Let $B_{2R}=B(x,2R)$ be the ball of radius $2R$ centered at $x$ and $v=log\,u$ for some positive function $u$ defined on $B_{2R}$. Denote by $v_{B_{2R}}=\frac{1}{w(B_{2R})}\int_{B_{2R}}v(x)w(x)\,dx$, where $w(B_{2R})=\int_{B_{2R}}w(x)\,dx$. Then \begin{align*} \frac{1}{w(B_{2R})}\int_{B_{2R}}e^{-p_0 v}w(x)\,dx\cdot\frac{1}{w(B_{2R})}\int_{B_{2R}}e^{p_0 v}w(x)\,dx\\ =\frac{1}{w(B_{2R})}\int_{B_{2R}}e^{(p_0 v-p_0 v_{B_{2R}})}w(x)\,dx\cdot\frac{1}{w(B_{2R})}\int_{B_{2R}}e^{(p_0 v_{B_{2R}}-p_0 v)}w(x)\,dx \end{align*} Can you kindly help me how to get the above equality. Thanking you.

  • $\begingroup$ There is just an equality. Is the question missing something? $\endgroup$ – deb Nov 13 '18 at 13:16
  • $\begingroup$ No the question is fine. But I am not getting how to prove. Can you please explain. Thanks. $\endgroup$ – Mathlover Nov 13 '18 at 13:19
  • $\begingroup$ Sorry, I'm not sure what the question is. Prove what? why the two sides are equal, post multiplying and dividing by a constant? I don't know if I'm missing something. $\endgroup$ – deb Nov 13 '18 at 13:24
  • $\begingroup$ I want to prove the above equality which looks trivial to you. If you kindly explain your argument more clearly, it will be very grateful for me. Thanks in advance. $\endgroup$ – Mathlover Nov 13 '18 at 13:26
  • $\begingroup$ it's multiplication by $e^{\alpha-\alpha}$, where $\alpha$ is the product above. $\endgroup$ – deb Nov 13 '18 at 13:28

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