Re: L^p norms
- From: The World Wide Wade <waderameyxiii@xxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 07 Oct 2006 14:26:23 -0700
In article <_SMVg.370$i84.346@trnddc01>,
"TCL" <tlim1@xxxxxxxxxxx> wrote:
Prove or disprove that to every positive function g on (0, infty) such that
g(p)-->infty as p-->infty, there exists a Lebesgue measurable function f on
(0,1) such that ||f||_p --> infty as p-->infty and ||f||_p <= g(p) for
sufficiently large p.
Note: Such an f , if exists, cannot be essentially bounded.
g >= 1 on some [p_0, oo). For each fixed n = 1, 2, ..., p ->
n^p*2^n/[g(p)^p] is bounded on [p_0, oo), so we can choose m_n >
0 such that m_n*n^p*2^n/[g(p)^p] < 1 on [p_0, oo). Making sure
the sum of the m_n's is < 1, choose pwdj subintervals I_n of
(0,1) of length m_n. Let X_n denote the characterisitic function
of I_n. Let f = sum(n=1,oo) n*X_n. Then int_(0,1) f^p =
sum(n=1,oo) n^p*m_n < sum(n=1,oo) [g(p)^p]/2^n = g(p)^p on [p_0,
oo).
.
- References:
- L^p norms
- From: TCL
- L^p norms
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