Let $\mu$ be a probability measure on a set $X$, i.e. $\mu(X)=1$, and let $f$ and $g$ be positive measurable functions on $X$. Show that if $fg\geq1$, then the integral of $f$ times the integral of $g$ is greater than or equal to $1$.

I'm not sure how to approach this. Since $f$ and $g$ are positive, the integral of $f$ and the integral of $g$ are equal to the $L_1$ norm of $f$ and the $L_1$ norm of $g$, but I'm not sure why $fg\geq 1$ implies that the product of their $L_1$ norms is greater than or equal to $1$. Considering that $fg$ is involved, I was thinking of using Holder's inequality, but I'm not sure what $p$ and $q$ to use.


This relies more on an arithmetic trick than on analysis. Since $fg \ge 1$ you have also that $\sqrt{fg} \ge 1$ so that $$1 \le \int_X \sqrt{fg} \le \left(\int_X f \right)^{1/2} \left( \int_X g \right)^{1/2}$$ by Holder's inequality. Now square both sides.

  • $\begingroup$ I don't understand your application of Holder's inequality. What are your $p$ and $q$? $\endgroup$ – Keshav Srinivasan Nov 8 '18 at 20:31
  • $\begingroup$ $2$ and $2$. $\mbox{}$ $\endgroup$ – Umberto P. Nov 8 '18 at 20:32
  • $\begingroup$ And how do we know that $\sqrt f$ and $\sqrt g$ are in $L_2$? $\endgroup$ – Keshav Srinivasan Nov 8 '18 at 20:35
  • $\begingroup$ You don't. Since $f$ and $g$ are positive the integrals are all defined, even if possibly infinite. $\endgroup$ – Umberto P. Nov 8 '18 at 20:39
  • $\begingroup$ OK, thanks for your answer. $\endgroup$ – Keshav Srinivasan Nov 8 '18 at 20:44

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