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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable, strictly convex function. Let $I\subset \mathbb{R}$ be a closed, bounded interval such that $f'(x) \neq 0$ on $I$.

Is $f$ strongly convex on $I$?

Note: If $f$ is twice differentiable the answer is yes.

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For this version, consider $f(x)=(1/4)(x-1)^4+x$ with first derivative $(x-1)^3+1$ positive in an interval $I$ about $1$, but with second derivative $3(x-1)^2$ which has no positive lower bound near $1$, so that $f$ isn't strongly convex on $I$.

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  • $\begingroup$ Excellent, thanks. I may have one more version coming. $\endgroup$ – Ashley Aug 8 '14 at 2:16
  • $\begingroup$ math.stackexchange.com/questions/890634/… $\endgroup$ – Ashley Aug 8 '14 at 2:36
  • $\begingroup$ A simpler example would be just $x^4+x$ on a small enough interval containing $0$. $\endgroup$ – coffeemath Aug 8 '14 at 2:40

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