Holder Inequality question...
 From: "James" <James545@xxxxxxxxx>
 Date: Sat, 14 May 2005 09:03:23 0400
Dear all, I am trying to prove the following question :
If f is in L^p(0,1) for some p, 1 <= p < infinity, that
lim y^[(1p)/p] * integral_[0,y] f(x)dx = 0.
y>0+
I'm not sure if my solution is valid, and I was wondering if someone could
comment : Well,
y^[(1p)/p] * integral_[0,y] f(x)dx <= y^[(1p)/p] * integral_[0,y] f(x)dx
= y^[(1p)/p] * integral_[0,y] f(x)*1dx <= y^[(1p)/p] * f_p 1_q =
y^[(1p)/p] * f_p (y)^(1/q) = y^[(1p)/p + 1/q] f_p = y^(11/p)
f_p
The last term goes to 0 as y > 0+ since f is in L^p.
One question I have is the following : When I write f_p, I must mean
(integral_[0,y] f^p)^(1/p). That is, the integral goes from 0 to y, not 0
to 1. The reason is that I want to apply Holder to integral_[0,y]
f(x)*1dx, and that integral goes from 0 to y. Am I right? This means
that f must be in L^p(0,y) for all y < 1. Is it true in general that if f
is in L^p(0,1), then f is in L^p(0,y) for all 0 < y < 1?
Thank you for your assistance,
James
.
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