Definition talk:Norm/Bounded Linear Functional
I think however this is refactored this page will look like a mess. The problem: a load (most things from Conway?) of functional analytic definitions are done in the context of Hilbert spaces only, where usually only a normed space is required. (a few like, Definition:Closed Linear Span only need a topology) This needs to be fixed before I can do a lot of work in functional analysis since I need to invoke these concepts for general normed or Banach spaces. It was first brought up last year on Definition talk:Hilbert Space#Improving Definition Page, for historical reference.
Then there's annoying technicalities. Since strictly speaking an inner product space is not a normed space, rather is readily and canonically associated with one, (an inner product space only has a vector space and an inner product. Then that inner product induces a canonical norm (there are others, just take constant multiples of this one for instance) which gives the topological stuff) I've been splitting foundational definitions into "Inner Product Space" (completeness is most often unnecessary) and "Normed Vector Space". Eventually there'll come a point where it's basically fine to treat an inner product space like a specific kind of normed vector space, of course, but I don't think the distinction should be blurred at the lowest level. It's one of these annoying cases where basically no mathematical text discusses this explicitly, either since it might confuse someone seeing the material for the first time, because third year+ students are trusted to make the connection themselves, or because it's such a trivial detail, so we have to fill the gap with little template. (do let me know if I'm wrong in this assessment) Indeed, if I was the one creating these pages, I probably would've just said "normed space" without contemplating it too much.
But I feel like the pages might then look a bit silly, basically the only difference between the subpage for Inner Product Spaces and the one for Normed Vector Spaces will be line saying "let $\norm \cdot_X$ be the inner product norm of $X$" and the rest of page will be almost word-for-word the same. It's vaguely similar to how metric spaces are very technically speaking not a topological space, (though saying this did make me double take and I got affirmation from this MSE post that I'm along the right lines here) but rather a set with a map that induces a topology in a natural way, so is readily associated with a topological space. We then treat metric spaces as "essentially a subclass of topological spaces" (just as we treat inner product spaces as "essentially a subclass of normed spaces") and apply definitions and theorems starting with "let $\struct {T, \tau}$ be a topological space" to metric spaces without worrying too much. But definitions in metric spaces often have nicer expressions than those in topological spaces, so splitting into cases is worthwhile. While inner products do add a lot of extra structure, it's not really enough for definitions to look much different. I'm wondering if anyone has a nicer idea on how to do this. Caliburn (talk) 16:37, 21 January 2022 (UTC)
- I feel your pain. As you say, I went through similar headscratching when setting up the initial metric spaces and topology pages.
- My advice is: don't worry about it looking silly. Each page is a standalone entity, and a person dropping in at random would quite possibly / probably need to be reminded of the distinction.
- If you feel ambitious, write a "warning" page or something, similar to what we sometimes do, explaining the above in a transcludable form, fully linkified and $\mathsf{Pr} \infty \mathsf{fWiki}$ed up, so it can then be "just" transcluded in as we like. (A transcluded page is IMO better than a template, but having said that, if implementing a template results in an even neater presentation and more streamlined way of developing this area of mathematics, go for the template option.)
- As I say, I don't have a dog in this fight, but the effort and care you are taking over this is seriously appreciated. If we end up highlighting something important that no other source manages to do, then we (collectively) have achieved something worthwhile. --prime mover (talk) 18:20, 21 January 2022 (UTC)
- Oh, and I meant to add: if we subsequently come up with a neater and altogether "better" way to present it, then that will then be attacked as and when. But my advice is: carry on as you are doing, this is great. --prime mover (talk) 18:21, 21 January 2022 (UTC)
Mostly done. I've changed the definition to insist $V \ne \set 0$, otherwise in Definition 2 and 3 we are taking the supremum over the empty set. (which is conventionally $-\infty$ and not what we want) Definitions 1 and 4 are still valid for the trivial $V = \set 0$. Anyone's welcome to change it back but I think it's a lot of awkwardness for a trivial case. I will change results as I go along. Caliburn (talk) 17:24, 22 January 2022 (UTC)
- I've commented on this from the context of Definition:Norm/Bounded Linear Functional/Normed Vector Space -- $\set 0$ is the trivial module, and for a vector space a module is specifically defined as being a unitary module, and from Trivial Module is Not Unitary it cannot be trivial. So I don't believe we need to specify that $V \ne \set 0$, and indeed it is confusingly misleading anyway, as the underlying abelian group that is $V$ is not necessarily a group whose identity is actually $0$ (the number). So I would insist that $V \ne \set 0$ should be removed.
- It is worth writing a transcluded subpage "Warning" or something, saying that by definition $V$ is necessarily not the trivial module, and transclude it into every definition page in sight, and if we do at this stage reference the $0$, we instead call it $e_V$ or something that reminds us of the general nature of the underlying structure of $V$ that it is in fact a general abelian group and not necessarily a number field. --prime mover (talk) 05:40, 11 July 2022 (UTC)
- Conventionally, $0$ or $0_V$ has denoted the identity of the underlying abelian group of a vector space $V$. This is the same as $0$ or $0_R$ denoting the additive identity of the underlying abelian group in a ring $R$.
- I have always understood $\set{0}$ to be a vector space.
- If I'm not mistaken, by definition of trivial module, $\set{0}$ is "a" trivial module and not "the" trivial module. Moreover, the statement of the theorem Trivial Module is Not Unitary explicitly excludes $\set{0}$ from the conclusion of the theorem that a trivial module is not unitary. In fact the theorem Trivial Module is Not Unitary shows that $\set{0}$ is the only unitary trivial module. That would explain the definition trivial vector space.
- So $\set{0}$ is not excluded as a vector space by definition. --Leigh.Samphier (talk) 01:36, 12 July 2022 (UTC)
- Of course the definition of the norm is a is a map from $V$ to the nonnegative reals:
- $\norm{\,\cdot\,}: V \to \R_{\ge 0}$
- So the supremum of the empty set $\sup \O$ is over the interval $\closedint {0} {\infty}$ (and not over the interval $\closedint {-\infty} {\infty} $), so equals $0$. Which is correct. So excluding the $\set{0}$ is technically unnecessary. Although an expalantion of this is warranted. --Leigh.Samphier (talk) 02:00, 12 July 2022 (UTC)
- Of course the definition of the norm is a is a map from $V$ to the nonnegative reals:
- Better still, add a comment to the pages that need it, that by convention the supremum is taken in the context of the totally ordered set of the nonnegative reals $\closedint {0} {\infty}$ so that the $\sup \O = 0$. So now the trivial vector space is included. You may want to add a simple theorem that the norm of the trivial linear functional from the trivial vector space with trivial norm to any other normed vector space is zero. Just to make this explicit --Leigh.Samphier (talk) 02:26, 12 July 2022 (UTC)
- The best and only rule of thumb, in cases like this, is to do what it's got in the text book. --prime mover (talk) 16:42, 12 July 2022 (UTC)