## Strictly convex at… [closed]

Another terminology question:

If a function $f(x)$ is strictly convex at $y$, does this mean:

a) $f'(y) = 0$

b) $f''(x) \geq 0$ and $f''(y) \neq 0$

c) $f''(y) > 0$

d) $f'(y) = 0$ and $f''(y)>0$

e) Something else (specify)

Thanks, L

EDIT: Where $f(x)$ is not differentiable at $x$, read $f'(x)$ as the subgradient of $f$ at $x$. i.e. $f'(x)=0$ means that for subgradient [a,b], $a \leq 0$, $b \geq 0$. Similarly for $f''$.

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convex functions need not be everywhere differentiable – Yemon Choi Aug 12 2011 at 4:00
Might I suggest math.stackexchange.com as a better place for this particular question? – Yemon Choi Aug 12 2011 at 4:02
Yemon, 1) I know. Maybe I asked in a not very sensible way. I will clarify it. 2) I asked this question here because I want the opinion of the kind of people who wrote the research I am reading - an AMS monograph. Which I am using for my research. I just wanted an opinion on an ill defined phrase from serious people. I assumed questions about research level maths for the purpose of research was exactly what this was site was about. – Lucas Aug 12 2011 at 4:38
Lucas, thanks for the clarification. In that case, I suggest you rewrite the question to add this extra context, and perhaps even the monograph you are reading. It is much easier for people to answer such questions if some context is given; terminology is not absolute, but depending on one's knowledge of the literature, it becomes possible to say more once the sources are known. – Yemon Choi Aug 12 2011 at 5:55
How about: there is a line $l(x) = ax+b$ such that $l(y)=f(y)$ and $l(x)<f(x)$ for all $x \ne y$. Don't mention derivatives at all! – Gerald Edgar Aug 12 2011 at 13:41