I would like to prove the following but I couldn't manage to do it. Let $a>b>0$ be two real numbers. Let $f$ be the function defined as:


$$\forall \sigma>0, ~\forall x\in\mathbb{R},~f\_\sigma(x):=\sigma e^{-x^2/\sigma^2}+\sqrt{\pi}x\text{erf}\left(\frac{x}{\sigma}\right).$$


**I would like to show that $f\_a-f\_b$ is log-concave.**


One can see $f\_\sigma$ as the function that, differentiated two times, gives the gaussian function (with a scaling factor) of mean $0$ and deviation $\sigma$. More formally, one can compute the second derivative and show that:
$$f\_\sigma''(x) = \frac{e^{-x^2/\sigma^2}}{2\sigma}.$$


Numerically, I am pretty sure that this statement is true. I tried to compute the second derivative of the $\log(f\_a-f\_b)$ but without sucess (the derivatives are too diffucult to analyse).


Any hints or solutions will be highly appreciated! Thank you.



Log-concavity of the difference of the second anti-derivative of Gaussians
This conjecture is true.


Indeed, for $h:=\ln(f\_a-f\_b)$ and real $x\ne0$ we have
\begin{equation}
 h'(x)=R(x):=\frac{F(x)}{G(x)},
\end{equation}
where
\begin{equation}
 F(x):=\sqrt{\pi }
 \left(\text{erf}\left(\frac{x}{a}\right)-\text{erf}\left(\frac{x}{b}\right)\right), 
\end{equation}
\begin{equation}
 G(x):= a e^{-x^2/a^2}+\sqrt{\pi } x\,
 \text{erf}\left(\frac{x}{a}\right)-b
 e^{-x^2/b^2}-\sqrt{\pi } x \, \text{erf}\left(\frac{x}{b}\right); 
\end{equation}
note that $G(0)=a-b>0$ and $G'(x)=\sqrt{\pi }
 \left(\text{erf}\left(\frac{x}{a}\right)-\text{erf}\left(\frac{x}{b}\right)\right)>0$ for real $x>0$, so that $G>0$ on $[0,\infty)$.


Since the function $h$ is smooth and even, it is enough to show that $R$ is decreasing on $(0,\infty)$. In what follows, $a>b>0$ and $x>0$, unless otherwise indicated.


Let
\begin{equation}
 R\_1(x):=\frac{F'(x)}{G'(x)}, 
\end{equation}
\begin{equation}
 R\_2(x):=\frac{F''(x)}{G''(x)}
 = -\frac{2 x \left(a^3 e^{x^2/a^2}-b^3
 e^{x^2/b^2}\right)}{a^2 b^2 \left(a e^{x^2/a^2}-b e^{x^2/b^2}\right)};
\end{equation}
note that $R\_2(x)$ is undefined at $x=x\_{a,b}$, where
\begin{equation}
 x\_{a,b}:=a b \sqrt{\frac{\ln a-\ln b}{a^2-b^2}}. 
\end{equation}


Then
\begin{equation}
\begin{aligned}
 H(t)&:=R'\_2(x)\frac{1}{2} a^3 b^3 \left(a e^{x^2/a^2}-b e^{x^2/b^2}\right)^2 \\ 
& =a^4 \left(b^2-2 t\right)-a b^5 e^{\left(\frac{1}{b^2}-\frac{1}{a^2}\right)t}+a^2 \left(b^4+4 b^2
 t\right)-a^5 b e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}-2 b^4 t, 
\end{aligned}
\end{equation}
where $t:=x^2>0$, and
\begin{equation}
 H'(t)\frac{ab}{a^2-b^2}=H\_1(u):=\frac{a^4}{u}-2 a^3 b+2 a b^3-b^4 u,
\end{equation}
where $u:=e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}>1$. The only root $u$ of the equation $H\_1(u)=0$ that may be positive is
\begin{equation}
 u\_{a,b}:=\frac{a b^3-a^3 b+\sqrt{a^6 b^2-a^4 b^4+a^2 b^6}}{b^4}. 
\end{equation}
Replacing now $e^{\left(\frac{1}{a^2}-\frac{1}{b^2}\right)t}$ and $e^{\left(\frac{1}{b^2}-\frac{1}{a^2}\right)t}$ in the expression for $H(t)$ by $u\_{a,b}$ and $1/u\_{a,b}$, respectively, we see that the condition $H'(t)=0$ for some real $t>0$ will imply
\begin{equation}
\begin{aligned}
 H(t)&=a^4 \left(-\frac{b^4}{-a^2+\sqrt{a^4-a^2 b^2+b^4}+b^2}+2 b^2-2
 t\right) \\ 
 &-a^2 b^2 \left(\sqrt{a^4-a^2 b^2+b^4}-4 t\right)-2
 b^4 t, 
\end{aligned}
\end{equation}
which is a rather simple algebraic expression, which is actually $<0$ (still assuming $a>b>0$ and $t>0$).
So, $H(t)<0$ at any critical point $t>0$ of $H$.
Also, $H(0)=-a (a - b)^2 b (a^2 + a b + b^2)<0$ and $H(\infty-):=\lim\_{t\to\infty}H(t)=-\infty$. So, $H(t)<0$ for all real $t\ge0$.


So, $R\_2$ is decreasing on $(0,x\_{a,b})$ and on $(x\_{a,b},\infty)$.


Also, $R\_1>0$ on $(0,x\_{a,b})$, $R\_1(x\_{a,b})=0$, $R\_1<0$ on $(x\_{a,b},\infty)$, $R\_1(0+)=\infty$, and $R\_1(\infty-)=-\infty$.


So, by [Table 1.1](https://eudml.org/doc/128522), $R\_1$ is decreasing on $(0,\infty)$. Also, $R'(0)=-\frac2{ab}<0$ and $R(\infty-)=-\infty$. So, again by [Table 1.1](https://eudml.org/doc/128522), $R$ is decreasing on $(0,\infty)$. $\quad\Box$


