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

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.