Let $\mu$ be a finite measure on a measurable space $(X,\Sigma)$. I want to prove that exists $C > 0$ so that $$ \| f\|_p \leq C \| f \|_{q,\infty}$$ when $1 \leq p < q < \infty$, where $$ \| f \|_{q,\infty} = \sup\limits_{\lambda > 0}\lambda \mu(\{x \in X: |f(x)| > \lambda \} )^{1/q}. $$ I think the key of the proof relies in the fact that $\mu$ is a finite measure, because it's not difficult to show a counterexample when you consider the Lebesgue measure on $\mathbb{R}$.
My best aproach consists of putting $$ \int_X |f|^p d\mu = \int_0^\infty \mu\{ |f| > t\} p t^{p-1} dt = \int_0^\lambda\mu\{ |f| > t\} p t^{p-1} dt + \int_\lambda^\infty \mu\{ |f| > t\} p t^{p-1} dt, $$ where $\lambda > 0$. For the first integral it's easy to see that it's $\leq \mu(X) \frac{\lambda^p}{p} = C_1 \lambda^p$, but I'm stuck with the second integral and I don't know how to continue.
Any idea of how to continue the proof? Do yo know any different approach to the solution? Thank you very much.
Best Answer
For the second integral, bound $\mu\{ |f| > t\}$ by $t^{-q}\left\lVert f\right\rVert_{q,\infty}^q$ and the integral can be therefore controlled by a constant times $\lambda^{-q+p}$. Then choose $\lambda$ appropriately.