Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
For completeness and for linking-purposes, I thought an entry p-norm was missing. So I created one.
I have added cross-links with the Idea-sections at sequence space and Lebesgue space and with the Examples-section at normed vector space. Created a stub for Minkowski’s inequality, so far containing essentially nothing but a pointer to Todd’s p-norms (toddtrimble), which I vote for copying to there.
Thanks, Urs. I didn’t bring that entry on my page to a form I was satisfied with, but I may come back to wrap it up, and then I’d consider copying the contents over to the nLab.
I added some material on $p \lt 1$.
There is a great deal on Minkowski's inequality already at Lebesgue space, which is where Urs's link above points to (the stub is at Minkowki’s inequality). Perhaps you can stock your stub by moving that material, Urs; if not, then I would put it at p-norm preferably to Lebesgue space.
Toby, thanks for alerting me of the material on Minkowski’s inequality over at Lebesgue space. I had indeed missed that, sorry. Now I have copied the proof given there to Minkowski’s inequality. It’s a bit redundant, but I think it does not hurt to have this duplication.
Over at Lebesgue space, is there not a lack of emphasis that one needs to divide out functions $f$ whose $\Vert f \Vert_p = 0$ before getting an actual norm?
To un-gray a link at Lebesgue space, I’ve written Hölder’s inequality. Now just about everything I was driving at in my web notes is incorporated into the nLab. There’s still some cross-linking to do.
Thanks, Todd!!
@Urs #4:
Good, I thought that you had rejected that in favour of what Todd had written or something.
Over at Lebesgue space, is there not a lack of emphasis that one needs to divide out functions $f$ whose $\Vert f \Vert_p = 0$ before getting an actual norm?
I suppose. The statement does say ‘(equivalence classes of)’ in an appropriate space, but it's rather brief.
Okay, I have made that a tad more explicit here.
This might be a good place to remind everybody that, when you use vertical bars (single or double) as delimiters in iTeX, you generally need to enclose them within braces (or use \left
and \right
) to ensure proper spacing. (This actually applies to any delimiters that are either undirected or op-directed.) Thus,
$$ |x| - \|x\| - [x[ - ]x] $$
produces
$|x| - \|x\| - [x[ - ]x]$(which is incorrect), but
$$ {|x|} - {\|x\|} - {[x[} - {]x]} $$
produces
${|x|} - {\|x\|} - {[x[} - {]x]}$(which is correct), and
$$ \left|x\right| - \left\|x\right\| $$
produces
$\left|x\right| - \left\|x\right\|$(which is also correct, but \left[x\right[
produces an error in iTeX).
Note that iTeX does not support \lvert
, \rvert
, \lVert
, and \rVert
, which are directed versions of |
and \|
(although it supports the synonyms \vert
and \Vert
for the undirected versions), nor does it support \mathopen
and \mathclose
(which is how you’re supposed to apply the proper directions to undirected and op-directed delimiters in real TeX).
In some cases, you need braces inside the delimiters too! Thus, even
$$ {|-|} - {\|-\|} - {[-[} - {]-]} $$
produces
$\left] {|-|} - {\|-\|} - {[-[} - {]-]}$(which is incorrect), but
$$ {|{-}|} - {\|{-}\|} - {[{-}[} - {]{-}]} $$
produces
${|{-}|} - {\|{-}\|} - {[{-}[} - {]{-}]}$(which is correct), while
$$ \left|-\right| - \left\|-\right\| $$
produces
$\left|-\right| - \left\|-\right\|$(which is correct, as is using \lvert
etc or \mathopen
and \mathclose
in real TeX.)
I usually put in the outer braces as a matter of good habit, even though they may not be necessary (depending on the surrounding characters). I usually don't bother with the inner braces, since they are needed so rarely, but I have used them in the past.
In principle, superscripts and subscripts should be inside the braces to get the correct heights, although it seems to me that this won't make a visible difference unless your delimiters are too small anyway. Thus,
$$ {|\frac{x}{y}|}^2 - {\Big|\frac{x}{y}\Big|}^2 - {\left|\frac{x}{y}\right|}^2 $$
produces
${|\frac{x}{y}|}^2 - {\Big|\frac{x}{y}\Big|}^2 - {\left|\frac{x}{y}\right|}^2$(which is in principle incorrect), and
$$ {|\frac{x}{y}|^2} - {\Big|\frac{x}{y}\Big|^2} - {\left|\frac{x}{y}\right|^2} $$
produces
${|\frac{x}{y}|^2} - {\Big|\frac{x}{y}\Big|^2} - {\left|\frac{x}{y}\right|^2}$(which is in principle correct, although I can only tell the difference in the first term, which one should not really use).
(If you read the previous comment before this comment appeared, then look at it again for additions.)
Thanks, yes, I am aware of this and I keep adding these extra braces in my edits. Did I not in this case? Sorry.
Toby might have also been tacitly addressing me (I also knew it but hadn’t followed through in all cases). In any case, I think I’ve fixed just about all of them.
My comment #9 is a general one, not addressed to anybody in particular. (I'm pretty sure that Urs is adding these braces most of the time, if not always, but I also know that I forget them occasionally.) Since I had to fix some, I put in the reminder, without worrying about who it was that left them out.
You should move your #9 to the HowTo so that next time you can send a reminder just by giving a pointer!
OK, I wrote HowTo#vbar.
Thanks!
At p-norm I moved the technical discussion of generalizations for $0 \leq p \lt 1$ out of the Idea-section, into a dedicated new subsection Generalizations.
Good idea, and it allows one to state the facts about these a little more leisurely.
1 to 19 of 19