As I understand it, the implicit function theorem states:

Let $F:\mathbb{R}^m\times \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a continuously differentiable function $\langle x_1,\ldots, x_m; y_1,\ldots y_n\rangle \mapsto \langle F_1, \ldots F_n\rangle$, let $\vec{p}=\langle \vec{a}, \vec{b}\rangle\in \mathbb{R}^m\times \mathbb{R}^n$ be a point in the domain of $F$, and let $\vec{q} = F(\vec{p})$ be its value.

If the Jacobian $\frac{\partial(F_1, \ldots F_n)}{\partial(y_1,\ldots y_n)}$ is invertible, then there exist neighborhoods $U \ni \vec{a}$ and $V \ni \vec{b}$ and a unique continuously-differentiable function $G: U\rightarrow \mathbb{R}^n$ such that:

  1. $G(\vec{a}) = \vec{b}$
  2. $\{\langle \vec{x}, G(\vec{x})\rangle : \vec{x}\in U\} = \{\langle \vec{x}, \vec{y}\rangle \in U\times V : F(\vec{x}, \vec{y}) = \vec{q}\}$ (Near $\vec{a}$, the graph of $G$ exactly consists of the points that make $F\equiv \vec{q}$)

Hence the implicit function theorem says that in the vicinity of a point, a relation $R \equiv \text{const.}$ corresponds to the graph of a function if a certain Jacobian, evaluated at that point, is invertible.

I am wondering if given a Jacobian, you can find a relation $R$ for which that Jacobian is the test that you would use to determine if $R$ is an implicitly-defined function. And is such a construction useful?

If it helps, I am especially interested in the 2x2 case $\partial(f,g)/\partial(u,v)$, where $f,g:\mathbb{R}^2\rightarrow \mathbb{R}$.

Edit: I think that the answer might be "Yes", but that the relation $R$ is an especially simple one. The reason is that with a typical Jacobian, you have a matrix of $n\times n$ functions each of which has the form $\mathbb{R}^n\rightarrow \mathbb{R}$. In contrast, for the implicit function theorem, you instead expect a matrix of functions of the form $\mathbb{R}^m\times \mathbb{R}^n\rightarrow \mathbb{R}$, where $m$ corresponds to the number of "input" variables and $n$ corresponds to the number of "output" variables. If you have $m=0$ input variables, then maybe the IFT collapses to just:

Zero-input IFT. Let $F:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a continuously differentiable function, let $\vec{y}$ be a point in its domain, and let $\vec{c}$ be its value. If the Jacobian $\frac{\partial F_i}{\partial x_j}$ is invertible, then there exists a neighborhood $V\ni \vec{y}$ and a constant $G(\cdot):1\rightarrow \mathbb{R}^n$ such that:

  1. $G(\cdot) = \vec{y}$
  2. The singleton set $\{\langle \cdot, \vec{y}\rangle\} = \{\langle \cdot, \vec{b}\rangle \in 1\times V : F(\vec{b})=\vec{c}) \}$. Together, these conditions imply that if the Jacobian is invertible, then $\vec{y}$ is the one and only point in the vicinity on which $F$ attains the value $\vec{c}$.

I think this is a statement of the inverse function theorem as a special case of the implicit function theorem?

  • $\begingroup$ Assuming I understand what you are asking, a function of the form $f(x,y) = Ax+By$ will suffice. $\endgroup$ – copper.hat Jun 15 '17 at 21:07

Suppose you have a Jacobian $J\equiv\partial(F_1,\ldots,F_n)/\partial(y_1,\ldots y_n)$ where $F:\mathbb{R}^n\rightarrow \mathbb{R}^n$.

We seek a relation $R(x_1,\ldots, x_m;\, y_1,\ldots y_m)$ for which $J$ is the test you would use to determine whether $R$ implicitly defines a function. That is, where the Jacobian matrix corresponds to the partial derivatives of $R$ with respect to the arguments which you want to be considered "output" arguments.

But in fact, $F$ itself can be considered a relation where $J$ is the appropriate test. This is a particularly simple kind of relation, because $F$ has no "input" arguments $x_1,\ldots, x_m$.

If $F$, considered as a relation, passes the implicit function theorem test at a point $\vec{y}$, the result of the implicit function theorem says simply that $F$ is invertible in a neighborhood of $\vec{y}$.


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