Theorem: Let $F: X\subseteq \mathbb{R}^n \rightarrow \mathbb{R}$ be of class of $C^1$ and let $a$ be a point of the level set $S=\{x\in\mathbb{R}^n \mid F(x)=c\}$. If $F_{x_n}(a)\neq 0$ then there is a neighborhood $U$ of $(a_1,a_1\dots a_{n-1}) \in \mathbb{R}^{n-1}$, a neighborhood $V$ for of $a_n\in\mathbb{R}$ and a function $f:U\subseteq\mathbb{R}^{n-1} \rightarrow V$ of class $C^1$ such that if $(x_1,x_2,\ldots,x_{n-1})\in U$ and $x_n \in V$ satisfy $F(x_1,x_2,\ldots,x_n)=c$ ,then $x_n = f(x_1,x_2,\ldots,x_{n-1})$ is representable as a function of $(x_1,\ldots,x_{n-1})$ .

I don't quite get what the theorem is trying to show. Can someone explain it? Is it just as simple as we just consider different part of the function?


Intuitively the implicit theorem is telling us that if a sufficiently smooth function $F$ defined on some Euclidean space (more generally on manifolds) satisfies a regularity condition then the equation $F(x_1,\cdots,x_n)=0$ can be locally "solved" for one of the coordinates.

For example, consider the map $F:\mathbb{R}^2\to\mathbb{R}$ defined $F(x,y)=x^2+y^2-1$. Note then that the equation $F(x,y)=0$ is nothing more than the usual locus for the unit circle $\mathbb{S}^1$. Consider the North pole of the cirlce, e.g. $(0,1)$, note then that $D_F(0,1)$ is nonsingular, and thus the IFT says that the equation $F(x,y)=0$ is locally solvable for one of the coordinates, say $y$. Of course this is true, locally around $(0,1)$ $\mathbb{S}^1$ one can solve the equation $x^2+y^2-1=0$ for $y$ to get $y=\sqrt{1-x^2}$.

Moreover, note that the validity of the theorem is further cemented by noting that while almost everywhere on $\mathbb{S}^1$ the equation $F(x,y)=0$ is locally solvable, there are two problem spots--the intersections with the $x$-axis. The reason why we might believe these to actually be problem spots is the second (easily deducible) interpretation of the IFT--if $F$ satisfies the regularity conditions at a point then the curve $F(x,y)=0$ locally looks like the graph of a (in this case) single-variable function. Going back to the previous paragraph, we see that locally at the North pole $\mathbb{S}^1$ does indeed look like the graph of $x\mapsto \sqrt{1-x^2}$. Moreover, now it's intuitively clear that the IFT should nto apply at the intersections with the $x$-axis since in any neighborhood of those points the curve $F(x,y)=0$ fails the "vertical line test" (i.e. isn't the graph of single-variable function). The reassuring part is that the IFT does not apply there since you can easily check, for example, that $D_F(1,0)$ is not invertible.

I hope that helps.

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    $\begingroup$ I saw this question a little bit ago (on my phone) and finally got around to a computer. This was exactly the example I would have used. $\endgroup$ – Matt Jan 9 '12 at 6:33
  • $\begingroup$ I think this is a fine explanation. However the part about $D_F(0,1)$ being invertible and $D_F(1,0)$ being non-invertible could perhaps be described more clearly: this Jacobian (assuming that is what you meant) is a nonsquare ($2\times 1$-) matrix. You are specifically considering one of its 'blocks', namely $\partial_yF$. So I would replace "solvable for one of the coordinates, say y" by " solvable for y". $\endgroup$ – wildildildlife Jan 9 '12 at 20:16

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