An inequality in $L^p$ spaces

Question

I have to prove the next statement.

Let $\Omega \subset \Bbb R^n$ be an open set and $h : \Omega \to \Bbb R$, $\rho : \Omega \to [0,\infty)$ be Lebesgue measurable functions with $\int_\Omega \rho dx=1 $. Prove that for each $p\in [1,\infty)$ $$ \left( \int_\Omega |h|\rho dx \right)^p \le \int_\Omega |h|^p\rho dx.$$

Now, my idea was to use Holder's inequality (so $q=\frac{p}{p-1}$)to get it directly, but I'm blocked here:

$$\left( \int_\Omega |h|\rho dx\right)^p \le \| h\|_p^p\|\rho \|_q^p=\int_\Omega |h|^p dx \left( \int_\Omega \rho^{\frac{p}{p-1}}dx\right)^{p-1}$$

Any hint or suggest?

Answer 1

Conceptually, this is a bad approach because any such argumentation would go through even if $\rho$ were not assumed to be positive. It is also a bad approach because there is no way to guarantee that $\rho$ will be $L^q$ anyway, $\rho$ could very well be $L^1$ but not in any $L^r$ for $r>1$, since the Lebesgue measure assigns arbitrarily small positive measures to sets.

Instead you should treat this in the abstract measure framework, thinking of $\rho dx$ as just an abstract measure (a probability measure, in this case). In that sense, you can think of integration against $\rho dx$ as intuitively being a convex combination of finitely many real numbers (approximating $h$ by simple functions makes this intuition precise). How does $x \mapsto x^p$ behave under convex combinations?

(Just for reference, the inequality you have written is a case of Jensen's inequality, which is proven along the lines of my hint above.)