Showing convexity, having trouble showing positive definiteness

Question

I am interested in showing the convexity of

$$-\log(-f(\pmb{x}))$$

for $f: \mathbb{R}^{n}\rightarrow \mathbb{R}^{-}$ and $f$ convex. If we let $\nabla f$ denote the column vector where the $i$th entry contains $\frac{\partial f}{\partial x_{i}}$, after some computation I arrived at

$$g(\pmb{x}) :=-\frac{\operatorname{Hess}(f(\pmb{x}))}{f(\pmb{x})} + \frac{\nabla f (\nabla f)^{T}}{f^{2}(\pmb{x})} $$

Now, taking the quadratic form $\pmb{v}^{T}g(\pmb{x})\pmb{v}$, the first term must be non-negative since $f$ is convex and negative. My problem is that I cannot figure out what to do with the second term. I have tried writing it out in the form

$$\sum_{i}\sum_{j}v_{i}v_{j}f_{i}f_{j}$$

but I have no idea how the hypothesis helps me with this in anyway. I also tried to write this out as

$$\sum_{i}\sum_{j} -\frac{v_{i}v_{j}f_{i,j}}{f} + \frac{v_{i},v_{j}f_{i}f_{j}}{f^{2}}$$

but I don't see any reasonable way to relate the mixed partials with the product of first derivatives.

Answer 1

Note that $a^Tbb^Ta=|b^Ta|^2\geq 0$. Use $a=v$ and $b=\Delta f$.

Answer 2

As noted in the comments, the term $\frac{\nabla f (\nabla f)^{T}}{f^{2}(\pmb{x})}$ is a an outer product of a vector with itself, so it's a positive semidefinite matrix of rank 1.

For an alternative proof that doesn't assume $f$ is differentiable, you can use this representation of $-\log$: $$ -\log(t) = \max_{s >0} -\log(s) - (t-s)/s $$ Thus we have the expression: $$ -\log(-f(\pmb{x})) = \max_{s >0} -\log(s) + f(\pmb{x})/s + 1 $$ which is convex because the maximum of convex functions is convex.