Talk:P-Product Metrics on Real Vector Space are Topologically Equivalent

It looks to me like the original author managed to do the hard part of this proof and then got stuck on the easy part, thinking zie had to prove more than was actually necessary. I replaced it with something far simpler that I think does work. However, I would really appreciate if other folks could check that I didn't do something totally stupid. --Dfeuer (talk) 18:32, 15 January 2013 (UTC)


 * There is nowhere where it is shown that $d_r \left({x,y}\right) \ge d_\infty \left({x, y}\right)$, or anything similar, as far as I can see. The step that shows why $d_r \left({x, y}\right) \ge d_{r+1} \left({x, y}\right)$ should be correct. --Anghel (talk) 19:06, 15 January 2013 (UTC)


 * Good point. I don't think that was proved in any of the material I erased either. --Dfeuer (talk) 19:09, 15 January 2013 (UTC)


 * But that follows trivially from the fact that the $n$th root is increasing, doesn't it? I'm thinking the whole proof of the chain of inequalities might well belong on a separate page to which this one need not link. Unless I'm thinking entirely bone-headedly, which is always a possibility. --Dfeuer (talk) 19:15, 15 January 2013 (UTC)


 * Yes, it does. So we just need to add that argument. Good job. --Anghel (talk) 19:34, 15 January 2013 (UTC)

I see your comment on "Nothing here seems to require the underlying factor space to be the real line; any metric space should do", but if you do see fit to expand it, please do it on another page. --prime mover (talk) 22:36, 15 January 2013 (UTC)


 * I have already posted this generalization for binary products on P-Product Metric Induces Product Topology. --abcxyz (talk) 02:38, 16 January 2013 (UTC)

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Yes, the proof here really doesn't need to deal with integers. But if you come up with a method that doesn't require the horrible derivative, I think it'd still be good to give that derivative its own page. I'm sure it must come up somewhere else. --Dfeuer (talk) 03:03, 16 January 2013 (UTC)


 * Proving $d_{\infty} \ge n^{-1/r} d_r$ is pretty similar to proving the special case $d_{\infty} \ge n^{-1} d_1$. --abcxyz (talk) 05:04, 16 January 2013 (UTC)


 * I don't have time tonight, but I'll try to work on it on Thursday. Do you have any ideas for a name for the function whose derivative is calculated here? &mdash;Dfeuer (talk) 05:50, 16 January 2013 (UTC)

There's no need for the general $d_r$-$d_t$ comparison. We can drop $d_t$ and just deal with $d_r$, $d_1$, and $d_\infty$. --Dfeuer (talk) 17:05, 16 January 2013 (UTC)


 * If you don't do it first, I'll strip out $d_t$. If that comparison has some value outside of proving this theorem (which it may), we should put it in its own page. --Dfeuer (talk) 17:09, 16 January 2013 (UTC)


 * We can also strip out $d_1$. I think that the comparison should have a page. --abcxyz (talk) 17:16, 16 January 2013 (UTC)


 * No problem. Can you come up with a title for the function whose derivative is currently used to prove it's decreasing? --Dfeuer (talk) 17:22, 16 January 2013 (UTC)


 * Don't worry, I saw that question. It seems reasonable to put it on Inequality of Hölder Means, but I don't know if it's the best place to put it. --abcxyz (talk) 17:43, 16 January 2013 (UTC)

Congrats, abcxyz. It looks like you've entirely eliminated the need for most of this proof. --Dfeuer (talk) 17:44, 16 January 2013 (UTC)


 * That's actually pretty much what I've been saying, anyway. --abcxyz (talk) 17:57, 16 January 2013 (UTC)

Now I see why the Hölder mean doesn't do it. Pesky $(1/n)^{1/p}$. If there's no standard name maybe we should make one up. Hölder sum? --Dfeuer (talk) 18:14, 16 January 2013 (UTC)


 * I think we can just state $n^{1/q} M_q \le n^{1/p} M_p$. No need for making up names, in my opinion. --abcxyz (talk) 18:19, 16 January 2013 (UTC)


 * The reason I would like a name (preferably a standard one) is that we can then have pages like Bounds for ____ Function, Derivative of _____ Function, and ______ Function is Increasing. The bounds theorem proves General Euclidean Metrics are Topologically Equivalent and P-Product Metrics are Topologically Equivalent. I don't know the applications of the other theorems. What's your alternative proposal? --Dfeuer (talk) 18:46, 16 January 2013 (UTC)


 * If you delete the existing proof, Dfeuer, then you're blocked indefinitely. --prime mover (talk) 19:28, 16 January 2013 (UTC)


 * I won't delete anything here without your permission. What I'm trying to do is find a solution that will satisfy your needs. We have here a proof that uses certain properties of a certain kind of REAL function. All I'm proposing is that we extract that function and its properties from the proof about general Euclidean metrics. My thinking at present is that the way to do this is to define a special case of p-norm for sequences of real numbers (and possibly even one for FINITE sequences of real numbers), give that definition its own page, and prove the necessary properties. We can even transclude it here, if we're careful. --Dfeuer (talk) 19:40, 16 January 2013 (UTC)


 * The existing proof is a valid proof. It can stay as it is. If you want to craft something else, then add it as an alternative proof. --prime mover (talk) 19:42, 16 January 2013 (UTC)


 * Even then, we still need to rename the first lemma and move the second one somewhere else (it doesn't actually play any role in this proof). Can you suggest appropriate names for each? --Dfeuer (talk) 20:10, 16 January 2013 (UTC)


 * No we don't, and no we don't. --prime mover (talk) 21:17, 16 January 2013 (UTC)

Why do you want a lemma on this page that's not used on this page? --Dfeuer (talk) 21:41, 16 January 2013 (UTC)


 * What lemma? --prime mover (talk) 21:45, 16 January 2013 (UTC)


 * The one comparing finite-p metrics to each other, with the derivative calculation. As abcxyz found, all we need is the first lemma, which shows how $d_\infty$ bounds $d_r$. The other lemma is interesting in its own right, but it doesn't help prove this theorem. --Dfeuer (talk) 22:01, 16 January 2013 (UTC)


 * As I said above, don't replace the existing proof (as it was before you and abcxyz started buggering about with it) with your new proof, however much of an improvement it is. Leave the old proof as an existing proof, and put your new stuff up as a new proof. --prime mover (talk) 22:18, 16 January 2013 (UTC)


 * Before we started messing with it, it was incomplete, and had a couple comments about how you didn't know how to proceed. It's not clear that it could be completed from that form. --Dfeuer (talk) 22:31, 16 January 2013 (UTC)


 * But abcxyz (?) effectively completed it by the trick of multiplying l&r by $n^{-1}$ yeah? --prime mover (talk) 23:06, 16 January 2013 (UTC)


 * I don't recall doing that, but never mind. Doesn't really matter who, anyway. --abcxyz (talk) 00:48, 17 January 2013 (UTC)

We started with your partial proof, Part A. I tried to complete it by adding Part B, but that still did not get all the way there. abcxyz then saw that combining Part B with his own Part C formed a proof. Part A is nowhere to be found therein. The way I saw in the middle to use Part A is tantamount to driving down a cul-de-sac and making a U-turn before continuing along the proper route—it used the same technique as Part C, but applied it only to a special case and then argued with great difficulty from there to the general one. There was simply no sense in doing that. --Dfeuer (talk) 23:28, 16 January 2013 (UTC)


 * How about this: Put all the humongous calculations (proving $p \le q \implies n^{-1/p} d_p \le n^{-1/q} d_q \le d_\infty \le d_q \le d_p$) in another page (maybe related to Hölder means?), which will contain (at least) $({1}): n^{-1/p} d_p \le d_\infty \le d_p$, $({2}): n^{-1/p} d_p \le n^{-1/q} d_q$, and $({3}): d_q \le d_p$. Then we can just invoke that theorem. --abcxyz (talk) 00:48, 17 January 2013 (UTC)


 * By the way, I don't think that the mentioned inequalities should be stated in terms of $p$-norms (or $p$-product metrics), because they apply more generally, i.e., the requirement $p \ge 1$ can be dropped (without affecting the proof). --abcxyz (talk) 01:12, 17 January 2013 (UTC)


 * I just came across the original version of this page by Prime.mover (see here). The second claim (i.e., $p \le q \implies n^{-1/p} d_p \le n^{-1/q} d_q$) is correct; in fact, it is the theorem proved in Inequality of Hölder Means. Thus, the original approach does result in a valid proof. I apologize for losing sight of the original version; I only viewed the current versions and did not investigate the matter deeply enough. Also, I think that the original theorem statement should be restored.


 * The Lipschitz equivalence of all the general Euclidean metrics follows easily from the inequality $n^{-1/p} d_p \le d_{\infty} \le d_q \le n^{1/q} d_{\infty} \le n^{1/q} d_p$. This argument seems to be where this page is headed, and obviously obtains the desired result (i.e., proves the existence of Lipschitz constants) much more quickly than the original approach does. However, the original approach (i.e., using $p \le q \implies n^{-1/p} d_p \le n^{-1/q} d_q \le d_\infty \le d_q \le d_p$), in fact, gives the smallest Lipschitz constants (which the new approach does not), and is therefore of interest.


 * That being said, I do think that the lemma should go into its own page (not transcluded inside this one; as I said earlier, the result holds more generally), and that the two approaches should be separated into two proofs/pages. On a less important note, it might be a good idea to have Proof 1 as the new approach and Proof 2 as the old one. This is simply because the new approach requires less background and viewers are more likely to read Proof 1 first. It would be less intimidating if readers clicked on a link and found a (short) proof of $d_{\infty} \le d_p \le n^{1/p} d_{\infty}$ instead of a (much longer) proof of $p \le q \implies n^{-1/p} d_p \le n^{-1/q} d_q$. If they wanted to get into the details of the smallest Lipschitz constants, then they could access the next proof, i.e., the one based on the original approach. --abcxyz (talk) 04:23, 17 January 2013 (UTC)


 * $p$-norms, for sure. Do you have some idea for proving that the $p$-norm is strictly decreasing without calculating the derivative? Maybe something with the binomial series? --Dfeuer (talk) 01:32, 17 January 2013 (UTC)


 * What do you mean by "$p$-norms, for sure"? I just explained why I do not think that the inequalities should be stated in terms of $p$-norms. (I don't have an answer to the second part of your post yet.) --abcxyz (talk) 04:23, 17 January 2013 (UTC)


 * I meant that the derivative is of a $p$-norm form (with a possibly-extended domain) rather than a Hölder mean. When I wrote that, I had somehow missed a huge section of your writing. The approach Wikipedia takes, which is admittedly awkward, is to call them $p$-norms even for $p<1$, when they are not actually norms. As for the long derivation, I have thought from the beginning that it was valuable—I never wanted to remove it; I just wanted to take the unnecessary integers out of it and then put it somewhere (possibly with $p$-norms or possibly not) where the magnitudes of the Lipschitz constants would be more relevant than they are in a proof about topological equivalence. However: as far as I could tell, the original proof never proved the smaller parts of the inequality, that is the parts to the right of the $d_\infty$. How do you complete those? --Dfeuer (talk) 04:52, 17 January 2013 (UTC)


 * A lot of my writing was posted after your post. That could have been why you missed a large part of it.
 * As this website does not define "$p$-norm" for $p < 1$, I would not suggest including "$p$-norm" in the page title. Maybe there is a better option than the Hölder mean, but I can't think of anything else at this moment.
 * The inequality $d_r \le d^{1/r} d_{\infty}$ is proven in P-Product Metric is Topologically Equivalent to Chebyshev Distance on Real Vector Space.
 * By the way, I think that "General Euclidean Metrics are Lipschitz Equivalent" might be a better name for this page. This way, the Lipschitz constants are more relevant to the page title. Comments? --abcxyz (talk) 05:08, 17 January 2013 (UTC)

I think if we do that we'll still need a page by this name linking to this proof and the proof that Lipschitz equivalence implies topological equivalence. I think the derivative calculation probably deserves its very own page. If as you say the inequality it proves gives an alternate proof of the Hölder sum inequality, maybe the thing to do is to use that as an alternate proof of Hölder, and then Hölder as a proof of the Lipschitz inequality we need? I haven't had time to sit down and puzzle out the Hölder connection. --Dfeuer (talk) 05:22, 17 January 2013 (UTC)


 * Of course we'll have a page by this name.
 * The second humongous calculation doesn't prove anything (yet).
 * In my opinion, we could use one or several of the inequalities involving Hölder means as a proof. --abcxyz (talk) 17:52, 17 January 2013 (UTC)