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I know that non negative second derivative means the function is convex but does positive second derivative means the function is strictly convex ? I could not find references to this and even my professor said he need to check and was not sure. Can someone point me to the proof or disprove this ? I saw some posts on the connection between the second derivative of a function and the fact that it is convex but did not see posts on the conditions for being strictly convex.

Tomer
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    See https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable: “If its second derivative is positive at all points then the function is strictly convex, but the converse does not hold.” – Martin R Jan 10 '23 at 19:00
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    It seems like pretty much the same proof as for nonnegative second derivative should work: suppose $a < b$ and $t \in (0, 1)$ and set $c := ta + (1-t)b$, Then by the mean value theorem, there exist $\xi_1 \in (a, c)$ and $\xi_2 \in (c, b)$ such that $f'(\xi_1) = \frac{f(c)-f(a)}{c-a}$ and $f'(\xi_2) = \frac{f(b)-f(c)}{b-c}$. Then $f''$ being strictly positive implies $f'(\xi_1) < f'(\xi_2)$, and from there straightforward algebra shows that this implies $f(c) < t f(a) + (1-t) f(b)$ as required. – Daniel Schepler Jan 10 '23 at 19:07
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    Does this answer your question? Second derivative positive $\implies$ convex – Kurt G. Jan 10 '23 at 19:30
  • Maybe - I rather have a post that address the strictly convex conditions directly. – Tomer Jan 10 '23 at 21:03
  • That duplicate contains a proof which is adapted to the strict convex case with zero effort. – Kurt G. Jan 11 '23 at 15:41

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Yes.

If you can find $x<y$ and $\lambda\in(0,1)$ such that $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$, then you can easily show that $f$ is linear on $(x,y)$, so its second derivative is $0$ on $(x,y)$. So if its second derivative is positive then you cannot find such $x,y$ and therefore it is strictly convex.

Note that the converse is not true. For instance $x\mapsto x^4$ is strictly convex, but its second derivative $x\mapsto12x^2$ is zero at $0$.

Will
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