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Let's say the $f(x)$ is a nonnegative function on the interval $[0,x_{max}]$.

$g(x) = \int_{0}^x f(t)\ dt$ where $x \leq x_{max}$

Why is g(x) quasi-convex if $f(x) \geq 0$ on the interval $[0,x_{max}]$?

Mike
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  • This is an exact problem from a take-home exam currently in course for Convex Optimization at Stanford. (I would know: I'm teaching the course.) Once the exams are all in (another day or two), I'll post the solution to this! – Nicholas Moehle Aug 14 '16 at 22:11

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Nonnegative has nothing to do with it. If you restrict to the case where $f\in C^1([0,x_{max}])$ the FTC becomes handy.

$g'(x)=f(x)$ and $g''(x)=f'(x)$.

So then $g$ is convex if...

  • My original notation was correct actually. Both $g(x)$ and $f(x)$ are functions of x. This is how the problem states the question when asking about quasi-convexity. – Mike Aug 14 '16 at 03:08
  • @ml123 Typically, you do not use the same letter for the dummy variable and the limits. –  Aug 14 '16 at 03:09
  • @ml123 See http://math.stackexchange.com/questions/109105/limit-of-integration-cant-be-the-same-as-variable-of-integration –  Aug 14 '16 at 03:12
  • My hint for this question states g(x) is convex if and only if f(x) is nondecreasing, but indicates that it is still quasi convex because f(x) $\geq$ 0. Is there something I'm missing about this aspect of the question? – Mike Aug 14 '16 at 03:20
  • I updated the question to better reflect my issue. Why is g(x) quasi-convex if $f(x) \geq 0$ on the interval $[0,x_{max}]$? – Mike Aug 14 '16 at 06:31