Skip to main content
1 of 3
F_G
  • 837
  • 5
  • 7

Function $W \mapsto f(W)$ is not convex.

Already the simplest case, take $n=1$ (matrices $W_i$ and vector $x$ become scalars) and fix $x=1$: then function $f$ becomes the product of the squares of its arguments, which is not convex unless there is only one factor in the product (for example $(x,y) \mapsto x^2 y^2$ is not convex, as can be checked by computing its Hessian matrix).

F_G
  • 837
  • 5
  • 7