Consider the function

$$f_{n}(x)=e^{-x^2}x^n.$$

My goal is to show that

$$ G(y):=\frac{(f_2*f_0)(y)}{(f_0*f_0)(y)}- \left(\frac{(f_1*f_0)(y) }{(f_0*f_0)(y)}\right)^2$$

is log-concave.

Let us first observe that indeed $G(y) \ge 0.$

This just follows from a Cauchy-Schwarz

$$(f_1*f_0)(y) \le \sqrt{(f_2*f_0)(y)(f_0*f_0)(y)}$$

so everything is well-defined.

Usually one can say a lot when convolutions are involved about log-concavity due to standard theorems see wikipedia

but this combination looks a bit tricky.

Consider the function

$$f_{n}(x)=e^{-x^2}x^n.$$

My goal is to show that

$$ G(y):=\frac{(f_2*f_0)(y)}{(f_0*f_0)(y)}- \left(\frac{(f_1*f_0)(y) }{(f_0*f_0)(y)}\right)^2$$

is log-concave.

Let us first observe that indeed $G(y) \ge 0.$

This just follows from a Cauchy-Schwarz

$$(f_1*f_0)(y) \le \sqrt{(f_2*f_0)(y)(f_0*f_0)(y)}$$

so everything is well-defined.

Usually one can say a lot when convolutions are involved about log-concavity due to standard theorems see wikipedia

but this combination looks a bit tricky.

Addendum I should add that I am in particular very interested in theoretical insights why this particular expression has to be log-concave.