Talk:Even Functions in 2-Lebesgue Space form Closed Subspace of 2-Lebesgue Space

I think it should be $\map {\LL^2} \R$ throughout, since $\map {L^2} \R$ should denote the $L^p$ space with $p = 2$. Caliburn (talk) 20:50, 4 July 2022 (UTC)
 * We also might want to specifically declare when the ambient measure space is dropped to write $\R$ in favour of $\struct {\R, \map {\mathrm {Leb} } \R, \lambda}$ or $\struct {\R, \map \BB \R, \lambda}$. Eg. in some cases we might want the $\sigma$-algebra to be only the Borel sets, or we might want the measure space to be complete so we'll have to use all Lebesgue measurable sets. We should therefore pick a default and then specify when it strays from that. Caliburn (talk) 21:32, 4 July 2022 (UTC)


 * Now I am thinking that instead I shoulder refer to the $L^p$ space. I am working with normed vector spaces, while Lebesgue space is equipped with a seminorm, and I don't think we have enough material to negotiate our way. The source refers to $L^p$ as the space of "Lebesgue-measurable" of "Lebesgue-integrable" functions, but its introduction is sketchy so I would rather stay on the safer side. As for the ambient measure space, on one hand I would check what physicists do in quantum mechanics. That would give us a hint which measure spaces can be collapsed to just one symbol. On the other hand, at this moment I am staring at a book stating that $\map {L^p} {\R, \AA_B, m_B} = \map {L^p} {\R, \AA_L, m_L} \equiv \map {L^p} {\R, \rd x}$ which is usually written as $\map {L^p} \R$ since in the integral null sets are irrelevant and $L^p$ are equivalence classes and can be defined in terms of Riemann integrals without recourse to measure theory.--Julius (talk) 23:54, 5 July 2022 (UTC)


 * The sketchy part is that you're talking about pointwise behaviour of elements of $L^p$, which is not usually permissible afaik. The abuse of notation conflating a function with its equivalence class is usually just about integral properties and almost-everywhere equalities with some indeterminate, you wouldn't be talking about $\map f 0$ for $f \in L^2$. It'd have to be something like "almost-everywhere even", otherwise given an even representative of an equivalence class you could just change a value to make it not even but it's still the same object under this identification. When I did some stuff on Navier-Stokes, they used weak continuity to talk about behaviour of a null set. Caliburn (talk) 09:47, 6 July 2022 (UTC)


 * I think just replacing this with $\LL$ it's all good. Introducing the A.E. identification just makes it murky. Caliburn (talk) 09:52, 6 July 2022 (UTC)