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EDIT!! The problem originally described (see below) has been reduced to the correctness of a simple extension of an argument from Rudin's PMA. Feel free to skip to the proposed solution, below.

As part of a much needed calculus review, I've been working through Rudin's Principles of Mathematical Analysis. Besides the following, I've found the prospective on the material therein to be rather illuminating -- which leads me to believe I've missed something, in this case.

The proof of a lemma (10.38) to be used in the proof of Poincare's Lemma (10.39) cites the following fact: (to paraphrase)

$V=\{(x,y)|x\in\Bbb{R^{p-1}, y\in\Bbb{R}}\}; p>1$; V convex open. $U=\{x|(x,y)\in V \text{ for some y}\}$. It follows that there is a continuously differentiable function on U, let's call it $\alpha$, such that $(x,\alpha(x))\in V.$

This is one of very few proof details in Rudin explicitly left as an exercise. The hint is that $\alpha$ may be taken to be a constant if V is a ball; it should also be noted that the same $\alpha$ notation is conspicuously used in the proof of the existence of partitions of unity on compact sets.

Now, ignoring the hints and using inf, sup, and handling the special case of extended values, I think I can find a continuous $\alpha$ easily enough, but the resulting function is not necessarily smooth.

I've seen a totally different solution involving countable open coverings, compact exhaustions, and smooth partitions of unity on open sets, but, although this solution makes use of the hints, I'm apprehensive about this scheme because it relies on so much machinery not mentioned in the text or exercises elsewhere in the book. It would be very uncharacteristic of the author if this were the intended solution (I say this having worked hundreds of other exercises from his books).

With that in mind, can someone propose a solution scheme that does not rely on so much beyond-the-text machinery?

EDIT!! In particular, after examining Rudin's proof of the existence of partitions of unity on compact sets, I believe I have found an extension of his argument that works in this case. Can anyone verify this? (See below for my proposed answer).

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    $\begingroup$ I suspect that $$\alpha(x) = \frac{\int_{V_x} ydy}{\int_{V_x} dy}$$ where $V_x = \{y\mid (x,y) \in V\}$ will work. Convexity of $V$ should provide all you need to prove it differentiable. $\endgroup$ – Paul Sinclair Mar 20 '16 at 18:57
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    $\begingroup$ @PaulSinclair: $V$ is not necessarily bounded. $\endgroup$ – copper.hat Mar 20 '16 at 19:09
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    $\begingroup$ This may not help, and is not really elementary, but if a (finite) convex function is differentiable on an open set, then it is continuously differentiable on the set (from Rockafellar). $\endgroup$ – copper.hat Mar 20 '16 at 19:38
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    $\begingroup$ I suspect that one can find a piecewise affine function (tessellate $U$ into small 'triangles') that lies in $V$ first and then do some sort of local convolution to smooth the function near the 'boundaries'. Still, not elementary and rather messy. $\endgroup$ – copper.hat Mar 20 '16 at 21:52
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    $\begingroup$ The answer here: math.stackexchange.com/questions/339645/… was partially helpful, but I'm still wondering about the use of locally finite covers and creating a partition of unity wrt it. Since this (to me) doesn't follow obviously from anything in Rudin. $\endgroup$ – Jason Mar 24 '16 at 0:27
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This potential answer was too long to comment, and it seemed inappropriate to so dramatically change the original post. Could any of you commenting/following the thread verify that I'm on the right track? Thanks!

From the prior exercises in chapter 10, it is clear that partitions of unity on compact sets can be chosen to be smooth. So, If V is bounded, its closure is compact, and the hints make it obvious that a smooth partition of unity on it's projection $\bar{U}$ does the trick.

If V is unbounded, the intersection of its closure with the balls of natural radii centered at the origin admit a solution, as above. The question is, can we patch the solutions together with tools from the text, rather than building all new machinery?

Take the balls of rational radii centered at points with rational coordinates, $B_{r,s}$; the balls cover $\bar{V}$. For each intersection of $\overline{B_{n,0}}$ with $\bar{V}$, collect the finitely many $B_{r,s}; r<1/2; s \in B_{n,0}-B_{n-1,0}$ required for a covering.

Notice that, in Rudin's partition of unity proof the functions are defined inductively: that is to say, by adjoining new sets proceeding from n-1 to n above, it seems we need not change any of the partition functions previously defined:

  • $\phi_i(x)$ is associated with one member of the countable covering, say $B_{r,s}$ and is 1 on it's projection to $\mathbb{R^{p-1}}$ but zero outside the projection of $B_{2r,s}$; it is assumed smooth;
  • $\psi_{i+1}=(1-\phi_1)...(1-\phi_i)\phi_{i+1}$ hence $\psi_{i}+...+\psi_{1}=1-(1-\phi_i)...(1-\phi_1)$

If this in fact gives a locally finite smooth partition of unity on $\bar{U}$, then this affords a solution which naturally extends the cases where V is a ball or V is bounded.

The major simplification here compared with the generally accepted answer, if this is correct, comes from the fact that the partition functions do not "pile up" on $\partial V$ as is the case with the more general smooth partitions of unity on open sets from, say, Munkres.

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    $\begingroup$ +1: Nice. I was hoping for something more 'constructive' :-). $\endgroup$ – copper.hat Mar 26 '16 at 19:32
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    $\begingroup$ Does that mean you're of the opinion that this approach works? (Thanks for all the feedback! I always worry when I disagree with a generally accepted answer to a question.) $\endgroup$ – entprise Mar 26 '16 at 19:56
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    $\begingroup$ Thx for your hard work on this. Sorry for my confusion, but could you elaborate on what $B_{n,0}$ is? $\endgroup$ – Jason Apr 8 '16 at 2:48
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    $\begingroup$ Ball of radius n centered at the origin. $\endgroup$ – entprise Apr 13 '16 at 21:38

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