# Is this function strongly convex? approaching boundary case

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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• hint: try graphing the 2nd derivative over $(-\pi,2\pi)$ and looking at its sign.
– Zim
Jun 23 at 10:06
• If the inequality holds in $(\pi,2\pi)$ then it holds in $[\pi,2\pi]$ by continuity. Jun 23 at 10:12
• @geetha290krm How to tell if it holds in $(\pi, 2\pi)$? Jun 23 at 19:45