}</math> at <math>x_0</math> 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 <math>g</math> implies the following bounds on <math>f</math>:
:<math>\frac{1}{2}\|x\| \le \|f(x)\| \le \frac{3}{2} \|x\|</math>.
The second bound follows from
:<math>\|f(x)\| \le \|f(x) - x\| + \|x\| = \|g(x)\| + \|x\| \le \frac{3}{2} \|x\|</math>,
and the first bound follows from
:<math>\|f(x)\| \ge |\|f(x) - x\| - \|x\|| = \left| \|g(x)\| - \|x\| \right| \ge \frac{1}{2} \|x\|</math>.

Now for the differentiability at <math>0</math>. We have, by substitution of limits (as <math>f</math> is continuous and <math>f(0) = 0</math>):
:<math>\begin{align}
\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{align}</math>
where the last expression converges to zero due to the differentiability of <math>f</math> at <math>0</math> with differential the identity, and the sandwhich criterion applied to the expressions
:<math>\frac{\|\mathbf h - f(\mathbf h)\|}{\frac{3}{2} \|\mathbf h\|}</math>
and
:<math>\frac{\|\mathbf h - f(\mathbf h)\|}{\frac{1}{2} \|\mathbf h\|}</math>. <math>\blacksquare </math>

== The implicit function theorem ==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
'''Theorem''':

:Let <math>f: \mathbb R^n \to \mathbb R</math> be a continuously differentiable function, and consider the set
::<math>S := \{(x_1, \ldots, x_n) \in \mathbb R^n | f(x_1, \ldots, x_n) = 0\}</math>.
:If we are given some <math>y \in S</math> such that <math>\partial_n f(y) \neq 0</math>, then we find <math>U \subseteq \mathbb R^{n-1}</math> open with <math>(y_1, \ldots, y_{n-1}) \in U</math> and <math>g: U \to S</math> such that
::<math>y = g(y_1, \ldots, y_{n-1})</math> and <math>\{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\} \subseteq S</math>,
:where <math>\{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\}</math> is open with respect to the subspace topology of <math>U</math>.

:Furthermore, <math>g</math> is a differentiable function.
</blockquote>

'''Proof''':

We define a new function
:<math>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))</math>.
The differential of this function looks like this:
:<math>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}</math>
Since we assumed that <math>\partial_n f(y) \neq 0</math>, <math>F'(y)</math> is invertible, and hence the inverse function theorem implies the existence of a small open neighbourhood <math>\tilde V \subseteq \mathbb R^n</math> containing <math>y</math> such that ''restricted to that neighbourhood'' <math>F</math> is itself invertible, with a differentiable inverse <math>F^{-1}</math>, which is itself defined on an open set <math>\tilde U</math> containing <math>F(y)</math>. Now set first
:<math>U := \{(x_1, \ldots, x_{n-1}) | (x_1, \ldots, x_{n-1}, 0) \in \tilde U \}</math>,
which is open with respect to the subspace topology of <math>\mathbb R^{n-1}</math>, and then
:<math>g: U \to \mathbb R, g(x_1, \ldots, x_{n-1}) := \pi_n(F^{-1}(x_1, \ldots, x_{n-1}, 0))</math>,
the <math>n</math>-th component of <math>F^{-1}(x_1, \ldots, x_{n-1}, 0)</math>. We claim that <math>g</math> has the desired properties.

Indeed, we first note that <math>F^{-1}(x_1, \ldots, x_{n-1}, 0) = (x_1, \ldots, x_{n-1}, g(x_1, \ldots, x_{n-1}))</math>, since applying <math>F</math> leaves the first <math>n-1</math> components unchanged, and thus we get the identity by observing <math>F(F^{-1}(x)) = x</math>. Let thus <math>(z_1, \ldots, z_{n-1}) \in U</math>. Then
:<math>\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}</math>.
Furthermore, the set
:<math>\{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\}</math>
is open with respect to the subspace topology on <math>S</math>. Indeed, we show
:<math>\{(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</math>.
For <math>\subseteq</math>, we first note that the set on the left hand side is in <math>S</math>, since all points in it are mapped to zero by <math>f</math>. Further,
:<math>F(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) = (z_1, \ldots, z_{n-1}, 0) \in \tilde U</math>
and hence <math>\subseteq</math> is completed when applying <math>F^{-1}</math>. For the other direction, let a point <math>(x_1, \ldots, x_n)</math> in <math>S \cap \tilde V</math> be given, apply <math>F</math> to get
:<math>F((x_1, \ldots, x_n)) = (x_1, \ldots, x_{n-1}, 0) \in \tilde U</math>
and hence <math>(x_1, \ldots, x_{n-1}) \in U</math>; further
:<math>(x_1, \ldots, x_{n-1}, g(x_1, \ldots, x_{n-1})) = (x_1, \ldots, x_n)</math>
by applying <math>F</math> to both sides of the equation.

Now <math>g</math> is automatically differentiable as the component of a differentiable function. <math>\blacksquare </math>


Informally, the above theorem states that given a set <math>\{x \in \mathbb R^n | f(x) = 0\}</math>, one can choose the first <math>n-1</math> coordinates as a "base" for a function, whose graph is precisely a local bit of that set.

== Licensing == 
Content obtained and/or adapted from:
* [https://en.wikibooks.org/wiki/Calculus/Inverse_function_theorem,_implicit_function_theorem Inverse function theorem, implicit function theorem, Wikibooks: Calculus] under a CC BY-SA license