Mixture of Gaussian is not log-concave I've encountered the statement:

For $p\in(0,1),$ the location mixture of standard univariate normal densities
$f(x)=p\phi(x)+(1-p)\phi(x-\mu)$ is log-concave if and only if $\Vert\mu\Vert \leq 2.$

For the life of me, I can't figure out how to prove this or find a reference. Could you point me to a reference or help proving this?

Since $f(x)$ is differentiable everywhere on $\mathbb{R}$, $g(x) := \log f(x)$ is concave if and only if $g''(x) \leq 0$ for all $x \in \mathbb{R}$. So the problem actually boils down to evaluate $g''(x)$ and look for the condition of $\mu$ such that $g''(x) \leq 0$.  Along the way, we will use the following properties (easy-to-check) of $\phi(x)$:
\begin{align}
\phi'(x) = -x\phi(x), \quad \phi''(x) = (x^2 - 1)\phi(x). \tag{1}
\end{align}
Now directly evaluation yields (I used $(1)$ to simplify, details omitted here)
\begin{align}
g''(x) = -\frac{p^2\phi^2(x)+(1 - p)^2\phi^2(x-\mu) - (\mu^2 - 2)\times p\phi(x) \times(1 - p)\phi(x - \mu)}{f^2(x)}.
\end{align}
Let $a = p\phi(x) > 0, b = (1 - p)\phi(x - \mu) > 0$, then the above expression implies that $g''(x) \leq 0$ for all $x$ if and only if for all $a > 0, b > 0$:
\begin{align}
a^2 + b^2 - (\mu^2 - 2)ab \geq 0, 
\end{align}
which implies for all $a > 0, b > 0$,
\begin{align}
\mu^2 \leq 2 + \frac{a^2 + b^2}{ab}.
\end{align}
Therefore,
\begin{align}
\mu^2 \leq 2 + \min_{a > 0, b > 0}\frac{a^2 + b^2}{ab} = 2 + 2 =4,
\end{align}
giving $|\mu| \leq 2$ as desired.
