In S.Boyd's lecture:
And in his vedio, he said: You are allowed one positive eigenvalue in the Hessian of log-concave function.
http://web.stanford.edu/class/ee364a/videos/video04.html (at 1:03:30)
I am confused about this? Why it is true? (Through this inequality)
The proof is similar to the proof of Corollary 3.18 in [Avriel, Mordecai, et al. Generalized Concavity]:
Suppose that $\nabla^{2}f(x)$ has two (or more) positive eigenvalues. Let $e_{1}$ and $e_{2}$ be orthogonal eigenvectors associated with positive eigenvalues. Let $T$ denote the subspace spanned by set of vectors orthogonal to $\nabla f(x)$ and let $E$ denote the subspace spanned by $e_{1}$ and $e_{2}$. Then $T \cap E$ must contain at least one non-zero vector $v = \alpha e_{1} + \beta e_{2}$, where $\alpha$ and $\beta$ are not both zero. By definition, $f(x) \nabla^{2} f(x) \preceq \nabla f(x) \nabla f(x)^{T}$ means, for all $z$:
$$ z^{T} \nabla^{2} f(x) z \leq \frac{(z^{T} \nabla f(x))^{2}}{f(x)}.$$
Since $e_{1}$ and $e_{2}$ are orthogonal eigenvectors with positive eigenvalues, we find that the quadratic from $v^{T} \nabla^{2} f(x)v$ is positive, whereas $v^{T} \nabla f(x) = 0$, contradicting the above inequality.
