How is the Cauchy-Schwarz inequality used in the proof of Lyapunov's criterion in the book "Analysis and Geometry of Markov Diffusion Operators"

Question

Let $(E,\mu,\Gamma)$ be a full Markov triple (see definition below), $J\in\mathcal A$ with $J\ge1$ and $g\in\mathcal A_0$. In the proof of Theorem 4.6.2 of the book "Analysis and Geometry of Markov Diffusion Operators", it's claimed that $$\Gamma\left(\frac{g^2}J,J\right)=\frac{2g}J\Gamma(g,J)-\frac{g^2}{J^2}\Gamma(J)\le\Gamma(g)\tag1.$$ The authors claim that the equality holds by the diffusion property of $\Gamma$, while the second inequality holds by the Cauchy-Schwarz inequality for $\Gamma$ (which states that ${\Gamma(f,h)}^2\le\Gamma(f)\Gamma(h)$.

I don't get how the inequality follows by the Cauchy-Schwarz inequality. Could anybody explain this to me?

Answer 1

Let $a:=\frac gJ$, $b:=\sqrt{\Gamma(J)}$, $c:=\sqrt{\Gamma(g)}$. By Cauchy--Schwarz, $\Gamma(g,J)\le bc$. So, the inequality in question reduces to $2abc-a^2b^2\le c^2$, or to the obvious inequality $(ab-c)^2\ge0$.