Thursday, March 07, 2013

in medias res: a three-star rating .. (UPDATED).

// initially posted: 2013-03-07 @ 13:34EEST

today i took the double dual $(V^*)^{**}$ of a certain dual Banach space $V^*$.
it was rather confusing, but seemed like a good idea at the time ..

// updated: 2013-03-08 @ 13:30EEST

i think i've either jumped off the deep end or gotten nostalgic for topics i learned during my student days. in the last two weeks i've gone to the library four times to read the books on functiοnal analysιs by rudin and by dunford-&-schwartz.

more and more i am impressed by the latter volume, but that's more an artifact of my current mathematical necessities; lately, you see, i have become wont of more gory details regarding some seeming familiar Banach spaces. this sounds like pure folly ..

.. but i want to understand better the space $L^\infty(\mathbb{R}^n)$ and its Banach dual! [1]
there is already a handy characterisation, but it uses words like "finitely additive" and "set function" .. which, admittedly, worry me. the awesome thing about D&S is that they have an entire chapter written in terms of these things. apparently, most of the usual measure and integration theory runs analogously through.

more to the point, i don't have to re-prove everything myself; these guys are lifesavers!
on a related note, these two passages from D&S caught my eye [2].
  1. Theorem III.5.13 & 14 (Алексaндров). [If $\mu$ is] a bounded, regular, complex-valued, [finitely-]additive set function defined on a field $\Sigma$ of subsets of a compact topological space $S$, [t]hen $\mu$ is countably additive ... [Moreover] there is a unique regular, countably additive extension to the $\sigma$-field determined by $\Sigma$.

    i'm probably too naive, but it's hard for me to appreciate the difference between finitely additive measures and the usual (countably additive) ones. in light of this theorem, though, there are two points that come to mind:

    (A) once the set function is assumed regular --- i.e. that it has "good limits" for sets that fit the topology of the space --- then on compacta, there is no difference between finitely- and countably- additive. on the other hand, set functions that are strictly outside of $L^1(\mathbb{R}^n)$ must therefore lack these nice limit-properties, which means that they will be hard to work with!

    (B) this seems to be one of the rare criteria for checking whether a bounded linear functional of $L^\infty(\mathbb{R}^n)$ is actually a Lebesgue integrable function .. or, more precisely, part of a criterion. the theorem guarantees that we can work with measures as usual, so the Radon-Nikodym theorem would apply in this setting.

  2. some parts of the book provide suggestions and instructions for how to proceed. for example, on pg. 122, III.3.6 it reads:

    The reader will more easily perceive the significance of the somewhat complicated conditions (ii) and (iii) in the following [Theorem 6], if [s]he reads the statement and proof of Theorem 7 after the statement of Theorem 6 but before its proof.

    wait .. so why didn't they just switch the order of theorems 6 and 7? (-:

// updated: 2013-03-08 @ 17:30EEST

the further i read on, the more uneasy i feel. some of these constructions are just .. creepy: i can't think of another way to put it.

one result is outright unnerving:
Theorem IV.6.18-19: If $S$ is a compact Hausdorff space, then there exists a(nother) totally disconnected, compact Hausdorff space $S_1$ and an embedding $i: S \hookrightarrow S_1$ so that $i(S)$ is dense in $S_1$ and that induces an isometric isomorphism $$ i_*: L^\infty(S,\mu) \,\to\, C(S_1). $$
to get a sense of what i mean, one can take as nice of a space $S$ as possible and extend it to some $S_1$ so that $S$ is dense, yet incredibly scattered within $S_1$ .. to the extent that characteristic functions $\chi_A$ of subsets $A$ in $S$ extend to continuous functions in $S$.

the real kicker is that the extension is also a characteristic function of some subset in $S_1$ as well (IV.9.10). roughly speaking, then, the continuous extension is not done through interpolating between the values $0$ and $1$, but by sufficiently separating the preimage sets in $S_1$ between these values .. and none of the subsets $A$ are chosen in advance!

in short, $S_1$ must be a weird, messed-up space.

.. i think i'm going to stop reading for today.

[1] .. and in case you're keeping count, no: $[L^1_\mu(\mathbb{R}^n)]^{**} \cong [L^\infty_\mu(\mathbb{R}^n)]^*$ is only two stars. (i might get to the third one in a later update.)

[2] the square brackets []'s indicate my reformatting.

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