# What type of convex constraint is defined by SQRT?

Let $A$ be an $n \times n$ positive semidefinite matrix and $\forall k, x_k \in \mathbb{R}^n$. The distance with respect to this matrix is defined as

$\|x_i -x_j\|_A := \sqrt{(x_i-x_j)^TA(x_i-x_j)}$.

Now, suppose we have the constraint $$\sum_{(x_i,x_j) \in D} \|x_i -x_j\|_A = \sum_{(x_i,x_j) \in D} \sqrt{(x_i-x_j)^TA(x_i-x_j)} \ge 1$$ I know (by reading in a paper), that this is a $\textbf{convex constraint in } \mathbf{A}$, but cannot verify that (because of the sqrt).

Can anyone help me please or give a hint? Why is it convex in $A$? Is it? More importantly, what type of convex constraint is that (linear, quadratic, SOC, SDP?)

• Any thoughts can be helpful. Oct 24, 2013 at 4:10
• How is $D$ defined? Oct 24, 2013 at 4:18
• $D$ is just a set of pairs. Some domain. Oct 24, 2013 at 4:28
• This is equivalent to asking if $\|x\|_A$ is a concave function of $A$. Oct 24, 2013 at 4:36
• Maybe should compute the Hessian and prove it is negative semi-definite. Oct 24, 2013 at 5:48