Talk:Twice Differentiable Real Function with Positive Second Derivative is Strictly Convex

I don't think the is right. $f'$ strictly increasing only gives $f'' \ge 0$ not $f' > 0$, eg. $\map {f'} x = x^3$ at $0$. So the correct result I think is if $f > 0$ then $f$ is strictly convex, and if $f$ is strictly convex then $f \ge 0$. (and you have $f$ suitably differentiable is convex $f \ge 0$ from Second Derivative of Convex Real Function is Non-Negative, so this means you can't distinguish between strict/non-strict convexivity this way if $f = 0$ somewhere) I can't find the result $f$ increasing + differentiable implies $f' \ge 0$, though, only the converse. If I'm not mistaken I will review these pages when I get time tomorrow. Caliburn (talk) 23:24, 30 January 2022 (UTC)


 * Please do -- I confess to being careless here. Feel free to offer up counterexamples to the other direction. There are indeed subtleties which often get glossed over. Would be good to provide the definitive answer to this. --prime mover (talk) 23:38, 30 January 2022 (UTC)

Title should include "Strictly positive"
Please notice that positive in just means the non-negative. This theorem requires strictly positive. --Usagiop (talk) 22:13, 6 January 2023 (UTC)


 * Works for me, go for it. Nice one. --prime mover (talk) 23:14, 6 January 2023 (UTC)