Talk:Young's Inequality for Convolutions

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)