The Unapologetic Mathematician

The Inverse Function Theorem

At last we come to the theorem that I promised. Let $f:S\rightarrow\mathbb{R}^n$ be continuously differentiable on an open region $S\subseteq\mathbb{R}^n$, and $T=f(S)$. If the Jacobian determinant $J_f(a)\neq0$ at some point $a\in S$, then there is a uniquely determined function $g$ and two open sets $X\subseteq S$ and $Y\subseteq T$ so that

• $a\in X$, and $f(a)\in Y$
• $Y=f(X)$
• $f$ is injective on $X$
• $g$ is defined on $Y$, $g(Y)=X$, and $g(f(x))=x$ for all $x\in X$
• $g$ is continuously differentiable on $Y$

The Jacobian determinant $J_f(x)$ is continuous as a function of $x$, so there is some neighborhood $N_1$ of $a$ so that the Jacobian is nonzero within $N_1$. Our second lemma tells us that there is a smaller neighborhood $N\subseteq N_1$ on which $f$ is injective. We pick some closed ball $\overline{K}\subseteq N$ centered at $a$, and use our first lemma to find that $f(K)$ must contain an open neighborhood $Y$ of $f(a)$. Then we define $X=f^{-1}(Y)\cap K$, which is open since both $K$ and $f^{-1}(Y)$ are (the latter by the continuity of $f$). Since $f$ is injective on the compact set $\overline{K}\subseteq N$, it has a uniquely-defined continuous inverse $g$ on $Y\subseteq f(\overline{K})$. This establishes the first four of the conditions of the theorem.

Now the hard part is showing that $g$ is continuously differentiable on $Y$. To this end, like we did in our second lemma, we define the function

$\displaystyle h(z_1,\dots,z_n)=\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{x=z_i}\right)$

along with a neighborhood $N_2$ of $a$ so that as long as all the $z_i$ are within $N_2$ this function is nonzero. Without loss of generality we can go back and choose our earlier neighborhood $N$ so that $N\subseteq N_2$, and thus that $\overline{K}\subseteq N_2$.

To show that the partial derivative $\frac{\partial g^i}{\partial y^j}$ exists at a point $y\in Y$, we consider the difference quotient

$\displaystyle\frac{g^i(y+\lambda e_j)-g^i(y)}{\lambda}$

with $y+\lambda e_j$ also in $Y$ for sufficiently small $\lvert\lambda\rvert$. Then writing $x_1=g(y)$ and $x_2=g(y+\lambda e_j)$ we find $f(x_2)-f(x_1)=\lambda e_j$. The mean value theorem then tells us that

\displaystyle\begin{aligned}\delta_j^k&=\frac{f^k(x_2)-f^k(x_1)}{\lambda}\\&=df^k(\xi_k)\left(\frac{1}{\lambda}(x_2-x_1)\right)\\&=\frac{\partial f^k}{\partial x^i}\bigg\vert_{x=\xi_k}\frac{x_2^i-x_1^i}{\lambda}\\&=\frac{\partial f^k}{\partial x^i}\bigg\vert_{x=\xi_k}\frac{g^i(y+\lambda e_j)-g^i(y)}{\lambda}\end{aligned}

for some $\xi_k\in[x_1,x_2]\subseteq K$ (no summation on $k$). As usual, $\delta_j^k$ is the Kronecker delta.

This is a linear system of equations, which has a unique solution since the determinant of its matrix is $h(\xi_1,\dots,\xi_n)\neq0$. We use Cramer’s rule to solve it, and get an expression for our difference quotient as a quotient of two determinants. This is why we want the form of the solution given by Cramer’s rule, and not by a more computationally-efficient method like Gaussian elimination.

As $\lambda$ approaches zero, continuity of $g$ tells us that $x_2$ approaches $x_1$, and thus so do all of the $\xi_k$. Therefore the determinant in the denominator of Cramer’s rule is in the limit $h(x,\dots,x)=J_f(x)\neq0$, and thus limits of the solutions given by Cramer’s rule actually do exist.

This establishes that the partial derivative $\frac{\partial g^i}{\partial y^j}$ exists at each $y\in Y$. Further, since we found the limit of the difference quotient by Cramer’s rule, we have an expression given by the quotient of two determinants, each of which only involves the partial derivatives of $f$, which are themselves all continuous. Therefore the partial derivatives of $g$ not only exist but are in fact continuous.

November 18, 2009 - Posted by | Analysis, Calculus

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