I have the following function:
$f(\mathbf{x}) = \sqrt{\mathbf{u(x)^T Q_1 u(x)} + \mathbf{v(x)^T Q_2 v(x)}}, \quad \mathbf{Q_1, Q_2} \text{ real, symmetric, positive definite.} $
and I am querying whether it is convex. I believe it is but would appreciate a verification.
Here are my thoughts.
- Quadratic forms, $\mathbf{z^TAz}$, are strictly convex iff A is positive-definite, as can be seen by diagonalising A under an affine transformation.
- Consider $\mathbf{z} + t \mathbf{y}, \;\; \forall \;\mathbf{z, y} \in \mathcal{R}^N, \; t \in \mathcal{R} $. Then for $A$ having same properties as $Q_1, Q_2$, $$ \sqrt{ (\mathbf{z} + t \mathbf{y})^T \mathbf{A} (\mathbf{z} + t \mathbf{y})} = \sqrt{ \mathbf{y^TAy} \; t^2 + 2\mathbf{z^TAy} \; t+ \mathbf{z^TAz}}$$ is a strictly convex function since the square root of a quadratic univariate polynomial with positive leading coefficient is strictly convex.
Is it an obvious step I am missing to combine the two quantities under the root now with this second property?
The epigraph $t\geq f(x)$ is even conic quadratic representable:
$$ \begin{array}{l} t \geq \sqrt{p^2+q^2}, \\ p \geq \|F_1u\|_2, \\ q \geq \|F_2v\|_2, \\ \end{array} $$
where $Q_i=F_i^TF_i$. So yes, it is convex.