1
$\begingroup$

Consider a ratio $\int_A f(x,a)dx/\int_B f(x,a)dx$ where $A, B \subset [0,1]$ and $a \in R$. Suppose for any x' \in A and $x \in B$, f(x',a)/f(x,a) > f(x',b)/f(x,b). Then can we say that $\int_A f(x,a)dx/\int_B f(x,a)dx>\int_A f(x,b)dx/\int_B f(x,b)dx$? Thank you very much.

  • 0
    Thank $y$ou for clarification.2011-03-28

1 Answers 1

0

Fix $y\in B$ therefore for any $x\in A$ $f(x,a)>f(x,b)\frac{f(y,a)}{f(y,b)}$ taking integral over $A$ implies $\int_{A}f(x,a)dx>\int_{A}f(x,b)dx\left(\frac{f(y,a)}{f(y,b)}\right).$ Thusf(y,b)\int_{A}f(x,a)dx>f(y,a)\int_{A}f(x,b)dx . Now it is time to take integral over $B$, hence\int_{B}f(y,b)dy\int_{A}f(x,a)dx>\int_{B}f(y,a)dy\int_{A}f(x,b)dx , consequently $\frac{\int_{A}f(x,a)dx}{\int_{B}f(y,a)dy}>\frac{\int_{A}f(x,b)dx}{\int_{B}f(y,b)dy}.$