# Log-concavity of function

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.

Direct calculations show that $$(f_2*f_0)(y)=\frac{1}{4} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} \left(y^2+1\right),$$ $$(f_1*f_0)(y)=\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} y,$$ $$(f_0*f_0)(y)=\sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}}$$ for all real $$y$$, so that $$G$$ is the constant $$1/4$$ and hence log concave.
• @Martinique : I doubt that a "behind-the-expression" reasoning exists, in particular given that $G$ is a constant and thus just "barely" log concave. Jan 14, 2020 at 17:36