# Hölder inequality on probability space

I just found the following Hölder inequality on a space $\Omega$ with $\lambda(\Omega)=1$ that I don't understand:

It says. Let $(u,v)$ be conjugate exponents, then we get

$$||f||_2 \le ||f||_1^{\frac{1}{2u}} ||f||_{v+1}^{\frac{v+1}{2v}}.$$

Does anybody see what is going on here and how to derive this result?

• Assume without loss of generality that $f\geqslant0$ and let $g=f^{1/u}$ and $h=f^{1+1/v}$ then $f^2=gh$ hence $$\int f^2=\int gh\leqslant\left(\int g^u\right)^{1/u}\left(\int h^v\right)^{1/v}.$$ The LHS is $\|f\|_2^2$, identifying the factors in the RHS yields the desired result. – Did Oct 25 '15 at 13:04
• @did: may I suggest you post your answers as such, and not as comments? – Martin Argerami Oct 25 '15 at 18:24
• @MartinArgerami I would rather the OP transforming the hints in them into some full answer that they would post below. – Did Oct 25 '15 at 18:42
• But that will likely not happen, and this will be one more of the thousands of unanswered question in the site, where an answer appears in the comments. – Martin Argerami Oct 25 '15 at 20:14