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alexandræ · <i>L<sup>p</sup></i> spaces measure "how close" to being integrable a function is (mostly reportative)

Lp spaces measure "how close" to being integrable a function is (mostly reportative)

april 28th, 2023

    integrability is usually quite close to summability. indeed, according to riemann integration, positive functions can be bounded by below and above by step/rectangular functions, with everdecreasing widths that tend to naught, a bit like so :


    if we can bound a function f by above with a single step (or, equivalently, finitely many steps), like ε·1[-nn] (which is ε on [-nn], 0 elsewhere) for some ε > 0 and n ≥ 0, then the integral of f over ℝ is bounded by above by the integral of ε·1[-nn], which is 2εn < ∞ ; we say such functions are both "bounded" (vertical bound) and "compactly supported" (horizontal bound). well, according to the lebesgue measure anyway, otherwise it can still yield infinite integrals : with the counting measure, any bounded compactly supported f such that |f -1(ℝ\{0})| = ∞ is not integrable, which is the case of most usual functions over ℝ.

    the usual (lebesgue) measure λ1, thankfully, does have the property that all bounded compactly supported functions have finite integrals. these are the most integrable functions one could think of, but they're obviously not the only ones. notably, for all ε > 0, here are a few examples :
  • x ⟼ 1/x1+ε is integrable over [1, +∞),
  • x ⟼ 1/(x log1+ε(x)) over [e, +∞),
  • x ⟼ 1/(x log(x) log(log(x))1+ε) over [ee, +∞),
  • x ⟼ 1/(x log(x) log(log(x)) log(log(log(x)))1+ε) over [eee, +∞),
  • basically x ⟼ 1/(x log(x) log(log(x)) ··· logk(x)1+ε) over [e↑↑k, +∞), regardless of k ≥ 1.
    (we'll call these functions ɕk for each k ≥ 1, standing for the way i call them : the "cauchy border functions" (ɕ looking like "c" and "∂" merged together) to the set of non-integrable functions, while always being integrable themselves).
    these exemples come from applications of the cauchy condensation test for series convergence (which you can see in the very wikipedia page), but since these are decreasing and positive functions over these intervals, their series bound their integrals by above, so if their series converge, so do their integrals. however, this doesn't go the other way around. indeed, although we were able to get functions with horizontal asymptotes that were integrable, we can also get functions with vertical asymptotes that are integrable ! we can even show one stems from the other, from a purely geometrical perspective : you can rotate the graph of, let's say, 1/x² for x ≥ 1 by 90° anticlockwise, to get the graph of 1/√x for -1 ≤ x < 0 and yet, the area under the curve only increments by 1 (because you have to add the area of [-1,0]×[0,1], which is 1) with this process, so it remains finite ; actually, with this very line of reasoning, we can argue [-1,0)1/(-x)1/k) = 1 + [1,+∞)1/xk. we can notice powers have been literally inverted when the horizontal asymptote has been rotated to a vertical asymptote, and that's where powers seem to come into play into seeing "how close" to being integrable a function might be. it's not like we could count on topological closure to give us a satisfying grasp of that anyway...

    indeed, we can notice that, if we call Ʒ the set of integrable functions from U ⊊ ℝ to ℝ, let's say, without loss of generality, Ʒ ⊂ ℝℝ\{0}, and Ʒc = ℝℝ\{0}\Ʒ the non-integrable functions over that same set, we have that cl(Ʒ) ⊈ Ʒ, because of all the non-integrable functions in the closure of Ʒ we've shown just before. therefore, Ʒ isn't closed, and Ʒc isn't open. to obliterate all hope, there's a very simple way to generate a sequence of non-integrable functions that tend to an integrable function : take f defined over ℝ\{0}, (f/n)n ≥ 0 tends to 0 almost everywhere, which is integrable. therefore, Ʒ and Ʒc are neither open nor closed. moreover, we can even show that cl(Ʒ) = cl(Ʒc) = ℝℝ\{0}, since if we take f0 non-integrable f1 integrable, we have (f1 + f0/n)n ≥ 0 a sequence with values in Ʒc yet tends to f1, and (1[-n,n]·min{f0,n})n ≥ 0 with values in Ʒ but tends to f0. du coup, pour toute fonction f ∈ ℝℝ\{0}, we have d(f,Ʒ) = d(fc) = 0 in that sense... just as i said, not very satisfying. that's why we'll use Lp norms instead.

    for f : U ⊆ ℝm ⟶ ℝn, the Lp norm is defined over a measurable U ⊆ ℝm as (U |f(x)|pdµ(x))1/p, with µ usually being the lebesgue measure λm over ℝm, and |·| being a norm over ℝn (they're all equivalent). in other words, it evaluates whether |f(x)|p is integrable in the most usual sense possible. when |f| is bounded over a measurable set V ⊆ U, then f is Lp(V, 𝔅V, λm) if one of two possibilities are satisfied : either V is bounded, or |f(x)| = O(ɕk(x)) for some k ≥ 1 and ε > 0 as |x| → ∞ (warning : this is sufficient, but not necessary !). when |f| is not bounded over V' a compact subset of U, since ɕk is strictly decreasing and positive over [e↑↑k, +∞), we can invert ɕk and we'll get an exact formula to go from one integral to the other like so : k)ɕk -1(x) dλm = [e↑↑k,+∞)ɕk(x) dλm + ɕk(e↑↑k)·(e↑↑k), and we generalize it to any power of the integrand to help evaluate whether it's Lp or not, like so :
[(0,ɕk(e↑↑k)p]ɕk -1(x)1/p dλm = [e↑↑k,+∞)ɕk(x)p dλm + ɕk(e↑↑k)p·(e↑↑k).
hence, for f to be Lp(V'), either |f| is bounded over V', or around all x0 where there's a vertical asymptote in f, we have that |f| = O(1/x1/p) or |f| = O(ɕk -1(xx0)1/p) for some ε > 0 and n ≥ 1 which is, once again, sufficient, but not necessary. but because of that, the greater p is, the quicker you'll have to reach your vertical asymptotes to be Lp, although you can converge much slower to 0 when reaching unbounded parts of your domain of definition (and its closure set, because much like the drawing measure, ∫cl(D)f = ∫Df). therefore, for vertical asymptotes, the greater p is, the more being Lp means it's "very" integrable over all bounded sets, for it "resists" to all Lx norms for 1 ≤ x ≤ p. however, in terms of horizontal asymptotes, it can spread out much further and, in a way, "less and less" integrable, strictly speaking.

    in other words, the larger p is, for a function f to be Lp, the more |f| can spread horizontally, tending to 0 slower and slower to ±∞, whereas |f|'s vertical asymptotes have to get tighter and tighter, geometrically approaching half-lines. here's a visualisation of Lp functions, where p goes from 1 to 5 :
    we can now motivate the exageration of this trend to the extent of L : the L norm is what we sometimes call the essential supremum, ie the function's supremum outside of a neglectable subset (ie of measure 0). hence, even though sup(1) = 1, the set ℚ is neglectable, so esssup(1) ≤ 0, meaning esssup(1) = 0 since 1 is positive. for a continuous function, esssup = sup, but otherwise it can very much not be, just like the latter example.

    you could argue this asymmetry is quite weird, and you'd be right. we'd basically want to say that a function is more and more integrable if it's Lp over bounded sets with large enough p, and Lq over unbounded sets with smaller q. although, we'd want p and q to be linked with each other with some kind of relation. one such possible relation, where the larger p is, the smaller q is, with 1 ≤ pq ≤ ∞, is 1/p + 1/q = 1. when this very relation is met, we actually a name for it : we say that p is conjugate to q, and vice-versa. this has a name of its own, because it's of great importance in functional analysis : the "dual" space of Lp, in terms of continuous linear functions instead of simply linear functions, is Lq, when p is conjugate to q and vice-versa. the hölder inequality also states that, for all f ∈ Lp and g ∈ Lq, ||fg||L1 ≤ ||f||Lp·||g||Lq, so with a bunch of norm operators you might get some lipschitzian shenanigans, and we know lipschitzian + linear ⇒ continuous even with infinitely many dimensions so it's pretty based ngl.

tl;dr. we could argue that f is "more" integrable for larger values of p ≥ 1, if f is Lp over bounded sets, and Lq over unbounded sets, with q = ∞ when p = 1, or 1/p + 1/q = 1 otherwise.
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