# prove $\log \left[ {\sum\limits_{i = 1}^M {{\varepsilon _i}{{\left[ {Q\left( {{a_i} + {b_i}\sqrt u } \right)} \right]}^2}} } \right]$ is convex

I am having difficulties to prove $$\log \left[ {\sum\limits_{i = 1}^M {{\varepsilon _i}{{\left[ {Q\left( {{a_i} + {b_i}\sqrt u } \right)} \right]}^2}} } \right]$$ is convex for non-negative a, b,u. Where, $$Q\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\int_x^\infty {{e^{ - \frac{{{v^2}}}{2}}}dv}$$.

I know the following properties of the above function:

1. $${{{\left[ {Q\left( {{a_i} + {b_i}\sqrt u } \right)} \right]}^2}}$$ is convex.

2. $$\sum\limits_{i = 1}^M {{\varepsilon _i}} =1$$

3. The Range of the log function is $$(0,0.5]$$

For $$a\ge0$$ and $$u\ge0$$, let $$q(u):=\ln Q(a+\sqrt u).$$ Then $$q_2(t):=q''(u)\frac{8 \sqrt{2 \pi } t^3 e^{\frac{1}{2} (a+t)^2} Q(a+t)^2}{a t+t^2+1} =2 Q(a+t)-\frac{\sqrt{\frac{2}{\pi }} t e^{-\frac{1}{2} (a+t)^2}}{a t+t^2+1},$$ where $$t:=\sqrt u\ge0$$. Next, $$q_2'(t)=-\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{1}{2} (a+t)^2} (a t+2)}{\left(a t+t^2+1\right)^2}<0$$ and $$q_2(\infty-)=0$$. So, $$q_2>0$$ and hence $$q''>0$$. So, $$q$$ is convex. So, $$Q(a+b\sqrt u)$$ is log convex in $$u\ge0$$ for any $$a,b\ge0$$.
Now the desired result follows for any positive $$\varepsilon_i$$'s in view of the well-known fact that the sum of log-convex functions is log convex.
• Thank you for you reply, Its really helpful but the last line I did not get yet. Did you mean sum of logconvex is logconvex for any positive $\epsilon_i$? But in my case the sum is inside the log function. And can you give me any ref. regarding last line that this is applicable for my case where the sum is inside log function. Again thank you very much. Nov 12, 2019 at 22:04
• @hasan : Here are details: I showed that $\ln Q(a+\sqrt u)$ is convex in $u\ge0$; that is, $Q(a+\sqrt u)$ is log convex in $u\ge0$. So, $\varepsilon_i Q(a_i+b_i\sqrt u)$ is log convex in $u\ge0$ for each $i$, if $a_i\ge0,b_i\ge0,\varepsilon_i>0$. So, $\sum_i \varepsilon_i Q(a_i+b_i\sqrt u)$ is log convex in $u\ge0$; that is, $\ln\sum_i \varepsilon_i Q(a_i+b_i\sqrt u)$ is convex in $u\ge0$, as desired. I have also added references about log-convex functions and their sum. Nov 12, 2019 at 22:57