Let $\mu(x) < \infty$, and define, for $f \in L^1$, the average of $\int f d\mu$ to be $\frac1{\mu(x)} \int f d\mu$. Show that for $1 \le p < r < \infty$, these holds $$(f|f|^p)^{1/p} \le (f|f|^r)^{1/r}.$$ In particular, if $\mu(x) = 1$ (i.e. $\mu$ is a probability measure), then $||f||_p \le ||f||_r$.
I really don't know how to solve this question.
I think that $(f|f|^p)^{1/p} = (f^{1/p}|f|) \ge (f^{1/r}|f|)$ for $p<r$. Isn't it? Could you give any hint please?
Have you studied Holder's inequality. All you have to do is think of $|f|^{p}$ as the product of $|f|^{p}$ constant function $1$ and apply Holder's inequality to the exponents $\frac r p$ and $\frac r {r-p}$. Let $p_1=\frac r p$ and $q_1=\frac r {r-p}$. Then $p_1>1,q_1>1$ and $\frac 1 {p_1} +\frac 1 {q_{1}}=1$. Therefore $\int |f|^{p} \leq (\int ({|f|^{p}}^{p_1}))^{1/p_1} (\int 1)^{1/q_1}$. Now just plug in the values of $p_1$ and $q_1$ and take p-th roots on both sides.