Let $(X,\mathcal{A},\mu)$ be a finite measure space with $\mu(X) = 1 $
Let $f,g$ be two measurable non-negative real valued functions $X \to (0,\infty)$ such that $f(x)g(x) \geq 1$
Show that $1 \leq \int_X f d\mu \cdot \int_X g d\mu $
My attempt:
Assume that both $f$ and $g$ are integrable otherwise the statement holds trivially as the integrals cannot be $-\infty$ (by non-negativity of $f$ and $g$). Define $a(x) = f(x)^{\frac{1}{2}}$ and $b(x) = g(x)^{\frac{1}{2}}$
Then by our assumption of the integrability of $f $ and $g$ we have that $a,b \in \mathcal{L}^2$
Hence taking $p = q = 2$ and applying Holder's inequality we have the following:
$\int_X a(x)b(x) d\mu \leq (\int_X a(x)^2 d\mu)^{\frac{1}{2}} \cdot(\int_X b(x)^2 d\mu)^{\frac{1}{2}} $
Putting everything back into terms of $f$ and $g$ simplifies to:
$\int_X (f(x)g(x))^{\frac{1}{2}} d\mu \leq (\int_Xf(x) d\mu)^{\frac{1}{2}} \cdot (\int_Xg(x) d\mu)^{\frac{1}{2}} $
Considering the convex function, $\theta \to \theta^{\frac{1}{2}},$ and applying Jensen's inequality to our left hand side, namely $(\int p(x) d\mu )^{\frac{1}{2}} \leq \int p(x)^{\frac{1}{2}} d\mu$ We are left with:
$(\int_X f(x)g(x) d\mu)^{\frac{1}{2}} \leq (\int_Xf(x) d\mu)^{\frac{1}{2}} \cdot (\int_Xg(x) d\mu)^{\frac{1}{2}} $
And so: $\int_X f(x)g(x) d\mu \leq \int_Xf(x) d\mu \cdot \int_Xg(x) d\mu $
And as $f(x)g(x) \geq 1$ and $\mu(X) = 1$ the result follows.
This feels awfully long and cumbersome, does anyone know a few tips or another way to provide a better solution? Thanks!
Best Answer
Since $fg\geq 1$, you have $\sqrt{f(x)\cdot g(x)} \geq 1$ for all $x$, and so $$ 1 \leq \int_X \sqrt{f}\cdot \sqrt{g} d\mu \leq \sqrt{\int_X f d\mu }\cdot \sqrt{\int_X g d\mu } $$ where the second inequality is Cauchy–Schwarz; squaring both sides gives the inequality.