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Another terminology question:

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

a) $f'(y) = 0$, or equally, $y = \arg \min_y f(y)$

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

c) 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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    What is the derivative/subgradient of a subgradient?2011-08-12
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    How about: there is a line $l(x) = ax+b$ such that $l(y)=f(y)$ and $l(x) for all $x\ne y$. Do not mention derivatives!2011-08-12
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    @GEdgar: there's however the following problem (see my answer below). A function is convex if the set of subdifferentials is non-empty. You propose that $f$ is strictly convex at $y$ if there is an element in its subdifferentials that is strictly less than $f$ (except at $y$). But one can alternatively propose that the requisite property is that all subdifferentials be strictly less than $f$ except at $y$.2011-08-12

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