# Holder's inequality

• November 26th 2010, 09:45 PM
noob mathematician
Holder's inequality
With s>1 and r>s. How can I show that $L^s \supset L^r$, for expectation with respect to a probability measure?
• November 27th 2010, 08:38 AM
Moo
Hello,

Show what you've done. Basically, you just have to apply Holder's inequality to $E[X^s]$, with $X^r$ as f and 1 as g. If X is in $L^r$ then $E[X^r]<\infty$, that's how it'll work.
Give it a try.
• November 28th 2010, 01:24 AM
noob mathematician
Holder inequality: $E|XY|\le (E|X|^p)^{1/p}(E|Y|^q)^{1/q}$
Let q=r/s
with r.v. $|Y|^s \text{ and } 1_\Omega$
Then $E|Y|^s\le (E|Y|^r)^{s/r}<\infty$?

Not very sure though
• November 28th 2010, 04:23 AM
Moo
This is exactly it ! (Clapping)

Now you have to talk a little :

Let's take $Y\in L^r$. Then $E[|Y|^r]<\infty$. - your calculations - . Thus $Y\in L^s$, which implies that $L^r\subset L^s$ : any element of the first belongs to the second.