proving inequality involving lebesgue measurable function on [0,1]

Question

Just by looking at the inequality, I can tell that if $\displaystyle \int_0^1 f(x)dx \geq 1$ then $\displaystyle \int_0^1 \frac{dx}{f(x)} \leq 1$. The same thing vice versa. How do I prove that, though?? Thanks for the help ahead of time.

Answer 1

Let us note that since $f>0$, then $\sqrt{f}$ and $\frac{1}{f}$ are defined. Now, we have that $$1 = \int_0^1 1 = \int_0^1 \frac{\sqrt{f}}{\sqrt{f}}$$ By Schwarz, $$\leq \left(\int_0^1 f\right)^{\frac{1}{2}}\left(\int_0^1 \frac{1}{f} \right)^{\frac{1}{2}}$$ Complete the proof by taking squares.

Answer 2

I like to use Jensen's inequality. The function $\phi(x) = 1/x$ is convex, and $m([0,1])=1.$ Jensen's inequality implies that

$$\int_0^1 1/f dm = \int_0^1 \phi(f) dm \geq \phi \left(\int_0^1 f dm \right)=\frac{1}{\int_0^1 f dm}.$$

Note, both this and the Holder's inequality argument generalize to show the following problem in Rudin (chapter 3, number 11): If $0 < f,g$ on a measure space $\Omega$ with $\mu(\Omega)=1$ so that $fg \geq 1$ almost everywhere, then $(\int_{\Omega}fd\mu) (\int_{\Omega}g d\mu) \geq 1$.