The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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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 [...]

    Pingback by The Implicit Function Theorem « The Unapologetic Mathematician | April 15, 2011 | Reply


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