# Holder's inequality

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

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

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

Now you have to talk a little :

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