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My prof mentioned that the sum of strictly convex and convex functions is strictly convex, Im having trouble swallowing that, is it accurate?

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  • $\begingroup$ Sure, if $a_i < b_i$ then $\sum_i a_i < \sum_i b_i$. Apply this to the inequality that characterizes strictly convex functions. $\endgroup$ – copper.hat Mar 9 '13 at 23:39
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    $\begingroup$ @copper.hat: Isn't the analogy more like "if $a\lt b$ and $c\le d$, then $a+c\lt b+d$"? $\endgroup$ – joriki Mar 10 '13 at 0:16
  • $\begingroup$ @joriki: You are correct, I missed the 'and convex' part. Will delete & correct the first comment. $\endgroup$ – copper.hat Mar 10 '13 at 0:34
  • $\begingroup$ My earlier comment was inaccurate, as @joriki pointed out. The relevant inequality is that if $a<b$ and $c_i \le d_i$, then $a+\sum_i c_i < b + \sum_i d_i$. $\endgroup$ – copper.hat Mar 10 '13 at 0:36
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Assume $f$ is convex and $g$ is stricly convex. Let $0<\theta < 1$. Then calculate that $$ \begin{align} (f+g)(\theta x + (1-\theta) y) &= f(\theta x + (1-\theta) y) + g(\theta x + (1-\theta) y) \\ &< \theta f(x) + (1-\theta) f(y) + \theta g(x) + (1-\theta) g(y) \\ &= \theta (f+g)(x) + (1-\theta) (f+g)(y) \end{align} $$ where there is strict inequality because the inequality is strict in one case (and not necessarily strict in the other case).

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