Does this integral inequality hold?

Does the following inequality hold?

∫R3|f1(x)|2g2(x)dx≤∫R3|f1(x)||f2(x)|√g1(x)g2(x)dx\int_\mathbb{R^3} |f_1(x)|^2 g_2(x)dx \leq
\int_\mathbb{R^3} |f_1(x)| |f_2(x)| \sqrt{g_1(x)g_2(x)}dx

where
gi(x)=∫R3|fi(y)|2|x−y|dy, fi∈C∞0(R3), i=1,2.g_i(x)= \int_\mathbb{R^3} \frac{|f_i(y)|^2}{|x-y|} dy, \ f_i \in C_0^\infty(\mathbb{R^3}), \ i=1,2.

Any advice would be appreciate.

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1 Answer
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Suppose [suppf1]∩[suppf2]=∅[\operatorname{supp} f_1] \cap [\operatorname{supp} f_2]=\emptyset, then you will see that
∫R3|f1(x)f2(x)|√g1(x)g2(x) dx=0\begin{align}
\int_{\mathbb{R}^3} |f_1(x)f_2(x)|\sqrt{g_1(x)g_2(x)}\ dx = 0
\end{align}
but
∫R3∫R3|f1(x)|2|f2(y)|2|x−y| dxdy≠0.\begin{align}
\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|f_1(x)|^2|f_2(y)|^2}{|x-y|}\ dxdy \neq 0.
\end{align}

  

 

Thank you very much.
– user298344
2 days ago