Inequality in the proof of "Sobolev inequalities implies isoperimetric inequalities"

Question

I'm trying to understand the proof on pages 24-25 of the book of Juha Heinonen "Lectures on analysis on metric space" for the equivalence between Sobolev inequalities $\|u\|_{n/(n-1)}\leq I_n\|\nabla u\|_1$ for $\mathbb{R}^n$, a constant $I_n>0$ and for all smooth, compactly supported functions $u$ on $\mathbb{R}^n$, and the isoperimetric inequalities $|U|^{(n-1)/n}\leq I_n|\partial U|$ for each closed, smooth submanifold $U$ of $\mathbb{R}^n$. Here, $|\cdot|$ is the volume and the surface area respectively.

In the proof I need to use the following inequality: $\int_0^{\infty}F(t)^{\alpha} dt \geq\left(\frac{1}{\alpha}\int_0^{\infty}F(t)t^{1/\alpha -1}dt\right)^{\alpha}$ where $F(t)$ is a decreasing function of $t$ and $0<\alpha\leq1$.

My idea is to use Hölder's inequalities for $p=\alpha/(\alpha-1)$ and $q=\alpha$ but I was not able to complete the proof... Can somebody help me?

Answer 1

What is the meaning of $1/\alpha-1? $ Either $\frac{1}{\alpha}-1$ or $\frac{1}{\alpha-1}-1?$

Assuming that $1/\alpha-1=\frac{1}{\alpha}-1$ and that $a=F(0+)$ is finite, by dividing by $a$ both sides of the inequality to be proven, we can assume $a=1$.This implies $F^{\alpha} >F$ and enough is to prove with $p=1/\alpha>1$ $$\int_0^{\infty}F(t)dt>\left(\int_0^{\infty}pt^{p-1}F(t)dt\right)^{1/p}.(*)$$ Now we write the decreasing function $F(t)=\int_t^{\infty}\mu(dx)$ where $\mu$ is a probability. By Fubini, (*) becomes $$\int_0^{\infty}x\mu(dx)>\left(\int_0^{\infty}x^p\mu(dx)\right)^{1/p}.(**)$$ Finally, $p\mapsto k(p)=\log\int_0^{\infty}x^p\mu(dx)$ is convex by Holder inequality. Therefore $k'$ is increasing and $k(0)=0$ since $\mu$ is a probability. As a consequence

$$k(p)=\int_0^pk'(t)dt\leq pk'(p), \ \frac{d}{dp}\frac{k(p)}{p}=\frac{pk'(p)-k(p)}{p^2}<0.$$ Since $p>1$ we have $\frac{k(p)}{p}\leq k(1),$ which is equivalent to (**).