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Usually for function of two variable checking the convexity of a constraint is quite simple since we just need to compute the Hessian. In this case, I am running into a problem where my constraint is a function of four non negative variable ${w_1},{w_2},\alpha ,\eta $ that is

$$w_1^2 + w_2^2 - w_2^2\alpha + 1 \ge \eta $$

$$f\left( {{w_1},{w_2},\alpha ,n} \right) = w_1^2 + w_2^2 - w_2^2\alpha + 1 - n \ge 0$$

The Hessian is

$$H = \left[ {\begin{array}{*{20}{c}} 2&0&0&0\\ 0&{2 - 2a}&{ - 2{w_2}}&0\\ 0&{ - 2{w_2}}&0&0\\ 0&0&0&0 \end{array}} \right]$$

And the corresponding eigen value is

$\begin{array}{*{20}{c}} {{\lambda _1} = 0}\\ {{\lambda _2} = 2}\\ {{\lambda _3} = \sqrt {{a^2} - 2a + 4w_2^2 + 1} - a + 1}\\ {{\lambda _4} = 1 - \sqrt {{a^2} - 2a + 4w_2^2 + 1} - a} \end{array}$

How to check for the convexity of such function (convex, concave or neither)?

Rodrigo de Azevedo
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Tuong Nguyen Minh
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1 Answers1

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You can compute the Hessian in any dimension.

You can also see that by fixing $w_2=\alpha=1$, $\eta=2$ you get the constraint $w_1^2\geq 1$ which is not convex, so your original constraint is not convex either.

Michal Adamaszek
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  • Oh, thank you but what would be the exact verdict in this case ? Is it convex, concave, or not convex and concave ? – Tuong Nguyen Minh Aug 11 '20 at 07:55
  • Neither because you can just as well get a section with $w_2^2\leq const$ which is not concave. Or you can see that your Hessian has a diagonal element which changes sign depending on $\alpha$. – Michal Adamaszek Aug 11 '20 at 08:04
  • I am sorry but in the case that $\alpha$ is a fixed positive constant would this constraint be consider as convex ? – Tuong Nguyen Minh Aug 11 '20 at 08:54
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    @TuongNguyenMinh Just take $\alpha=2$ and you have the expression $w_1^2-w_2^2$ which is neither convex nor concave. – Michal Adamaszek Aug 11 '20 at 09:12