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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.

  1. Quadratic forms, $\mathbf{z^TAz}$, are strictly convex iff A is positive-definite, as can be seen by diagonalising A under an affine transformation.
  2. 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?

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    $\begingroup$ What are $u$ and $v$? $\endgroup$ – Berci Jun 29 '18 at 9:23
  • $\begingroup$ Good question, in my specific case they are affine transformations of x, i,e, $u(x) = A_1 x$, and $v(x) = A_2 x$, but I'm guessing you are hinting at the real possibility that the answer very much hinges on their form. In my linear case it is easier to show, but in the more general case for u and v, it may or may not be convex? $\endgroup$ – Attack68 Jun 29 '18 at 9:27
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    $\begingroup$ If $u$ and $v$ are linear, you have $u(x)^\top Q_1 u(x) + v(x)^\top Q_2 v(x) = x^\top D x$ for a certain symmetric, positive semidefinite matrix $D$. $\endgroup$ – gerw Jun 29 '18 at 9:40
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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.

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