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$$f(x) = |x|^{3/2}, x \in \mathbb{R}$$ If you take the 2nd derivative you get $$ f''(x) = \frac{3}{4\sqrt{|x|}}$$ If you go by the requirement that $f$ is strongly convex when $$f''(x) \geq m \gt 0$$ Then $f$ is strongly convex.

However, if you go by the requirement that $f$ is strongly convex when

$$g(x)=f(x)−\frac{m}{2} ||x||^2$$ is convex, $∀x$ and some $m>0$ $$g(x) = |x|^{3/2} - \frac{m}{2} ||x||^2$$ $g$ is clearly not convex since $\frac{d^2}{dx^2}\frac{m}{2}||x||^2 > \frac{d^2}{dx^2}|x|^{3/2}$ for large enough $x$.

Please let me know where my mistake is!

Thanks

user5046480
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  • @Ian Huh, $$g''(x)=f''(x)-m$$ ? For instance $$\frac{\mathrm d^2}{\mathrm dx^2} \left[\frac{m\lVert x\rVert^2}2\right]=m$$ since $\lVert x\rVert^2=x^2$. So isn't $$f''(x)\geq m$$ equivalent to $$g''(x)\geq0$$ ? – Maximilian Janisch Sep 29 '20 at 14:56
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    @MaximilianJanisch I misread the OP, either that or it got edited after my comment, not sure which. You're correct. I think what I said underlies OP's confusion, however. Namely, the first requirement doesn't hold here specifically because the $m$ that lower-bounds $f''$ is not universal; it has to go to zero at infinity, so it can't be constant everywhere. This breaks strong convexity without breaking strict convexity. – Ian Sep 29 '20 at 15:07
  • That makes sense! I wasn't appreciating the need for the constant $m$. Since by my interpretation I was essentially testing for strict convexity. Thank you for your answer! In terms of edits, I replaced a few statements where I had "convex" instead of "strongly convex." Sorry for the confusion! – user5046480 Sep 29 '20 at 15:15
  • @Ian I think your comment should be an answer – supinf Sep 29 '20 at 16:44

1 Answers1

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It is true that $f''>0$ where it is defined, and in fact $f$ is still convex on an interval containing $0$ even though $f''(0)$ is not really defined. (In a somewhat loose sense you can say $f''(0)=+\infty$.)

However, the first requirement you mentioned actually does not hold because there is no single number $m>0$ such that $f'' \geq m$ everywhere, since $f''$ goes to zero at infinity. Thus $f$ is not strongly convex, but it is nevertheless strictly convex.

Ian
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