# Estimate on average with weight

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.

• There is just an equality. Is the question missing something? – deb Nov 13 '18 at 13:16
• No the question is fine. But I am not getting how to prove. Can you please explain. Thanks. – Mathlover Nov 13 '18 at 13:19
• 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. – deb Nov 13 '18 at 13:24
• 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. – Mathlover Nov 13 '18 at 13:26
• it's multiplication by $e^{\alpha-\alpha}$, where $\alpha$ is the product above. – deb Nov 13 '18 at 13:28