Inverse function theorem

To prove existence, it can be assumed after an affine transformation that ${\displaystyle f(0)=0}$ and ${\displaystyle f^{\prime }(0)=I}$, so that ${\displaystyle a=b=0}$.

By the mean value theorem for vector-valued functions, for a function ${\displaystyle u:[0,1]\to \mathbb {R} ^{m}}$, ${\textstyle \|u(1)-u(0)\|\leq \sup _{0\leq t\leq 1}\|u^{\prime }(t)\|}$. Setting ${\displaystyle u(t)=f(x+t(x^{\prime }-x))-x-t(x^{\prime }-x)}$, it follows that

${\displaystyle \|f(x)-f(x^{\prime })-x+x^{\prime }\|\leq \|x-x^{\prime }\|\,\sup _{0\leq t\leq 1}\|f^{\prime }(x+t(x^{\prime }-x))-I\|.}$

Now choose ${\displaystyle \delta >0}$ so that ${\textstyle \|f'(x)-I\|<{1 \over 2}}$ for ${\displaystyle \|x\|<\delta }$. Suppose that ${\displaystyle \|y\|<\delta /2}$ and define ${\displaystyle x_{n}}$ inductively by ${\displaystyle x_{0}=0}$ and ${\displaystyle x_{n+1}=x_{n}+y-f(x_{n})}$. The assumptions show that if ${\displaystyle \|x\|,\,\,\|x^{\prime }\|<\delta }$ then

${\displaystyle \|f(x)-f(x^{\prime })-x+x^{\prime }\|\leq \|x-x^{\prime }\|/2}$.

In particular ${\displaystyle f(x)=f(x^{\prime })}$ implies ${\displaystyle x=x^{\prime }}$. In the inductive scheme ${\displaystyle \|x_{n}\|<\delta }$ and ${\displaystyle \|x_{n+1}-x_{n}\|<\delta /2^{n}}$. Thus ${\displaystyle (x_{n})}$ is a Cauchy sequence tending to ${\displaystyle x}$. By construction ${\displaystyle f(x)=y}$ as required.

To check that ${\displaystyle g=f^{-1}}$ is C¹, write ${\displaystyle g(y+k)=x+h}$ so that ${\displaystyle f(x+h)=f(x)+k}$. By the inequalities above, ${\displaystyle \|h-k\|<\|h\|/2}$ so that ${\displaystyle \|h\|/2<\|k\|<2\|h\|}$. On the other hand if ${\displaystyle A=f^{\prime }(x)}$, then ${\displaystyle \|A-I\|<1/2}$. Using the geometric series for ${\displaystyle B=I-A}$, it follows that ${\displaystyle \|A^{-1}\|<2}$. But then

${\displaystyle {\|g(y+k)-g(y)-f^{\prime }(g(y))^{-1}k\| \over \|k\|}={\|h-f^{\prime }(x)^{-1}[f(x+h)-f(x)]\| \over \|k\|}\leq 4{\|f(x+h)-f(x)-f^{\prime }(x)h\| \over \|h\|}}$

tends to 0 as ${\displaystyle k}$ and ${\displaystyle h}$ tend to 0, proving that ${\displaystyle g}$ is C¹ with ${\displaystyle g^{\prime }(y)=f^{\prime }(g(y))^{-1}}$.

The proof above is presented for a finite-dimensional space, but applies equally well for Banach spaces. If an invertible function ${\displaystyle f}$ is C^k with ${\displaystyle k>1}$, then so too is its inverse. This follows by induction using the fact that the map ${\displaystyle F(A)=A^{-1}}$ on operators is C^k for any ${\displaystyle k}$ (in the finite-dimensional case this is an elementary fact because the inverse of a matrix is given as the adjugate matrix divided by its determinant). [1][7] The method of proof here can be found in the books of Henri Cartan, Jean Dieudonné, Serge Lang, Roger Godement and Lars Hörmander.

Here is a proof based on the contraction mapping theorem. Specifically, following T. Tao,[8] it uses the following consequence of the contraction mapping theorem.

Basically, the lemma says that a small perturbation of the identity map by a contraction map is injective and preserves a ball in some sense. Assuming the lemma for a moment, we prove the theorem first. As in the above proof, it is enough to prove the special case when ${\displaystyle a=0,b=f(a)=0}$ and ${\displaystyle f'(0)=I}$. Let ${\displaystyle g=f-I}$. The mean value inequality applied to ${\displaystyle t\mapsto g(x+t(y-x))}$ says:

${\displaystyle |g(y)-g(x)|\leq |y-x|\sup _{0<t<1}|g'(x+t(y-x))|.}$

Since ${\displaystyle g'(0)=I-I=0}$ and ${\displaystyle g'}$ is continuous, we can find an ${\displaystyle r>0}$ such that

${\displaystyle |g(y)-g(x)|\leq 2^{-1}|y-x|}$

for all ${\displaystyle x,y}$ in ${\displaystyle B(0,r)}$. Then the early lemma says that ${\displaystyle f=g+I}$ is injective on ${\displaystyle B(0,r)}$ and ${\displaystyle B(0,r/2)\subset f(B(0,r))}$. Then

${\displaystyle f:U=B(0,r)\cap f^{-1}(B(0,r/2))\to V=B(0,r/2)}$

is bijective and thus has an inverse. Next, we show the inverse ${\displaystyle f^{-1}}$ is continuously differentiable (this part of the argument is the same as that in the previous proof). This time, let ${\displaystyle g=f^{-1}}$ denote the inverse of ${\displaystyle f}$ and ${\displaystyle A=f'(x)}$. For ${\displaystyle x=g(y)}$, we write ${\displaystyle g(y+k)=x+h}$ or ${\displaystyle y+k=f(x+h)}$. Now, by the early estimate, we have

${\displaystyle |h-k|=|f(x+h)-f(x)-h|\leq |h|/2}$

and so ${\displaystyle |h|/2\leq |k|}$. Writing ${\displaystyle \|\cdot \|}$ for the operator norm,

${\displaystyle |g(y+k)-g(y)-A^{-1}k|=|h-A^{-1}(f(x+h)-f(x))|\leq \|A^{-1}\||Ah-f(x+h)+f(x)|.}$

As ${\displaystyle k\to 0}$, we have ${\displaystyle h\to 0}$ and ${\displaystyle |h|/|k|}$ is bounded. Hence, ${\displaystyle g}$ is differentiable at ${\displaystyle y}$ with the derivative ${\displaystyle g'(y)=f'(g(y))^{-1}}$. Also, ${\displaystyle g'}$ is the same as the composition ${\displaystyle \iota \circ f'\circ g}$ where ${\displaystyle \iota :T\mapsto T^{-1}}$; so ${\displaystyle g'}$ is continuous.

It remains to show the lemma. First, the map ${\displaystyle f}$ is injective on ${\displaystyle B(0,r)}$ since if ${\displaystyle f(x)=f(y)}$, then ${\displaystyle g(y)-g(x)=x-y}$ and so

${\displaystyle |g(y)-g(x)|=|y-x|}$,

which is a contradiction unless ${\displaystyle y=x}$. (This part does not need the assumption ${\displaystyle g(0)=0}$.) Next we show ${\displaystyle f(B(0,r))\supset B(0,(1-c)r)}$. The idea is to note that this is equivalent to, given a point ${\displaystyle y}$ in ${\displaystyle B(0,(1-c)r)}$, find a fixed point of the map

${\displaystyle F:{\overline {B}}(0,r')\to {\overline {B}}(0,r'),\,x\mapsto y-g(x)}$

where ${\displaystyle 0<r'<r}$ such that ${\displaystyle |y|\leq (1-c)r'}$ and the bar means a closed ball. To find a fixed point, we use the contraction mapping theorem and checking that ${\displaystyle F}$ is a well-defined strict-contraction mapping is straightforward. Finally, we have: ${\displaystyle f(B(0,r))\subset B(0,(1+c)r)}$ since

${\displaystyle |f(x)|=|x+g(x)-g(0)|\leq (1+c)|x|.\square }$

As might be clear, this proof is not substantially different from the previous one, as the proof of the contraction mapping theorem is by successive approximation.

The inverse function theorem can be used to solve a system of equations

${\displaystyle {\begin{aligned}&f_{1}(x)=y_{1}\\&\quad \vdots \\&f_{n}(x)=y_{n},\end{aligned}}}$

i.e., expressing ${\displaystyle y_{1},\dots ,y_{n}}$ as functions of ${\displaystyle x=(x_{1},\dots ,x_{n})}$, provided the Jacobian matrix is invertible. The implicit function theorem allows to solve a more general system of equations:

${\displaystyle {\begin{aligned}&f_{1}(x,y)=0\\&\quad \vdots \\&f_{n}(x,y)=0\end{aligned}}}$

for ${\displaystyle y}$ in terms of ${\displaystyle x}$. Though more general, the theorem is actually a consequence of the inverse function theorem. First, the precise statement of the implicit function theorem is as follows:[9]

• given a map ${\displaystyle f:\mathbb {R} ^{n}\times \mathbb {R} ^{m}\to \mathbb {R} ^{m}}$, if ${\displaystyle f(a,b)=0}$, ${\displaystyle f}$ is continuously differentiable in a neighborhood of ${\displaystyle (a,b)}$ and the derivative of ${\displaystyle y\mapsto f(a,y)}$ at ${\displaystyle b}$ is invertible, then there exists a differentiable map ${\displaystyle g:U\to V}$ for some neighborhoods ${\displaystyle U,V}$ of ${\displaystyle a,b}$ such that ${\displaystyle f(x,g(x))=0}$. Moreover, if ${\displaystyle f(x,y)=0,x\in U,y\in V}$, then ${\displaystyle y=g(x)}$; i.e., ${\displaystyle g(x)}$ is a unique solution.

To see this, consider the map ${\displaystyle F(x,y)=(x,f(x,y))}$. By the inverse function theorem, ${\displaystyle F:U\times V\to W}$ has the inverse ${\displaystyle G}$ for some neighborhoods ${\displaystyle U,V,W}$. We then have:

${\displaystyle (x,y)=F(G_{1}(x,y),G_{2}(x,y))=(G_{1}(x,y),f(G_{1}(x,y),G_{2}(x,y)),}$

implying ${\displaystyle x=G_{1}(x,y)}$ and ${\displaystyle y=f(x,G_{2}(x,y)).}$ Thus ${\displaystyle g(x)=G_{2}(x,0)}$ has the required property. ${\displaystyle \square }$

In differential geometry, the inverse function theorem is used to show that the pre-image of a regular value under a smooth map is a manifold.[10] Indeed, let ${\displaystyle f:U\to \mathbb {R} ^{r}}$ be such a smooth map from an open subset of ${\displaystyle \mathbb {R} ^{n}}$ (since the result is local, there is no loss of generality with considering such a map). Fix a point ${\displaystyle a}$ in ${\displaystyle f^{-1}(b)}$ and then, by permuting the coordinates on ${\displaystyle \mathbb {R} ^{n}}$, assume the matrix ${\displaystyle \left[{\frac {\partial f_{i}}{\partial x_{j}}}(a)\right]_{1\leq i,j\leq r}}$ has rank ${\displaystyle r}$. Then the map ${\displaystyle F:U\to \mathbb {R} ^{r}\times \mathbb {R} ^{n-r}=\mathbb {R} ^{n},\,x\mapsto (f(x),x_{r+1},\dots ,x_{n})}$ is such that ${\displaystyle F'(a)}$ has rank ${\displaystyle n}$. Hence, by the inverse function theorem, we find the smooth inverse ${\displaystyle G}$ of ${\displaystyle F}$ defined in a neighborhood ${\displaystyle V\times W}$ of ${\displaystyle (b,a_{r+1},\dots ,a_{n})}$. We then have

${\displaystyle x=(F\circ G)(x)=(f(G(x)),G_{r+1}(x),\dots ,G_{n}(x)),}$

which implies

${\displaystyle (f\circ G)(x_{1},\dots ,x_{n})=(x_{1},\dots ,x_{r}).}$

That is, after the change of coordinates by ${\displaystyle G}$, ${\displaystyle f}$ is a coordinate projection (this fact is known as the submersion theorem). Moreover, since ${\displaystyle G:V\times W\to U'=G(V\times W)}$ is bijective, the map

${\displaystyle g=G(b,\cdot ):W\to f^{-1}(b)\cap U',\,(x_{r+1},\dots ,x_{n})\mapsto G(b,x_{r+1},\dots ,x_{n})}$

is bijective with the smooth inverse. That is to say, ${\displaystyle g}$ gives a local parametrization of ${\displaystyle f^{-1}(b)}$ around ${\displaystyle a}$. Hence, ${\displaystyle f^{-1}(b)}$ is a manifold. ${\displaystyle \square }$ (Note the proof is quite similar to the proof of the implicit function theorem and, in fact, the implicit function theorem can be also used instead.)

More generally, the theorem shows that if a smooth map ${\displaystyle f:P\to E}$ is transversal to a submanifold ${\displaystyle M\subset E}$, then the pre-image ${\displaystyle f^{-1}(M)\hookrightarrow P}$ is a submanifold.[11]

The inverse function theorem can also be generalized to differentiable maps between Banach spaces X and Y.[20] Let U be an open neighbourhood of the origin in X and ${\displaystyle F:U\to Y\!}$ a continuously differentiable function, and assume that the Fréchet derivative ${\displaystyle dF_{0}:X\to Y\!}$ of F at 0 is a bounded linear isomorphism of X onto Y. Then there exists an open neighbourhood V of ${\displaystyle F(0)\!}$ in Y and a continuously differentiable map ${\displaystyle G:V\to X\!}$ such that ${\displaystyle F(G(y))=y}$ for all y in V. Moreover, ${\displaystyle G(y)\!}$ is the only sufficiently small solution x of the equation ${\displaystyle F(x)=y\!}$.

There is also the inverse function theorem for Banach manifolds.[21]

The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with constant rank near a point can be put in a particular normal form near that point.[22] Specifically, if ${\displaystyle F:M\to N}$ has constant rank near a point ${\displaystyle p\in M\!}$, then there are open neighborhoods U of p and V of ${\displaystyle F(p)\!}$ and there are diffeomorphisms ${\displaystyle u:T_{p}M\to U\!}$ and ${\displaystyle v:T_{F(p)}N\to V\!}$ such that ${\displaystyle F(U)\subseteq V\!}$ and such that the derivative ${\displaystyle dF_{p}:T_{p}M\to T_{F(p)}N\!}$ is equal to ${\displaystyle v^{-1}\circ F\circ u\!}$. That is, F "looks like" its derivative near p. The set of points ${\displaystyle p\in M}$ such that the rank is constant in a neighborhood of ${\displaystyle p}$ is an open dense subset of M; this is a consequence of semicontinuity of the rank function. Thus the constant rank theorem applies to a generic point of the domain.

When the derivative of F is injective (resp. surjective) at a point p, it is also injective (resp. surjective) in a neighborhood of p, and hence the rank of F is constant on that neighborhood, and the constant rank theorem applies.

If it is true, the Jacobian conjecture would be a variant of the inverse function theorem for polynomials. It states that if a vector-valued polynomial function has a Jacobian determinant that is an invertible polynomial (that is a nonzero constant), then it has an inverse that is also a polynomial function. It is unknown whether this is true or false, even in the case of two variables. This is a major open problem in the theory of polynomials.

When ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} ^{m}}$ with ${\displaystyle m\leq n}$, ${\displaystyle f}$ is ${\displaystyle k}$ times continuously differentiable, and the Jacobian ${\displaystyle A=\nabla f({\overline {x}})}$ at a point ${\displaystyle {\overline {x}}}$ is of rank ${\displaystyle m}$, the inverse of ${\displaystyle f}$ may not be unique. However, there exists a local selection function ${\displaystyle s}$ such that ${\displaystyle f(s(y))=y}$ for all ${\displaystyle y}$ in a neighborhood of ${\displaystyle {\overline {y}}=f({\overline {x}})}$, ${\displaystyle s({\overline {y}})={\overline {x}}}$, ${\displaystyle s}$ is ${\displaystyle k}$ times continuously differentiable in this neighborhood, and ${\displaystyle \nabla s({\overline {y}})=A^{T}(AA^{T})^{-1}}$ (${\displaystyle \nabla s({\overline {y}})}$ is the Moore–Penrose pseudoinverse of ${\displaystyle A}$).[23]