Is there any convex tight upper bound (major) for $f(\mathbf{x},Y) = \mathbf x^T Y \mathbf x$?

Asked Jan 09 '21 at 18:34

Active Jan 09 '21 at 18:37

Viewed 37 times

In a minimization problem, I have this expression in my objective function

$$f(\mathbf{x},Y) = \mathbf x^T Y \mathbf x,$$ for $(\mathbf x,Y)\in \Omega = \{Y\succ 0,x\neq0\}$.

So the function is biconvex, but not convex as a whole, (one eigenvalue is always negative in the Hessian). I was trying to majorize it. i.e. to find a function $F$ tight around a point $(\mathbf x_0,Y_0)$ such that: $$F(\mathbf x_0,Y_0,\mathbf x,Y) = f(\mathbf x_0,Y_0),$$ $$F(\mathbf x_0,Y_0,\mathbf x,Y) \geq f(\mathbf x,Y),$$ $$\forall\; (\mathbf x_0,Y_0),(\mathbf x,Y)\in \Omega.$$ I know I can use a biconvex alternating procedure but, I think, if I could making it by majorization-minimization I could prove convergence.

edited Jan 09 '21 at 18:37

Asaf Karagila

393,674

asked Jan 09 '21 at 18:34

NSan

Do you want convexity of $F$ w.r.t. $(x,Y)$? – gerw Jan 09 '21 at 19:10
Yes, sorry I didn't explain it in the body, $F$ must be convex w.r.t. $(x,Y), \forall (x_0,Y_0) \in \Omega$ – NSan Jan 09 '21 at 19:23
I have a similar problem, did you find anything useful about this? – justmyfault Aug 18 '21 at 18:23

Is there any convex tight upper bound (major) for $f(\mathbf{x},Y) = \mathbf x^T Y \mathbf x$?

0 Answers0