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Given the function $f(x,y)=x^2y-3xy+y^2$ how can i find the subset of $R^2$ where $f$ is convex?
I already know there are saddle points in $(0,0)$ and $(3,0)$ and a local minimun in $(\frac 32, \frac 98)$.

Alex_28
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  • Where is the Hessian positive semidefinite? – copper.hat Nov 15 '23 at 19:48
  • The Hessian is $\begin{pmatrix} 2y & 2x-3\2x-3 & 2\end{pmatrix}$. Using Sylvester's criterion, since the principal minors are $2$, $2y$ and $-4x^2+12x+4y-9$, the Hessian is positive semideifinite is and only if $-4x^2+12x+4y-9>=0$, am i right? – Alex_28 Nov 15 '23 at 22:42
  • And $y\ge 0$. Also, keep in mind the question is a little ambiguous. A function is not convex at a single point. It depends on the domain. If the domain is the convex set described by the above constraints, then it is convex on that domain. If the domain is the plane, I would say that it is convex on the interior of that set. – copper.hat Nov 15 '23 at 23:22
  • Thank you. The question that i was given is "Find a region of the plane where the function is convex " so i just assumed that the domain was $R^2$. Why is it that if the domain is the plane then the function is convex only on the interior of the set? – Alex_28 Nov 16 '23 at 11:13
  • You are correct, the domain of definition is the entire plane. This may be splitting hairs. The answer depends on the intent of the question. Suppose $C$ is the (closed, convex) region described by the inequalities. Then I would say that $f$ restricted to the domain $C$ is convex. However, in the context of the plane as the whole domain, $f$ is not convex in a neighborhood of the boundary, so in this case I would say that $f$ is convex on the interior of $C$. – copper.hat Nov 16 '23 at 18:48
  • But why is $f$ convex only on the interior of $C$? – Alex_28 Nov 17 '23 at 11:36
  • Convexity is not a point thing. – copper.hat Nov 17 '23 at 14:38

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