For $\Omega \subset \mathbb R^n$ and a measure $\gamma$ such that $\gamma(\Omega)=1$, I saw a result which states that $$\left(\int |x-y|^p d\gamma \right)^{\frac{1}{p}} \leq diam(\Omega)^{\frac{p}{p-1}}\left(\int |x-y| d\gamma \right)^{\frac{1}{p}} $$
I was trying to prove this for $p=2$ using Holders inequality, since $1=\frac{1}{2}+\frac{1}{2}$ we have $$\left(\int |x-y|^2 d\gamma \right) \leq \left(\int (|x-y|^{1/2})^2 d\gamma \right)^{\frac{1}{2}} \left(\int (|x-y|^{3/2})^2 d\gamma \right)^{\frac{1}{2}} $$ But this doesn't give the desired inequality. Can anyone please figure out where I am making the mistake?
You have $\mathrm{diam}(\Omega)=\sup\limits_{x,y\in \Omega}|x-y|$.
Hence
$$\left(\int\limits_{\Omega} |x-y|^p d\gamma\right)^{1/p} \leq \sup\limits_{x,y\in \Omega}|x-y|^{\frac{p-1}{p}}\left(\int\limits_{\Omega} |x-y| d\gamma\right)^{1/p}\\=\mathrm{diam}(\Omega)^{\frac{p-1}{p}}\left(\int\limits_{\Omega} |x-y| d\gamma\right)^{1/p} $$