Is it possible for a function to be in LpL^p for only one pp?

I’m wondering if it’s possible for a function to be an Lp space for only one value of p[1,) (on either a bounded domain or an unbounded domain).

One can use interpolation to show that if a function is in two Lp spaces, (e.g. p1 and p2,with p1p2 then it is in all p1pp2).

Moreover, if we’re on a bounded domain, we also have the relatively standard result that if fLp1 for some p1[1,), then it is in Lp for every pp1 (which can be shown using Hölder’s inequality).

Thus, I think that the question can be reduced to unbounded domains if we consider the question for any p>1.

Intuitively, a function on an unbounded domain is inside an Lp space if it decrease quickly enough toward infinity. This makes it seem like we might be able to multiply the function by a slightly larger exponent. At the same time, doing this might cause the function to blow up near zero. That’s not precise/rigorous at all though.

So I’m wondering if it is possible to either construct an example or prove that this can’t be true.

Answer

Robert’s and joriki’s examples are of course nice and explicit, but you can get examples on any subset of Rn with infinite measure. Here’s how:

Take a function f that is in Lp but not in Lq for q>p (on the unit ball B around zero, say). Now take a sequence x=(xn) that is in p but not in q for q<p (there are standard examples for both of these things). Now take disjoint balls Bn of volume 1 (disjoint from B) and consider g=f+xn[Bn] where [Bn] denotes the characteristic function of Bn. Obviously, is in L^q if and only if q = p. If q \lt p then \|x\|_q = \infty and if q \gt p then \|f\|_{q}^q = \infty.

I leave it to you to make that explicit and to modify it when your domain is not all of \mathbb{R}^n.

Attribution
Source : Link , Question Author : user1736 , Answer Author : t.b.

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