# 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^[(1-p)/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^[(1-p)/p] * integral_[0,y] f(x)dx <= y^[(1-p)/p] * integral_[0,y] |f(x)|dx

= y^[(1-p)/p] * integral_[0,y] |f(x)*1|dx <= y^[(1-p)/p] * ||f||_p ||1||_q =

y^[(1-p)/p] * ||f||_p (y)^(1/q) = y^[(1-p)/p + 1/q] ||f||_p = y^(1-1/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)*1|dx, 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

.

**Follow-Ups**:**Re: Holder Inequality question...***From:*José Carlos Santos

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