Dear All
Could you please help prove two integral inequalities in the picture uploaded to this message?
Best regards
Feng Qi (F. Qi)
https://qifeng618.wordpress.com
Suppose for simplicity that \sigma has a density, d\sigma=pdt. The first inequality is equivalent to 2(Lp)(L(t^2p))^2\geq (L(tp))^2L(t^2p)+(Lp)(L(tp))(L(t^3p)) (L denotes the Laplace transform), or using the Laplace convolution \ast,
2L(p\ast(t^2p)\ast(t^2p))\geq L((tp)\ast(tp)\ast(t^2p)+p\ast(tp)\ast(t^3p)).
So, this is valid if
2p\ast(t^2p)\ast(t^2p)\geq (tp)\ast(tp)\ast(t^2p)+p\ast(tp)\ast(t^3p).
(And the same approach for the second inequality).
Best regards,
Adolf Mirotin
Denote e^{xt}d\sigma(t)=d\omega(t) and the problem reduces to the case x=0.
Dear Feng Qi,
Here is an explanation to Gevorg Grigorian's comment. For a fixed x consider such a function \omega that e^{xt}d\sigma(t)=d\omega(t). In other words \omega(t) = \int_0^t e^{xs}d\sigma(s).
Then, for a fixed x your original problem reduces to the same problem for \omega(t) with x=0.
Best, Vladimir
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