I am finding Rudin's proofs of these theorems very non-intuitive and difficult to recall. I can understand and follow both as I work through them, but if you were to ask me a week later to prove one or the other, I couldn't do it.

For instance, the use of a contraction mapping in the inverse function theorem seems to require one to memorize, at the very least, a non-obvious (at least to me) function (viz. $\phi(\mathbf{x}) = \mathbf{x} + \mathbf{A}^{-1}(\mathbf{y}-\operatorname{f}(\mathbf{x}))$) and constant (viz. $\lambda^{-1} = 2 \Vert \mathbf{A}^{-1}\Vert$), where $\mathbf{A}$ is the differential of $\operatorname{f}$ at $\mathbf{a}$.

The implicit function theorem proof, while not as bad, also requires one to construct a new function without ever hinting as to what the motivation is.

I searched the previous questions on this site and haven't found this addressed, so I figured I'd ask. I did finnd this proof to have a much more intuitive approach to the inverse function theorem, but would like to see what proofs are preferred by others.

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    $\begingroup$ I find Rudin hard going for new stuff. For implicit function theorem, etc., I like 'Functional analysis', by Kantorovich & Akilov. But these things are matters of taste. I like that K&A give formulae with explicit constants, etc. $\endgroup$
    – copper.hat
    Jun 30 '13 at 22:51
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    $\begingroup$ $\phi(\mathbf{x}) = \mathbf{x} + \mathbf{A}^{-1}(\mathbf{y}-\operatorname{f}(\mathbf{x}))$ is (almost) Newton's method. $\endgroup$ Jun 30 '13 at 22:57
  • $\begingroup$ I think Newton's method is a better proof in that it makes some details explicit that are otherwise lost. For example, if $f$ map some closed set $C$ into itself and the initial guess is in this set, then the solution is also. Ultimately most proofs I have seen rely on the contraction map. $\endgroup$
    – copper.hat
    Jun 30 '13 at 23:04
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    $\begingroup$ @AnonSubmitter85 I learned these things from Edwards' Advanced Calculus, it looks like Rudin is not much different. I have seen an argument which avoids the contraction mapping argument in favor of some epsilon/delta type argument, however, I think it was less general and it didn't offer the error estimate which comes with the contraction technique. $\endgroup$ Jul 1 '13 at 1:48
  • $\begingroup$ @copper.hat Can you give a reference for a proof using Newton's method? $\endgroup$
    – littleO
    Jan 28 '17 at 13:18

Suppose you want to find the inverse of the mapping $F: \mathbb{R}^n \rightarrow \mathbb{R}^n$ near a point $x_o$ where $F'(x_o)$ is invertible. The derivative (Jacobian matrix) provides an approximate form for the map $F(x) = F(x_o)+F'(x_o)(x-x_o)+\eta$. If you set $y = F(x)$ and ignore the error term $\eta$ then solving for $x$ gives us the first approximation to the inverse mapping. $$ x = x_o+[F'(x_o)]^{-1}(y-F(x_o)). $$ Then, you iterate. The technical details are merely to insure this iteration does indeed converge to the inverse mapping, but at the start, it's just using the derivative to linearize the problem.

I don't know if this helps or not, but really the approach is almost brute force, to invert $F(x)=y$ what do you do? You solve for $x$. We can't do that abstractly for $F$ so instead we solve the next best thing, the linearization. Then the beauty of the contraction mapping technique completes the argument.

  • $\begingroup$ I just re-worked through the proof and this actually helps tremendously. $\endgroup$ Jul 2 '13 at 23:39
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    $\begingroup$ @AnonSubmitter85 good to hear, I spent about 5 hours going through this proof and its surrounding theory with one of my students. Btw, the motivation for that weird function construction for the implicit mapping theorem proof is simply that of convenience. You could go through a similar argument by solving the linearized level equation for $y$ to prove the implicit mapping theorem... but, why bother when we can instead just borrow the result from the inverse mapping theorem. I think some motivation could be made from how we parametrize surfaces which are graphs in calculus III... $\endgroup$ Jul 2 '13 at 23:59
  • $\begingroup$ I like the "inverting the linearization" intuition but the map $\varphi(x) = x + [F'(x_0)]^{-1} ( y - F(x) )$ used by Rudin is slightly different from what your argument yields. Is there a better explanation than convenience for the replacements $x_0 \mapsto x$ that he makes? $\endgroup$ May 18 '18 at 17:17
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    $\begingroup$ I would guess there is more than one way to set-up this proof. I think of $x_o$ as fixed whereas apparently Rudin takes $y$ as fixed in his construction of $\phi(x)$. This answer is not a complete argument, it really just offers intuition, the complete argument is about 3 pages in Edward's Advanced Calculus of Several Variables. Maybe I'll get a chance to prove it properly next semester... $\endgroup$ May 18 '18 at 21:06

The other answer takes care of the Inverse Function Theorem case. Therefore, I'll give "another" proof of the Implicit Function Theorem. (Note that I inverted the order of $\mathbb{R}^n$ and $\mathbb{R}^m$, compared to Rudin's)

Consider the commutative diagram

enter image description here

where $F(x,y)=\big(x,f(x,y)\big)$. By the Inverse Function Theorem, it follows that locally we have

enter image description here

Due to how $F$ is defined, the commutativity of the diagram above is obvious. Due to how $F$ is defined and due to the fact that it is invertible in the given neighbourhood, the existence of $g$ is clear, and $g=\pi_2 \circ F^{-1} \circ i$ is also clear. Since $g=\pi_2 \circ F^{-1} \circ i$, $g$ is $C^1$. The rest of the Implicit Function Theorem follows by using the composition of the diagonal arrows.

As to why take the function $F$ as is, just note that the proof is basically finding a function $F$ making everything commute. The choice of $F$ is the natural one (actually, practically imposed one) to make that happen.

  • $\begingroup$ I liked this answer, and wanted to understand each part of the diagram. I guess $i(x) =(x,0)$. Why is $i \times \pi_2$ part of the composition of the diagonal arrow in the square? Further, isn't $i \times \pi_2 = (i(x),g(x)) = (x,0,g(x)) \in \mathbb R^n \times \mathbb R^m \times \mathbb R^m$? $\endgroup$ Dec 3 '17 at 19:38
  • $\begingroup$ @AnthonyPulido Yes, $i(x)=(x,0)$ or $i(x)=(x,c)$ for whatever $c$ you are taking in $f(x,y)=c$ in the statement of the theorem (it suffices $c=0$, and most texts suppose so). For the next part, sorry for the ambiguity, but the correct interpretation is $i \times (\pi_2 \circ F^{-1} \circ i)$, not $(i \times \pi_2) \circ ~etc$. $\endgroup$
    – Aloizio Macedo
    Dec 3 '17 at 19:44
  • $\begingroup$ Thanks. Do we not have the same problem, though? Recall that $g = \pi_2 \circ F^{-1} \circ i$. Then $i \times g(x) = (i(x),g(x)) = (x,0,g(x)) \in \mathbb R^{n + m + m}$. $\endgroup$ Dec 3 '17 at 20:03
  • $\begingroup$ @AnthonyPulido Yes, maybe I should have used an $\iota$ (or simply $\mathrm{Id}$) in the product map, since that "$i$" in the product map $i \times g$ is the identity. I can fix these notational issues (if there are any more, please tell me), but it will require some time since those diagrams are on images and not directly mathjax. I'll try to update it later today. $\endgroup$
    – Aloizio Macedo
    Dec 3 '17 at 20:07
  • $\begingroup$ Thanks, again. Please take your time updating the diagram. It isn't urgent, at least for me. Just after I posted I noticed that it should probably be the identity. My question is now, why write it that way? Isn't $F^{-1} \circ i$ enough? $\endgroup$ Dec 3 '17 at 20:22

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