Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rudin Lp question

  1. Dec 2, 2007 #1
    This is problem 3.9 (from second edition of daddy rudin). (This isn't homework).

    Suppose f is Lebesgue measurable on (0,1), and f is not essentially bounded.

    Is it true that for every function g on [tex](0,\infty)[/tex] such that
    [tex]g(p) \rightarrow \infty[/tex] as [tex]p \rightarrow \infty[/tex]

    one can find an f such that
    [tex]\|f\|_{p} \rightarrow \infty[/tex] as [tex]p \rightarrow \infty[/tex] with[tex]\|f\|_{p} < g(p)[/tex] for all sufficiently large p?

    Any hints are appreciated.

    I know that given any function f, the set of p for which [tex]\|f\|_{p} < \infty[/tex] can be any connected subset of [tex](0,\infty)[/tex].

    Hence, given any r with 1 < r < [tex]\infty[/tex], we can find a fuction such that the [tex]\|f\|_{p} < \infty[/tex] if p < r but that the norm is infinite for p >= r.

    But say we choose g(p)=p. The above mechanism fails to find a suitable f, since we can only choose some r finite, then g(p) < [tex]\|f\|_{p}[/tex] for all p > r.

    On the other hand, I don't know how to prove that such a function f can't exist. But every function I can think of that meets the above criteria always has a finite inflection point where the norm goes infinite.

    So I'm stumped.
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2
    I think this should be true. You just need to make [tex]f[/tex] have a singularity that is really 'weak'. For example, you could have the function look like [tex]\frac{1}{\log(x)}[/tex]. That one is in [tex]L^p[/tex] for all large [tex]p[/tex] but the [tex]L_p[/tex] norm is growing as a function of [tex]p[/tex]. Of course this isn't a full solution to the problem but it should give you some idea of what's going on.. I hope it helps.
  4. Dec 3, 2007 #3
    I think I'm getting closer. The hint was actually in a different problem.

    I think the function i need is like:


    I'll keep plugging away.
  5. Dec 3, 2007 #4
    Well, you won't get a specific function. The idea is that given any [tex]g[/tex] that specifies an asymptotic growth condition, you can find a function whose [tex]L^p[/tex] norms grow even slower than [tex]g[/tex]. You really don't need to construct an explicit one (it might be extremely hard if [tex]g[/tex] is really complicated).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook