How do I show that the hessian of the log barrier function is positive semi definite ? How do I show that $z^THz \ge 0$ for a non-zero z ? It looks extremely complicated. I forgot to mention that in the expression below, $f_i(x)$ is convex.
This is my attempt at a solution:
- $\triangledown f_i(x) \triangledown f_i(x)^T$ is positive semidefinite and $\dfrac{1}{f_i(x)^2}$ is just a positive multiplier. Thus the expression on the left is convex.
- $\triangledown^2 f_i(x)$ is non negative because it is the second derivative of a convex function. Since $f_i(x) $ must be $< 0$ (since it is used as a barrier for the inequality), then $\dfrac{1}{-f_i(x)}$ is positive.
Thus the expression is a sum of 2 positive semidefinite matrices, making it convex. Would this be correct ?