Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 < p < \infty$
I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:
$$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$
The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.
By Hölder's inequality, we obtain:
$$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$
where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:
$$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$
I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated.
Best Answer
You have shown that
$$1 \leq \left(\int f^pd\mu\right)\left(\int g^qd\mu\right)^\frac{p}{q}$$
for all $p,q>0$ with $\frac{1}{p} + \frac{1}{q} = 1$. Without loss of generality, you can assume that $p\ge q$. Then the function $[0,\infty)\ni x\mapsto x^\frac{p}{q}$ is convex and Jensen's inequality yields
$$ \left(\int g^q d\mu\right)^\frac{p}{q}\le \int (g^q)^\frac{p}{q} d\mu = \int g^p d\mu$$
which completes your proof.