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

Theorem:

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

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

Proof:

We first reduce to the case $f(x_{0})=0$, $x_{0}=0$ and $f'(x_{0})={\text{Id}}$. 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 $x_{0}$). We set

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

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-affine approximation by $h'(x_{0})^{-1}$ 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

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

it can be seen that $h^{-1}$ is an inverse of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} with all the required properties (which is a bit of a tedious exercise, but involves nothing more than the definitions).

Thus let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} be a function such that $f(0)=0$, $f$ is invertible at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(0) = \text{Id}} . We define

$g(x):=f(x)-x$.

The differential of this function is zero (since taking the differential is linear and the differential of the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \mapsto x} is the identity). Since the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is also continuously differentiable at a small neighbourhood of $0$, we find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta > 0} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial g}{\partial x_j}(x) < \frac{1}{2n^2}}

for all $j\in \{1,\ldots ,n\}$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in B_\delta(0)} . Since further Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(0) = f(0) - 0 = 0} , the general mean-value theorem and Cauchy's inequality imply that for $k\in \{1,\ldots ,n\}$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in B_\delta(0)} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |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,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \|g(x)\| \le |g_1(x)| + \cdots + |g_n(x)| \le \frac{1}{2} \|x\|} (triangle inequality),

and thus, we obtain that our preparatory lemma is applicable, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a bijection on ${\overline {B_{\delta }(0)}}$, whose image is contained within the open set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{B_{\delta/2}(0)}} ; thus we may pick Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U := 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{-1}} at $0$. Now we even prove the slightly stronger claim that the differential of $f^{-1}$ at $x_{0}$ 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 $f$:

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

The second bound follows from

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

and the first bound follows from

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

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

${\begin{aligned}\lim _{\mathbf {h} \to 0}{\frac {\|f^{-1}(\mathbf {h} )-f^{-1}(0)-\operatorname {Id} (\mathbf {h} -0)\|}{\|\mathbf {h} \|}}&=\lim _{\mathbf {h} \to 0}{\frac {\|f^{-1}(f(\mathbf {h} ))-f(\mathbf {h} )\|}{\|f(\mathbf {h} )\|}}\\&=\lim _{\mathbf {h} \to 0}{\frac {\|\mathbf {h} -f(\mathbf {h} )\|}{\|f(\mathbf {h} )\|}},\end{aligned}}$

where the last expression converges to zero due to the differentiability of $f$ at $0$ with differential the identity, and the sandwhich criterion applied to the expressions

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

and

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

The implicit function theorem

Theorem:

Let $f:\mathbb {R} ^{n}\to \mathbb {R}$ be a continuously differentiable function, and consider the set

$S:=\{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}|f(x_{1},\ldots ,x_{n})=0\}$.

If we are given some $y\in S$ such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial_n f(y) \neq 0} , then we find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U \subseteq \mathbb R^{n-1}} open with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (y_1, \ldots, y_{n-1}) \in U} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g: U \to S} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = g(y_1, \ldots, y_{n-1})} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\} \subseteq S} ,

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{(z_1, \ldots, z_{n-1}, g(z_1, \ldots, z_{n-1})) | (z_1, \ldots, z_{n-1}) \in U\}} is open with respect to the subspace topology of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U} .

Furthermore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is a differentiable function.

Proof:

We define a new function

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Since we assumed that $\partial _{n}f(y)\neq 0$, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(y)} is invertible, and hence the inverse function theorem implies the existence of a small open neighbourhood Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde V \subseteq \mathbb R^n} containing $y$ such that restricted to that neighbourhood Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} is itself invertible, with a differentiable inverse Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F^{-1}} , which is itself defined on an open set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde U} containing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(y)} . Now set first

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U := \{(x_1, \ldots, x_{n-1}) | (x_1, \ldots, x_{n-1}, 0) \in \tilde U \}} ,

which is open with respect to the subspace topology of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb R^{n-1}} , 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} -th component of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F^{-1}(x_1, \ldots, x_{n-1}, 0)} . We claim that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} has the desired properties.

Indeed, we first note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F^{-1}(x_1, \ldots, x_{n-1}, 0) = (x_1, \ldots, x_{n-1}, g(x_1, \ldots, x_{n-1}))} , since applying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} leaves the first Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n-1} components unchanged, and thus we get the identity by observing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(F^{-1}(x)) = x} . Let thus $(z_{1},\ldots ,z_{n-1})\in U$. Then

${\begin{aligned}f(z_{1},\ldots ,z_{n-1},g(z_{1},\ldots ,z_{n-1}))&=(\pi _{n}\circ F)(F^{-1}(z_{1},\ldots ,z_{n-1},0))\\&=\pi _{n}((F\circ F^{-1})(z_{1},\ldots ,z_{n-1},0))=0\end{aligned}}$.

Furthermore, the set

$\{(z_{1},\ldots ,z_{n-1},g(z_{1},\ldots ,z_{n-1}))|(z_{1},\ldots ,z_{n-1})\in U\}$

is open with respect to the subspace topology on $S$. Indeed, we show

$\{(z_{1},\ldots ,z_{n-1},g(z_{1},\ldots ,z_{n-1}))|(z_{1},\ldots ,z_{n-1})\in U\}=S\cap {\tilde {V}}$.

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 as the component of a differentiable function. $\blacksquare$

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

Licensing

Content obtained and/or adapted from:

• Inverse function theorem, implicit function theorem, Wikibooks: Calculus under a CC BY-SA license