1

How to prove that the following function is non-convex?

$$f(\mathbf{c}) = \left( \|\mathbf{V}\|_{2}^2 - \|\mathbf{A}\cdot\mathbf{c}\|_{2}^2 \right)^2$$

I am trying to do it by demonstrating that $f$ does not fulfil

$$f(\alpha x+\beta y) \leq \alpha f(x)+\beta f(y)$$

with the help of the triangle inequality. Unfortunately, I am stuck since hours in an endless loop of confusing computations that I can't manage to solve. Does anyone have any hints or shortcuts for a reasoning showing that $f$ is not convex?

Thank you in advance for any help.

Rodrigo de Azevedo
  • 1
SuperKogito
  • 123
  • What is $\mathbf V$? – Ma Joad Jul 09 '19 at 10:26
  • $\mathbf{V}$ is a precomputed constant representing a vector of measurements. – SuperKogito Jul 09 '19 at 10:30
  • 1
    Consider the case $n=1$, $V=1$, $A=I$. The resulting function looks like $(1-c^2)^2$. – daw Jul 09 '19 at 12:17
  • Thank you for the suggestion; would this explanation make sense though: for $n=1, V=1, A=I, \alpha=1, and~ \beta=0$ I can say that the inequality does not hold since $(1-\alpha^2x^2)^2 > \alpha(1-x^2)^2$ and consequently it is a not a convex optimization – SuperKogito Jul 09 '19 at 13:05
  • my bad that does not work either, the inequality holds since it is not strict :( Alternatively, I can compute $f'' = 12c^2-4$ and since $f''(x)>0$ does not hold for all x, then $f$ is not a convex function. Is that correct? However, in the case of complex input, this does not hold? – SuperKogito Jul 09 '19 at 15:24
  • 1
    Functions should have domains and codomains. It's confusing when some vectors are uppercase and others are lowercase. Using the definition of the $2$-norm and expanding would be wiser than using the definition of convexity. – Rodrigo de Azevedo Jul 09 '19 at 19:50

0 Answers0