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I have question about global optimum and negative (positive) semidefiniteness.

I know that the sufficient conditions (for KKT) is the concavity (convexity) of the objective function. (The constraints are affine.)

I know that if the Hessian is negative definite for all $x$, the function is strictly concave and the first order conditions give the global maximum. If the Hessian is positive definite for all $x$, the function is strictly convex and the first order conditions give the global minimum.

What happens if the function is negative semidefinite or positive semidefinite? If one of the leading principal minors is $0$, does this mean the "second order test" inconclusive? Is the stationary point still global max or min? Is the function concave (convex) but not strictly concave (strictly convex)?

For example: $$f(x,y)=\frac{-1}{ax-by}$$

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Let $C$ be an open convex subset of $\mathbb{R}^n$ and $f:C\rightarrow \mathbb{R}$ be $C^2$. If, for every $x\in C$, $Hess(f)(x)$ is symmetric $\geq 0$, then $f$ is convex over $C$.

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  • $\begingroup$ Note that the condition being required at every point rules out $f(x)=x^3$ whenever $C$ includes negative numbers. $\endgroup$
    – Ian
    Commented Jul 18, 2016 at 11:52

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