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.

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  • $\begingroup$ 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. $\endgroup$ Jun 23 at 10:12
  • $\begingroup$ @geetha290krm How to tell if it holds in $(\pi, 2\pi)$? $\endgroup$ Jun 23 at 19:45


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