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.
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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 $$\int_{A}f(x,a)dx>\int_{A}f(x,b)dx\left(\frac{f(y,a)}{f(y,b)}\right).$$ Thus$$f(y,b)\int_{A}f(x,a)dx>f(y,a)\int_{A}f(x,b)dx .$$ 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 $$\frac{\int_{A}f(x,a)dx}{\int_{B}f(y,a)dy}>\frac{\int_{A}f(x,b)dx}{\int_{B}f(y,b)dy}.$$ |
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