Convex matrix function

Question

Please give me some hints for the following problem:

Let $S = \{D \in R^{m \times n}, \|d_i\| \leq 1, i = 1, 2, \dots, n \}$. Find condition of $F \in R^{m \times m}$ such that the function: $ f(D) = trace(DD^TF)$ is convex.

Thank you so much.

Answer 1

$$f(D)=\mathop{\textrm{Tr}}(DD^TF)=\mathop{\textrm{Tr}}(D^TFD)=\sum_{i=1}^n d_i^TFd_i.$$ Each term in the summation a simple quadratic form with $F$ as the quadratic coefficient matrix. So $f$ is convex if $F$ is positive semidefinite.

There are a couple of ways to see this. One is to define $g(v)=v^TFv$, so that $f(D)=\sum_i g(d_i)$. Because $f$ is separable across the $d_i$s, $f$ is convex if and only if $g$ is. The Hessian of $g$ is $\nabla^2 g(v)=2F$, so $g$ is convex if and only if $F$ is positive semidefinite.

Alternatively, let $\bar{d}=\mathop{\textrm{vec}} D\in\mathbb{R}^{mn}$ be the vector formed by "stacking" the vectors $d_i$ on top of each other, and rewrite $f$ in terms of this vector. Then $$\nabla^2 f(\bar{d}) = 2\begin{bmatrix} F \\ & F \\ & & F \\ & & & \ddots \\ & & & & F \end{bmatrix}$$ This block diagonal matrix is positive semidefinite if and only if $F$ itself is positive semidefinite.

At first glance the set $S$ is not relevant, because $f$ is convex over all $\mathbb{R}^{m\times n}$. But if one is going to consider a restricted domain such as $S$, it is important that the set itself be convex as well. Fortunately, in this case, it is, being the intersection of $n$ norm balls.