Abstract Nonsense

Crushing one theorem at a time

The Inverse Function Theorem (Proof)


Point of Post: This is a continuation of this post.

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Inverse Function Theorem

Let f:U\to\mathbb{R}^m, where U\subseteq\mathbb{R}^n is open, we say that a\in U is a regular point of f if f is totally differentiable at a and \text{rk}\left(D_f(a)\right)=m. If f is differentiable at a and \text{rk}\left(D_f(a)\right)<m we say that a is a critical point. By a basic theorem of linear algebra we know that if m=n we have that a point a\in U, for which f is differentiable, is regular if D_f(a) is invertible (i.e. \det\text{Jac}_f(a)\ne 0) and critical otherwise (i.e. if \det\text{Jac}_f(a)=0). Thus, with this terminology in mind we state the inverse function theorem as follows:

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Theorem (Inverse Function Theorem): Let f:U\to\mathbb{R}^n, where U\subseteq\mathbb{R}^n is open, be C^1(U). Let then a be a regular point for f, then there exists a neighborhood V\subseteq U of a such that f(V)\subseteq f(U) is open, f_{\mid V}:V\to f(V) is a bijection, and the inverse \left(f_{\mid V}\right)^{-1}:f(V)\to V is  C^1(V) with D_{f^{-1}}(f(x))=D_f(x)^{-1}.

Proof: Since a is a regular point we have that D_f(a) is invertible, call it’s inverse M. With a little fiddling one can see that we can choose \delta>0 and r>0 such that \overline{B_\delta(a)}\subseteq U and

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\displaystyle \left\|D_f(a)-D_f(x)\right\|_\text{op}\leqslant\frac{1}{2m\|M\|_\text{op}}\qquad r\leqslant\frac{\delta}{2\|M\|_\text{op}}

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both hold. Now, for each y\in B_r(f(a)) we define A_y:\overline{B_\delta(a)}\to\mathbb{R}^n:x\mapsto x+M(y-f(x)). Note then that for each x\in \overline{B_\delta(a)}  one has that A_y could be equally well described as \mathbf{1}+M\circ f+\text{const} (where the constant in M(y) and \mathbf{1} is the identity function) and so, using the chain rule,

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D_{A_y}(x)=\mathbf{1}+D_M(f(x))\circ D_f(x)=\mathbf{1}+M\circ D_f(x)

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where we used the fact that \mathbf{1} and M are linear and the total derivative of a linear map is itself. Thus, we see that

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\begin{aligned}\left\|D_{A_y}(x)\right\|_\text{op}&=\left\|\mathbf{1}-M\circ D_f(x)\right\|_\text{op}\\ &=\left\|M\left(D_f(a)-D_f(x)\right)\right\|_\text{op}\\ &\leqslant \|M\|_\text{op}\left\|D_f(a)-D_f(x)\right\|_\text{op}\\ &\leqslant\frac{1}{2n}\end{aligned}

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We can easily deduce then by the multivariable mean value theorem that

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\displaystyle \left\|A_y(x_1)-A_y(x_2)\right\|=\left\|D_{A_y}(\xi)(x-y)\right\|\leqslant \|D_{A_y}(\xi)\|_{\text{op}}|x_1-x_2|\leqslant \frac{1}{2n}|x_1-x_2|

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Next note that for each x\in \overline{B_\delta(a)} one has that

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\displaystyle \begin{aligned}\left\|A_y(x)-a\right\| &=\|A_y(x)-A_y(a)\|+\|A_y(a)-a\|\\ &\leqslant \frac{1}{2}\|x-a\|+\|M(y-f(x)\|\\ &\leqslant \frac{\delta}{2}+\|M\|r\\ &\leqslant \delta\end{aligned}

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Thus, A_y is a contraction mapping and so by the Banach fixed point theorem there is a unique solution to A_y(x)=x in \overline{B_\delta(x)}. But, note that writing it out one finds that A_y(x)=x if and only if f(x)=y. Thus, there is a unique solution to y=f(x) for x\in\overline{B_\delta(a)}. So, define the map g:B_r(f(a))\to \overline{B_\delta(a)} by sending y to the unique x\in \overline{B_\delta(a)} such that A_y(x)=x. It is clear that g is an embedding (i.e. that it is a homemorphism onto its image). What we claim though is that g(B_r(f(a)) is open in \mathbb{R}^n. To see this let g(y)\in g(B_r(f(a)) be arbitrary. Since f is continuous we know there exists a neighborhood B_\varepsilon(g(y)) such that f(B_\varepsilon(g(y))\subseteq B_r(f(a)) but it’s clear to see that this implies B_\varepsilon(g(y))\subseteq g(B_r(f(a)). Thus, since evidently f(a)\in B_r(f(a)) we have that if V=g(B_rf(a)) and f(V)=B_r(f(a)) then V,f(V) are both open and f:V\to f(V) is a bijection with inverse equal to g.

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What we lastly claim is that g is differentiable and satisfies D_g(f(x))=D_f(x)^{-1}. So, let f(x) be any point of f(V). Let then w be some vector in \mathbb{R}^n with \varepsilon small enough magnitude such that y+\varepsilon w\in f(V). We note then that

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\displaystyle \left\|A_{f(x)}(x+g(f(x)+\varepsilon w)-g(f(x)))-A_y(x)\right\|\leqslant \frac{1}{2}\left\|g(f(x)+\varepsilon w)-g(f(x))\right\|

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from this it easily follows that

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\displaystyle \|g(f(x)+\varepsilon w)-g(f(x))\|\leqslant 2\varepsilon \|M\|_{\text{op}}\|w\|

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That said, since f is differentiable we know that f(x+v)-f(x)=D_f(x)(v)+h(v) where \displaystyle \lim_{v\to\bold{0}}\frac{h(v)}{\|v\|}=\bold{0}. Thus, putting it all together we see that

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\displaystyle \begin{aligned}\lim_{\varepsilon\to0}\frac{g(f(x)+\varepsilon w)-g(f(x)}{\varepsilon} &=\lim_{\varepsilon}D_f(x)^{-1}\left(\frac{\varepsilon w-h(g(f(x)+\varepsilon w)-g(f(x))}{\varepsilon}\right)\\ &=D_f(x)^{-1}\end{aligned}

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But, this left side precisely says that D_g(f(x)) exists and D_g(f(x))=D_f(x)^{-1}. \blacksquare

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Corollary: Let f be as in the statement of the inverse function theorem, but that f\in C^k(U) for some k\in\mathbb{N}\cup\{\infty\} then the guaranteed f^{-1} is also C^k.

Proof: This follows immediately from the fact that D_{f^{-1}}(f(x))=D_f(x)^{-1}. Indeed, since f\in C^k(U) it’s easy to see that the entries of D_f(x) for any x are C^{k-1}, but then the entries of \displaystyle D_{f^{-1}}f(x)=D_f(x)^{-1}=\frac{1}{\det(D_f(x))}\text{adj}(D_f(x)) are C^{k-1} (as can be checked since they are quotients of polynomials of C^{k-1} functions and so we can go back quickly to conclude that f^{-1} is C^k. \blacksquare

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References:

1.  Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.

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September 8, 2011 - Posted by | Analysis | , , , ,

4 Comments »

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  2. […] and so by construction we have that . The rest is just a literal statement of the inverse function theorem. […]

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