LEC CAL1 22

Implicit function theorem

Notation

If $\mathbf{x}\in {\mathbb{R}}^n$ and $\mathbf{y}\in {\mathbb{R}}^m$,

$$(\mathbf{x},\mathbf{y})\equiv (x_1,\dots ,x_n,y_1,\dots ,y_m)\in {\mathbb{R}}^{n+m}.$$

Secondly, every $A\in \mathcal{L}({\mathbb{R}}^{n+m},{\mathbb{R}}^n)$ can be split into two linear transformations $A_x$ and $A_y$, defined by

$$A_x\mathbf{h}\equiv A(\mathbf{h},0),A_y\mathbf{k}\equiv A(0,\mathbf{k})$$

for any $\mathbf{h}\in {\mathbb{R}}^n$, $\mathbf{k}\in {\mathbb{R}}^m$. Then $A_x\in \mathcal{L}({\mathbb{R}}^n,{\mathbb{R}}^n)$, $A_y\in \mathcal{L}({\mathbb{R}}^m,{\mathbb{R}}^n)$, and

$$A(\mathbf{h},\mathbf{k})=A_x\mathbf{h}+A_y\mathbf{k}.$$

Consider a system of $n$ equations in $n+k$ variables $\mathbf{f}(\mathbf{x},\mathbf{t})=0$, where $\mathbf{f}:{\mathbb{R}}^{n+k}\to {\mathbb{R}}^n$, $\mathbf{x}\in {\mathbb{R}}^n$, and $\mathbf{t}\in {\mathbb{R}}^k$. If $\mathbf{f}$ is the map

$$(\mathbf{x},\mathbf{t})\mapsto \left[\begin{array}{c}{f}_1(\mathbf{x},\mathbf{t})\\ ⋮\\ f_n(\mathbf{x},\mathbf{t})\end{array}\right],$$

then $\mathbf{f}(\mathbf{x},\mathbf{t})=0$ represents the system of equations

$$\begin{array}{c}{f}_1(\mathbf{x},\mathbf{t})=0,\\ ⋮\\ f_n(\mathbf{x},\mathbf{t})=0.\end{array}$$

Clearly, every system of $n$ equations in $n+k$ variables can be expressed in this manner.

Note that every such system of equations always represents a relation on ${\mathbb{R}}^n\times {\mathbb{R}}^k$, namely, the tuples $(\mathbf{x},\mathbf{t})$ which satisfy the system. The implicit function theorem tells us when such a relation is (locally) a function from ${\mathbb{R}}^k$ to ${\mathbb{R}}^n$, that is, when $\mathbf{x}$ can be determined uniquely as a function $\mathbf{\phi }$ of $\mathbf{t}$, in which case $\mathbf{f}$ is said to implicitly define $\mathbf{\phi }$.

The linear version of the implicit function theorem is as follows:

Theorem 191.1(Theorem).

If $A\in \mathcal{L}({\mathbb{R}}^{n+m},{\mathbb{R}}^n)$ and if $A_x$ is invertible, then there corresponds to every $\mathbf{k}\in {\mathbb{R}}^m$ a unique $\mathbf{h}\in {\mathbb{R}}^n$ such that $A(\mathbf{h},\mathbf{k})=0$. This $\mathbf{h}$ can be computed from $\mathbf{k}$ by the formula

$$\mathbf{h}=-(A_x{)}^{-1}{A}_y\mathbf{k}.$$

Proof
$A(\mathbf{h},\mathbf{k})=0$ iff $A_x\mathbf{h}+A_y\mathbf{k}=0$. Given $\mathbf{k}\in {\mathbb{R}}^m$, this can be solved uniquely for $\mathbf{h}$ iff $A_x$ is invertible, in which case $\mathbf{h}=-(A_x{)}^{-1}{A}_y\mathbf{k}$.

Theorem 191.2(Theorem).

Let $\mathbf{f}$ be a $C^1$ mapping of an open set $E\subseteq {\mathbb{R}}^{n+m}$ into ${\mathbb{R}}^n$. Let $(\mathbf{a},\mathbf{b})\in E$. Define $A\equiv {\mathbf{f}}^{\prime }(\mathbf{a},\mathbf{b})$. Assume:

• $\mathbf{f}(\mathbf{a},\mathbf{b})=0$;
• $A_x$ is invertible.

Then, there exist open sets $U\subseteq {\mathbb{R}}^{n+m}$ and $W\subseteq {\mathbb{R}}^m$, with $(\mathbf{a},\mathbf{b})\in U$ and $\mathbf{b}\in W$, having the property:

To every $\mathbf{y}\in W$ corresponds a unique $\mathbf{x}$ such that

$$(\mathbf{x},\mathbf{y})\in U\text{ and }\mathbf{f}(\mathbf{x},\mathbf{y})=0.$$

If this $\mathbf{x}$ is defined to be $\mathbf{g}(\mathbf{y})$, then

• $\mathbf{g}$ is a $C^1$ mapping of $W$ into ${\mathbb{R}}^n$;
• $\mathbf{g}(\mathbf{b})=\mathbf{a}$;
• $\mathbf{f}(\mathbf{g}(\mathbf{y}),\mathbf{y})=0$ for all $\mathbf{y}\in W$;
• ${\mathbf{g}}^{\prime }(\mathbf{b})=-(A_x{)}^{-1}{A}_y$.

Proof
Define $\mathbf{F}:E\to {\mathbb{R}}^{n+m}$ by

$$\mathbf{F}(\mathbf{x},\mathbf{y})=(\mathbf{f}(\mathbf{x},\mathbf{y}),\mathbf{y})((\mathbf{x},\mathbf{y})\in E).$$

Its derivative is given by the block matrix

$${\mathbf{F}}^{\prime }(\mathbf{x},\mathbf{y})=\left[\begin{array}{c}\underset{n\times (m+n)}{{\mathbf{f}}^{\prime }(\mathbf{x},\mathbf{y})}\\ \underset{m\times n}{0}\underset{m\times m}{I}\end{array}\right].$$

All the partial derivatives are clearly continuous. By Theorem 188.3, $\mathbf{F}$ is a $C^1$ mapping of $E$ into ${\mathbb{R}}^{n+m}$. Also, since $\det {\mathbf{F}}^{\prime }(\mathbf{a},\mathbf{b})=\det A_x\det I_{m\times m}=\det A_x\ne 0$, ${\mathbf{F}}^{\prime }(\mathbf{a},\mathbf{b})$ is invertible. It follows from the inverse function theorem that there exist open sets $U$ and $V$ in ${\mathbb{R}}^{n+m}$, with $(\mathbf{a},\mathbf{b})\in U$, $(0,\mathbf{b})\in V$, such that $\mathbf{F}$ is an injective mapping of $U$ onto $V$.

Let $W$ be the set of all $\mathbf{y}\in {\mathbb{R}}^m$ such that $(0,\mathbf{y})\in V$. Note that $\mathbf{b}\in W$. It is clear that $W$ is open since $V$ is open. If $\mathbf{y}\in W$, then $(0,\mathbf{y})=\mathbf{F}(\mathbf{x},\mathbf{y})$ for some $(\mathbf{x},\mathbf{y})\in U$. It follows that $\mathbf{f}(\mathbf{x},\mathbf{y})=0$ for this $\mathbf{x}$. The uniqueness of $\mathbf{x}$ follows from the injectivity of $\mathbf{F}$ on $U$. This proves he first part of the theorem.

For the second part, define $\mathbf{g}(\mathbf{y})$ for $\mathbf{y}\in W$ to be the unique $\mathbf{x}$ such that $(\mathbf{x},\mathbf{y})\in U$ and $\mathbf{f}(\mathbf{x},\mathbf{y})=0$. Then for $\mathbf{y}\in W$,

$$\mathbf{F}(\mathbf{g}(\mathbf{y}),\mathbf{y})=(0,\mathbf{y}).$$

If $\mathbf{G}$ is the mapping of $V$ onto $U$ that inverts $\mathbf{F}$, then $\mathbf{G}\in C^1$ by the inverse function theorem and $(\mathbf{g}(\mathbf{y}),\mathbf{y})=\mathbf{G}(0,\mathbf{y})$ for $\mathbf{y}\in W$.

Theorem 191.3(Theorem).

Let $f:X\to {\mathbb{R}}^n$, where $X\subseteq {\mathbb{R}}^{n+m}$ is open, be $C^1$. Let $Z$ be the zero set of $f$. Let $f^{\prime }(x,y)=[A_x{A}_y]$. Assume $A_x$ is invertible for all $(x,y)\in Z$. Then $M$ is a manifold of $\dim m$.

The requirement above is not nThe function $f:{\mathbb{R}}^3\to \mathbb{R}$ defined by $f(x,y,z)=x^2+y^2+z^2-1$ does not satisfy these hypotheses: $A_x=[2x]$ is not invertible for $x=0$. However, at these points, we can solve for $z$ or $y$ instead; all three of these cannot be simultaneously zero. So, the zero set of $f$ is a manifold.