Exercise in finite measure space.

Question

Let $(X,\mu)$ be a finite measure space and $f:X\to[0,\infty]$ be a measurable function. Define a function $\varphi:\mathbb{R}^+\to\mathbb{R}^+$ by $\varphi(r)=\left(\frac{1}{\mu(X)}\int_Xf^rd\mu\right)^\frac1r$. Prove that $\varphi(r)\leq\varphi(s)$ if $r<s$.

Answer 1

We have $$\varphi(r) = \left(\frac1{\mu(X)}\int_X f^r\ \mathsf d\mu\right)^{\frac1r} = \left(\int_X f^r \ \frac1{\mu(X)}\ \mathsf d\mu \right)^{\frac1r}. $$ Now, by Hölder's inequality, in general, for measurable functions $F, G$ and conjugate exponents $p,q$, that is, $\frac1p + \frac1q = 1$ where $1<p,q<\infty$, $$\int_X |FG|\ \mathsf d\mu \leqslant \left(\int_X \ |F|^p \mathsf d\mu\right)^\frac1p\left(\int_X |G|^q \ \mathsf d \mu\right)^\frac1q.$$ Set $F = f^r$, $G = \frac1{\mu(X)}$, $p=\frac sr$, and $q = \frac s{s-r}$. Then (as $f$ and $\mu$ are non-negative we can omit the absolute value signs) it follows that \begin{align} \phi(r) &= \left(\int_X FG\ \mathsf d\mu\right)^{\frac1r}\\ &\leqslant \left(\int_X F^{\frac sr}\ \mathsf d\mu\right)^{\frac r{rs}}\left(\int_X G^{\frac s{s-r}} \right)^{\frac{s-r}{rs}}\\ &=\left(\int_X f^s\ \mathsf d\mu \right)^{\frac 1s}\left(\int_X \left(\frac1{\mu(X)}\right)^{\frac s{s-r}} \ \mathsf dx\right)^{\frac{s-r}{rs}}\\ &=\left(\int_X f^s\ \mathsf d\mu \right)^{\frac 1s}\left(\mu(X)^{\frac s{r-s}}\mu(X) \right)^{\frac{s-r}{rs}}\\ &=\left(\frac1{\mu(X)}\int_X f^s\ \mathsf d\mu \right)^{\frac1s}\\ &=\varphi(s), \end{align} as desired.