Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f(X)\geq 0$ be a nonconvex $C^\infty$ function: $\mathbb R^3\to \mathbb R$.

Give any fixed $X_0$ such that $f(X_0)=\epsilon^+$, and the level set: ${L}=\{X\in \mathbb R^3:f(X)\leq \epsilon^+\}$ is a convex set.

And $\forall X_0$, $L$ is always a convex set.

Then what is the necessary and sufficient conditions, for $f(X)$ to be locally convex on ${L}$? (A mild sufficient condition would also be OK).

The sufficient condition better be easily (e.g. numerically) verifiable.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Equivalent to convexity is that the Hessian matrix is positive semidefinite for all $x \in L$. This can be done by checking if all eigenvalues of this symmetric matrix are nonegative. The eigenvalues in turn can be calculated by the zeros of the characteristic polynomial. They are real. In your special case $L\subset \mathbb{R}^3$, the characteristic polynomial is a polynomial of degree at most 3, that means the zeros can be found even analytically by Cardano's formula. The zeros would clearly depend on $X_0$.

Note that you assumed implicitely $\min f < \epsilon^+$ in order to avoid empty $L$.

share|improve this answer
    
Is there any sufficient condition which can be numerically verified? Since it is impossible to check all Hessians on $L$ –  LCFactorization Jan 17 at 14:05
1  
Yes, I'll edit my answer. –  Horst Grünbusch Jan 17 at 14:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.