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I have a question regarding the Implicit Function Theorem.

Prove that if $\mathbf{g}(s,t)= \begin{bmatrix} g_{1}(s,t) \\ g_{2} (s,t) \\ g_{3}(s,t) \end{bmatrix}$ and $\begin{vmatrix} \frac{\partial g_{1}}{\partial s} & \frac{\partial g_{1}}{\partial t} \\ \frac{\partial g_{2}}{\partial s} & \frac{\partial g_{2}}{\partial t} \end{vmatrix}(s_{0},t_{0}) \neq 0$, for some points $s_{0},\,t_{0}$, then $\exists \delta>0$ such that $\mathbf{g}[B_{\delta}(s_{0},t_{0})]$ (an open ball centered at $(s_{0},t_{0})$ of radius $\delta$) is the graph of some function.

I am quite comfortable with proving special cases of the Implicit Function Theorem, but in this case the dimension of the domain is smaller than the dimension of the codomain.

I would appreciate any help!

P.S. This is for test preparation.


Here is my attempt at the problem:

By the Inverse Function Theorem, $g_{1}$ and $g_{2}$ have inverses.

Now define a function $\mathbf{G} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{bmatrix} g_{1}^{-1}(x) \\ g_{2}^{-1}(y) \\ z \end{bmatrix}$.

$D\mathbf{G}(x_{0},y_{0},z_{0})$ can be found to be invertible, where $(s_{0},t_{0}) \mapsto (x_{0},y_{0},z_{0})$. Therefore, there exists a neighborhood around $(x_{0},y_{0},z_{0})$ for which this is also true.

Define $s$ and $t$ such that their respective spaces contain an origin (i.e. $(s,t)=(0,0)$). Now define a function $\phi$ that maps the space of the variables $x$ and $y$ to the space of the variable $z$:

\begin{equation} \mathbf{G}^{-1} \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} = \begin{bmatrix} \phi_{1}(z) \\ \phi_{2}(z) \\ z \end{bmatrix} \end{equation}

Now note that: $\mathbf{G} \begin{bmatrix} \phi_{1}(z) \\ \phi_{2}(z) \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}$. So the preimage can be redefined:

\begin{equation} \mathbf{G} \begin{bmatrix} \phi_{1}(z) \\ \phi_{2}(z) \\ z \end{bmatrix} = \mathbf{G} \begin{bmatrix} \mathbf{G}^{-1} \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}. \end{equation}

The same can be reasoned of

\begin{equation} \mathbf{G}^{-1} \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \end{equation}

and so $\phi_{1}(z)=x$ and $\phi_{2}(z)=y$. In vector form, that is $\phi(z)=\begin{bmatrix} x \\ y \end{bmatrix}$. Therefore there exists a function whose graph is some ball $g[B_{\delta}(s_{0},t_{0})]$.

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HINT: By the Inverse Function Theorem, locally you can invert the mapping $(s,t)\mapsto (g_1(s,t),g_2(s,t))$. Use this to show that locally the image of $\mathbf g$ can be written in the form $z=\phi(x,y)$.

EDIT: To provide a bit more detail, let $\mathbf G(s,t) = \begin{bmatrix} g_1(s,t)\\g_2(s,t)\end{bmatrix}$. By the given condition at $(s_0,t_0)$, there is a neighborhood $U$ of $(s_0,t_0)$ on which we have an inverse function $\mathbf G^{-1}\colon \mathbf G(U)\to U$. Now consider $\phi = g_3\circ\mathbf G^{-1}$.

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  • $\begingroup$ Thank you for the hint! I have tried to incorporate this into an edit of my original post (below the horizontal line). Am I on the right track? $\endgroup$ – Clubr Dec 4 '13 at 5:33
  • $\begingroup$ Right track, but wheels screeching :P The individual functions $g_1$ and $g_2$ do not have inverses. I'll edit slightly. $\endgroup$ – Ted Shifrin Dec 4 '13 at 13:05
  • $\begingroup$ Thanks Ted! I think I understand now. By the way, I am currently using your book, Multivariable Mathematics and I must say it is one of the best mathematics books I have read. Keep up the good work! :) $\endgroup$ – Clubr Dec 4 '13 at 19:24

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