While reading the paper of Gang Tian, "Kähler-Einstein metrics with positive scalar curvature". In the proof of Theorem 1.6, he pointed that if two Kahler metrics $\omega $ and $\omega'$ satisfies $\frac{1}{2}\omega\leq\omega'\leq 2\omega$ and $\int_M \omega^n=\int_M\omega'^n=1$, then we have $\lambda_{1,\omega'}\geq 2^{-n-1}\lambda_{1,\omega}$,where $$\lambda_{1,\omega}=\inf_{v\in C^1(M)\cap\{|\nabla v|\neq 0\}}\frac{\int_M |\nabla v|^2\omega^n}{\int_M v^2\omega^n-(\int_M v\omega^n)^2}.$$ How can this be proved? I have tried to estimate $$\frac{\int_M |\nabla v|^2\omega'^n}{\int_M v^2\omega'^n-(\int_M v\omega'^n)^2}\geq\frac{2^{-n-1}\int_M |\nabla v|^2\omega^n}{2^n\int_M v^2\omega^n-[2^{-2n}(\int_M v_+\omega^n)^2-2*2^{2n}\int_M v_+\omega^n\times\int_M v_-\omega^n+2^{-2n}(\int_M v_-\omega^n)^2]}\geq\frac{2^{-n-1}\int_M |\nabla v|^2\omega^n}{2^n\int_M v^2\omega^n-[2^{-2n}(\int_M v_+\omega^n)^2-2*2^{-2n}\int_M v_+\omega^n\times\int_M v_-\omega^n+2^{-2n}(\int_M v_-\omega^n)^2]}\geq\frac{2^{-n-1}\int_M |\nabla v|^2\omega^n}{2^n\int_M v^2\omega^n-2^{-2n}(\int_M v\omega^n)^2}.$$ But the last item can not be controlled by $\int_M v^2\omega^n-(\int_M v\omega^n)^2$, which can not imply the desired inequality. Whether some other better method can be valid?
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$\begingroup$ My immediate instinct, without going through your calculation, is to insert a factor $\int \omega^n$ to the first term in the denominator. $\endgroup$– Deane YangCommented Mar 17, 2014 at 16:50
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$\begingroup$ Sorry, I forgot it. And I add the normalize condition of the metrics. $\endgroup$– DanielCommented Mar 17, 2014 at 16:52
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$\begingroup$ Does it help if you write the denominator $\int_M (v-\int_M v \omega^n)^2\omega^n$, and then change the infinum to be over functions of zero average ? $\endgroup$– usernameCommented Mar 17, 2014 at 19:05
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$\begingroup$ It do not work. If $\inf_M v\omega^n=0$, we may not have $\inf_M v\omega'^n=0$. $\endgroup$– DanielCommented Mar 18, 2014 at 1:19
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$\begingroup$ Let $$ J_{\omega}(v)=\frac{\int_{M}\left|\nabla v\right|\omega^{n}}{\int_{M}v^{2}\omega^{n}-\left(\int_{M}v\omega^{n}\right)^{2}}. $$ Since $\int_{M}\omega^{n}=1$, we have \begin{eqnarray*} J_{\omega}(v) & = & \frac{\int_{M}\left|\nabla v\right|\omega^{n}}{\int_{M}\left(v-\int_{M}v\omega^{n}\right)^{2}\omega^{n}}\\ & = & J_{\omega}(v+C) \end{eqnarray*} for any constant $C$. So the constant term does not matter, as you can always define $w=v+C$ such that $\int_{M}w\omega^{\prime n}=\lambda\int_M v \omega^n$ for any convenient $\lambda$. $\endgroup$– usernameCommented Mar 18, 2014 at 9:27
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Note that if $$ J_\omega(v)=\frac{\int_M \left|\nabla v\right|^2 \omega^n}{\int_M v^2 \omega^n - \left(\int_M v\omega^n\right)^2}, $$ then due to the fact that $\int_M \omega^n=1$, $J_\omega(v)=J_\omega(v+C)$ for any constant $C$.
Similarly, $J_{\omega^\prime}(u)=J_{\omega^\prime}(u+C^\prime)$ for any $C^\prime$.
Thus you can always assume that $\int_M v\, \omega^n=1$, and that for any $\lambda$ that is convenient for your calculation, there is a $u=v+C$ such that $\int_M u \,\omega^{\prime n} = \lambda$.
