Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be strictly convex and differentiable. Is $f$ strongly convex when restricted to a closed and bounded interval $[a,b]$?

This is true if $f$ is smooth but am wondering if it still holds when $f$ doesn't have a second derivative.

Strong Convexity (Wiki)

  • $\begingroup$ I think it's false. What about linear functions? They're strictly convex but nowhere strongly convex (since the gradient is constant). Perhaps you might want to rule out this scenario though. $\endgroup$ – bartgol Aug 7 '14 at 22:14
  • $\begingroup$ linear functions aren't strictly convex. Strictly convex: $f(x_2)> f(x_1) +f'(x_1)(x_2-x_1)$ for $x_1 \neq x_2$. $\endgroup$ – Ashley Aug 7 '14 at 22:19
  • $\begingroup$ My bad. Too tired to think. I'll let someone else jump in. ;-) $\endgroup$ – bartgol Aug 7 '14 at 22:27

$f(x)=x^4$ is strictly convex, but not strongly convex on any interval including $0$ since its second derivative $12x^2$ won't be bounded above some positive constant near $0.$ [Note this example appeared on the wiki page you cited.]

  • $\begingroup$ Thank you. I've been focusing on the case where the interval does contain a minimizers/a point w horizontal tangent for the function. In that case, any smooth function will become strongly convex, but what about the case where there's no 2nd derivative? (I will probably make this new quesiton). $\endgroup$ – Ashley Aug 8 '14 at 1:46
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    $\begingroup$ Here it is math.stackexchange.com/questions/890613/… $\endgroup$ – Ashley Aug 8 '14 at 1:57
  • $\begingroup$ @Pilo The re-vamped question you reference in your comment looks interesting. $\endgroup$ – coffeemath Aug 8 '14 at 2:02
  • $\begingroup$ Thanks, yeah I've spent all day trying to figure it out. $\endgroup$ – Ashley Aug 8 '14 at 2:04

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