Talk:Young's Inequality for Convolutions

Page Changes, Theorem Changes, Creation of Corollaries
This should be an unnecessary statement, per the definitions of Lebesgue Spaces and a $p$-integrable function. That a function $f$ (or more properly/pedantically, the equivalence class of $f$) is in $L^p(\R^n)$ trivially implies that $\int |f|^p ~\mathrm d \mu < \infty$. So for this theorem, we have that $f$ is $p$-integrable  and $g$ is  $q$-integrable, that is to say  $\int |f|^p ~\mathrm d \mu < \infty$ and $\int |g|^q ~\mathrm d \mu < \infty$.--Mdibah (talk) 20:06, 14 October 2014 (UTC)


 * I don't understand why the condition on $p$, $q$, and $r$ was moved to the end of the theorem's statement. This is a necessary hypothesis and not a conclusion of the proof. I have moved this back to the introduction of $p$, $q$, and $r$, unless there is a stylistic rule that I am unknowingly violating.--Mdibah (talk) 20:06, 14 October 2014 (UTC)


 * I'll let someone else take this one on. --prime mover (talk) 20:24, 14 October 2014 (UTC)


 * The problem we got here is that the original proof what was here is not here any more it's something different. Which is bad. --prime mover (talk) 20:25, 14 October 2014 (UTC)


 * Looking back through the page histories, it doesn't look like there was ever a proof. Furthermore, the statement of the theorem c. Sep 2013 is a very specific case of the current statement, using $q=1$ and $p=r$. I guess I'll make a page for it as a corollary... --Mdibah (talk) 21:06, 14 October 2014 (UTC)


 * The point is, the statement that was the subject of this theorem was entered as it was based upon how it appears in the work cited. We strive to maintain accuracy to such cited works in order to provide a smooth journey for someone tracing this through. Therefore, however specific, idiosyncratic exemplic or plain lame a result is here, if it comes from a cited work it is preferable to write a completely separate page to put the most general and strongest result in, and reference the original page as a special case / example / corollary of that general case.


 * We had this argument when someone tried to replace all the real analysis work with general cases that encompassed the complex case as well; all well and good, but a reader who is studying real analysis and has not progressed to complex will have trouble following it. There's a similar ongoing thing that happens in topology: a result for metric spaces, proved specifically within the framework of metric spaces, is often casually discarded in favour of something for the general topological space, and thence something of value has been lost. --prime mover (talk) 21:46, 14 October 2014 (UTC)


 * I can see that making sense in the grand context here. PhD student in math, but new to the site. I'll move the current source to the new "Corollary 2" page. Would that be sufficent? Or should I swap the roles / page names for what are currently listed as the theorem & corollary 2? --Mdibah (talk) 22:42, 14 October 2014 (UTC)


 * I already did this: thought it better to JFDI than talk about it. :-) --prime mover (talk) 06:50, 15 October 2014 (UTC)

Infinity Cases, chance of error
Young's Inequality for Convolutions also extends to when $p$, $q$, and $r$ might be infinite, i.e. $1 \leq p,q,r \leq \infty$ and not just $p,q,r \in \R_{\geq 1}$. (see wikipedia for a statement with this.

Probably need to add the infinite case as a seperate section in the proof. Also, there are a couple times where denominators appear with $r-p$ or $r-q$. In the case that $r=p$ or $r=q$, we would be dividing by zero...I think it still works out if we interpret this as indicating the $\Vert \bullet \Vert_\infty$ norm, but my eyes are going a bit crossed. Might need to separate these out as another case...blerrgh. Might 3 cases actually: $r=p \neq q$, $r=q \neq p$ (follows from symmetry?), and $r=p=q$. Or maybe just show that things work for $p=r$, symmetry to get $p=q$, then $p=q=r$ follows? --Mdibah (talk) 23:08, 14 October 2014 (UTC)


 * Before we can post this up, first we have to define the $\Vert \bullet \Vert_\infty$ norm. Tedious but necessary. One of the thing that causes would-be contributors to this site to lose interest very quickly is the realisation that all results must be backed up by definitions of all the objects used, and by an establishment of all the results contributing to such. The definitions of these objects is often too tedious for them to bother with.


 * Our usual approach till now has been to plod through source works and progressively build up a body of definitions and proofs, and so working upwards towards the big exciting results that people are so keen on putting in place. But it takes time and dedication. --prime mover (talk) 05:16, 15 October 2014 (UTC)


 * Sorry, I see that $\Vert \bullet \Vert_\infty$ does seem to have been defined.


 * My point rests, though.


 * If the "conjugate exponents" definition were expanded so as to take on all the conditions under which YIFC is valid, then there would be no need to include all the special cases in the definition of the theorem, just state that $p, q, r$ are, etc. Then the special cases can all follow as every more simple and special-case corollaries. --prime mover (talk) 07:02, 15 October 2014 (UTC)