0
$\begingroup$

The definition for strongly convex is: A differentiable function $f$ is strongly convex if $$ f(y) \geq f(x)+\nabla f(x)^{T}(y-x)+\frac{\mu}{2}\|y-x\|^{2} $$ for some $\mu>0$ and all $x, y$.

And for this function: $$f(x) = sin(x),~~ x\in(\pi, 2\pi)$$

One way to see this is to calculate the second derivative, which is $-\sin (x)$ and that you cannot lower bound on the lower bound as $\sin (\pi)=\sin (2 \pi)=0$. But the region for this function does not contain $\pi$ and $2\pi$ which confuses me.

New contributor
Chengkun Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
3
  • $\begingroup$ hint: try graphing the 2nd derivative over $(-\pi,2\pi)$ and looking at its sign. $\endgroup$
    – Zim
    Jun 23 at 10:06
  • $\begingroup$ If the inequality holds in $(\pi,2\pi)$ then it holds in $[\pi,2\pi]$ by continuity. $\endgroup$
    – geetha290krm
    Jun 23 at 10:12
  • $\begingroup$ @geetha290krm How to tell if it holds in $(\pi, 2\pi)$? $\endgroup$
    – Chengkun Li
    Jun 23 at 19:45

0

Reset to default

Your Answer

Chengkun Li is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.