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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.

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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 $L$.

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

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