A function f(x) is said to be in Lp if the integral of |f(x)|p is finite. So if you know f is in Lp for some value of p, can conclude that f is also in Lq for some q? In general, no. But for two important special cases, yes. The special cases have to do with the domain of f.
If f is a function on a finite measure space and p > q, then every function in Lp is in Lq. In particular probability spaces are finite measure spaces, and so if a random variable’s pth moment exists, the qth moment exists.
If the domain of f is the integers with counting measure, f is a sequence and the “integral” of |f(x)|p is simply a sum. Now the relationship between Lp spaces is reversed: p > q, every sequence in Lq is in Lp.
There are two reasons a function could fail to be in Lp: singularities and fat tails. In a finite measure space, there are no tails; you only have to worry about singularities. On the integers, there are no singularities; you only have to worry about tails.
Now consider functions on the whole real line. Membership in Lp says nothing about membership in Lq because now you can have singularities and fat tails.
7 thoughts on “How Lp spaces nest”
What if you have a finite measure on the integers, e.g. mu(n)=(1/2)^abs(n)? If f is in a particular Lp, is it in all Lp? Is there really only one Lp?
Steve: the precise result is this. (According to Wikipedia, which is always right except when it’s wrong.) Let M be the set of all measures of subsets of your space. Then, if p<q, then Lp < Lq if M has a positive lower bound, and Lq < Lp if M has a finite upper bound.
So if your space is the positive integers with an exponentially decaying measure, then M has an upper bound but no positive lower bound, so bigger exponents mean smaller Lp spaces.
This is a matter of "singularities" rather than "fat tails": what you need to worry about is having your function get large on regions of small measure.
Thanks g! very interesting. Obviously all Lp spaces are the same for counting measure on a finite set. I wonder if there is a more interesting example where all Lp’s are the same.
Well, according to that same theorem all the Lp are the same iff our measure has both an upper and a lower bound on nonzero measures of sets. The upper bound means that the total measure is finite. The lower bound means that if we ignore stuff of zero measure we’ve just got a finite number of discrete bits, so it seems our space is “morally” a finite discrete space. But of course we can’t really ignore stuff of zero measure so let’s try to be more careful and see where the pathologies lie.
For every measurable A, we have mu(A) = 0 or b <= mu(A) = b, and all these sets are atoms: they have no subsets of smaller positive measure.
Now, a measurable function is constant almost everywhere on any atom. So, measure-theoretically, our space is “just like” a finite discrete space. Any function differs only on a set of measure 0 from a function that’s constant on each of our atoms.
So: no, there is no much more interesting example where all the Lp are the same. Unless I’ve goofed, which I often do.
Er, of course the inequality in the second paragraph was meant to be b <= mu(A) <= B.
Ah, I see what’s happened. A bunch of what I wrote disappeared because the software that handles comments thought it was an HTML angle-bracketed thing and stripped it out. Which is why, in addition to the inequality being wrong, it doesn’t make any sense. So let me try again, avoiding angle brackets. This is a replacement for the second paragraph of my comment beginning “Well, according to that same theorem”:
For every measurable A, we have mu(A) either zero or between b and B. Now either A is the union of two disjoint sets of strictly smaller measure, or not. If it isn’t, then in fact A is an atom: it has no subset of smaller positive measure. On the other hand, if it is the union of smaller things then they’re smaller by at least b. So we can repeat this recursively, and after a finite amount of splitting we end up with finitely many disjoint atoms. And we can do this starting with the whole space, which is therefore also the union of a finite number of disjoint atoms.
Thanks again g!