Is it true that $\int fg \, d\mu \leq \int f \, d\mu \int g \, d\mu$?

Question

Let $f,g$ be positive measurable functions on a bounded set $\Omega$. Is it true that $$\int fg \, d\mu \leq \int f \, d\mu \int g \, d\mu$$

It seems true to me based on discrete analysis, but I cannot be sure.

Answer 1

It seems false to me take $$f(x) = g(x) = x^3$$ on $\Omega = [0,1]$

Answer 2

Observe we have \begin{align} \int^1_0 x^2\ dx = \frac{1}{3} \end{align} but \begin{align} \left(\int^1_0 x\ dx\right)^2 = \frac{1}{4}. \end{align}

Additional: One might wonder: maybe there exists a constant $C$ such that \begin{align} \int fg d\mu \leq C\int f d\mu \int g\ d\mu. \end{align}

But if we take $f=g \geq0$ we get \begin{align} \|f\|_2 \leq C'\|f\|_1. \end{align} which is clearly false. Consider $f(x) = x^{-3/4}$, then we see that $\|f\|_2 = \infty$ but $\|f\|_1<\infty$.

Answer 3

This is false as has been pointed out. But there's a nice way to know that it's false just by looking at it - the claimed inequality lacks the correct homogeneity. In other words, the two sides scale differently under some change. In this case, the problem is the measure $\mu$; the left hand side depends linearly on the mass of $\mu$, while the right hand side does not.

This even tells us exactly what kind of functions to test against to show the inequality is false: take characteristic functions of sets of small measure. If $f = g = \chi_E$, then we would be asking to have

$$\mu(E) \le \mu(E)^2$$

which forces $\mu(E) \ge 1$ (or $\mu(E) = 0$). So for any measure in which there are sets of small measure, the inequality is false. This is why you perhaps didn't detect an issue with the inequality when considering the discrete case.