# The Unapologetic Mathematician

## The Implicit Function Theorem II

Okay, today we’re going to prove the implicit function theorem. We’re going to think of our function $f$ as taking an $n$-dimensional vector $x$ and a $m$-dimensional vector $t$ and giving back an $n$-dimensional vector $f(x;t)$. In essence, what we want to do is see how this output vector must change as we change $t$, and then undo that by making a corresponding change in $x$. And to do that, we need to know how changing the output changes $x$, at least in a neighborhood of $f(x;t)=0$. That is, we’ve got to invert a function, and we’ll need to use the inverse function theorem.

But we’re not going to apply it directly as the above heuristic suggests. Instead, we’re going to “puff up” the function $f:S\rightarrow\mathbb{R}^n$ into a bigger function $F:S\rightarrow\mathbb{R}^{n+m}$ that will give us some room to maneuver. For $1\leq i\leq n$ we define $\displaystyle F^i(x;t)=f^i(x;t)$

just copying over our original function. Then we continue by defining for $1\leq j\leq m$ $\displaystyle F^{n+j}(x;t)=t^j$

That is, the new $m$ component functions are just the coordinate functions $t^j$. We can easily calculate the Jacobian matrix $\displaystyle dF=\begin{pmatrix}\frac{\partial f^i}{\partial x^j}&\frac{\partial f^i}{\partial t^j}\\{0}&I_m\end{pmatrix}$

where ${0}$ is the $m\times n$ zero matrix and $I_m$ is the $m\times m$ identity matrix. From here it’s straightforward to find the Jacobian determinant $\displaystyle J_F(x;t)=\det\left(dF\right)=\det\left(\frac{\partial f^i}{\partial x^j}\right)$

which is exactly the determinant we assert to be nonzero at $(a;b)$. We also easily see that $F(a;b)=(0;b)$.

And so the inverse function theorem tells us that there are neighborhoods $X$ of $(a;b)$ and $Y$ of $(0;b)$ so that $F$ is injective on $X$ and $Y=F(X)$, and that there is a continuously differentiable inverse function $G:Y\rightarrow X$ so that $G(F(x;t))=(x;t)$ for all $(x;t)\in X$. We want to study this inverse function to recover our implicit function from it.

First off, we can write $G(y;s)=(v(y;s);w(y;s))$ for two functions: $v$ which takes $n$-dimensional vector values, and $w$ which takes $m$-dimensional vector values. Our inverse relation tells us that \displaystyle\begin{aligned}v(F(x;t))&=x\\w(F(x;t))&=t\end{aligned}

But since $F$ is injective from $X$ onto $Y$, we can write any point $(y;s)\in Y$ as $(y;s)=F(x;t)$, and in this case we must have $s=t$ by the definition of $s$. That is, we have \displaystyle\begin{aligned}v(y;t)&=v(F(x;t))=x\\w(y;t)&=w(F(x;t))=t\end{aligned}

And so we see that $G(y;t)=(x;t)$, where $x$ is the $n$-dimensional vector so that $y=f(x;t)$. We thus have $f(v(y;t);t)=y$ for every $(y;t)\in Y$.

Now define $T\subseteq\mathbb{R}^m$ be the collection of vectors $t$ so that $(0;t)\in Y$, and for each such $t\in T$ define $g(t)=v(0;t)$, so $F(g(t);t)=0$. As a slice of the open set $Y$ in the product topology on $\mathbb{R}^n\times\mathbb{R}^m$, the set $T$ is open in $\mathbb{R}^m$. Further, $g$ is continuously differentiable on $T$ since $G$ is continuously differentiable on $Y$, and the components of $g$ are taken directly from those of $G$. Finally, $b$ is in $T$ since $(a;b)\in X$, and $F(a;b)=(0;b)\in Y$ by assumption. This also shows that $g(b)=a$.

The only thing left is to show that $g$ is uniquely defined. But there can only be one such function, by the injectivity of $f$. If there were another such function $h$ then we’d have $f(g(t);t)=f(h(t);t)$, and thus $(g(t);t)=(h(t);t)$, or $g(t)=h(t)$ for every $t\in T$.

November 20, 2009 - Posted by | Analysis, Calculus

## 1 Comment »

1. […] we can go back and clean up not only the statement of the implicit function theorem, but its proof, as well. And we can even extend to a different, related statement, all using the inverse function […]

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