Timeline for 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

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Dec 18, 2019 at 3:25	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 1 character in body
Nov 13, 2019 at 17:22	comment	added	hasan		Thank you very much losif Pinelis. I appreciate it.
Nov 13, 2019 at 15:59	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 74 characters in body
Nov 12, 2019 at 22:57	comment	added	Iosif Pinelis		@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 22:47	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 124 characters in body
Nov 12, 2019 at 22:04	comment	added	hasan		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 18:22	history	answered	Iosif Pinelis	CC BY-SA 4.0