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In Pr. Boyd's optimisation book, it is stated that for twice differentiable function $f$ with convex domain, $f$ is convex if and only if $\nabla^2f(x) \succcurlyeq 0$ for all $x \in dom f$. The point here is that the Hessian matrix should be positive semi-definite. My question is that can we have this for functions with scalar inputs? If so, why doesn't this hold for $f(x)=x^3$. While the stated condition is necessary and sufficient, the second derivative is $\frac{d^2}{dx^2}f(x) = 6x$ which is zero for input zero, but this is not a minimiser point.

Green Falcon
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  • It doesn't need to be a minimizer point : a convex function has at most one point of local (therefore global) maximum, and can possibly have none. Besides, $6x$ is negative for $x < 0$, so $x^3$ doesn't qualify as a convex function. – Sarvesh Ravichandran Iyer Aug 25 '20 at 05:08
  • @TeresaLisbon but $x^3$ is unbounded below. – Green Falcon Aug 25 '20 at 05:10
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    The function you have given is not convex, so its properties cannot be compared with those of a convex function. However, the function is convex on $[0,\infty)$, and on this domain it is not unbounded below. – Sarvesh Ravichandran Iyer Aug 25 '20 at 05:11

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For functions $\Bbb R\to \Bbb R$ the condition $\nabla^2_xf\succcurlyeq 0$ reduces to $f''(x)\geqslant0$.

$x^3$ is in point of fact convex on $[0,\infty)$ (because $f''\geqslant 0$ there) and not convex in any larger interval (because $f''$ has some negative values there).