0
$\begingroup$

We consider $$ p^\star = \text{argmin}_{p\in \mathbb{R}^n} \quad 1/2f(p)^2- p\cdot g $$ where $f: \mathbb{R}^n\mapsto \mathbb{R}$ is a (smooth) 1-homogeneous and $g \in \mathbb{R}^n$ is constant.

We introduce $$ \pi^\star = \text{argmax}_{\pi | f(\pi)=1} \quad \pi\cdot g $$

Then, is this statement is true

$$ p^\star = \left(\pi^\star \cdot g\right) \pi^\star $$

?

EDIT: proposed solution: Ok let assume $f(p) \neq 0$.

We have

$$ p^\star = argmin_{p\in\mathbb{R}^n} 1/2f(p)^2-(p\cdot g) $$ $$ =argmin_{p\in \mathbb{R}^n} \min_{p|f(p)=1} 1/2f(p)^2-f(p)(p/f(p)\cdot g) $$ $$ =argmin_{p\in \mathbb{R^n}} 1/2f(p)^2-f(p) \max_{\pi|f(\pi)=1}(\pi\cdot g) $$

Let denote $\max_{\pi|f(\pi)=1}(\pi\cdot g) = \pi^\star$. $$ p^\star = argmin_{p\in \mathbb{R}^n} 1/2f(p)^2-f(p)(\pi^\star\cdot g) $$

then it is not difficult to see that $$ p^\star = (\pi^\star\cdot g)\pi^{\star} $$

what do you think ?

I am not looking for a proof that would work with precise statement. I want at least to know if I didn't have any reasoning mistakes like this step: $$ argmin_{p\in \mathbb{R}^n} h= argmin_{p\in \mathbb{R}^n} \min_{p|h(p)=1} h $$ but I ignore some extreme case where the result doesn't hold.

?

  • 0
    How would you like to cope with $f(p)=0$ (argmax over empty set), when argmin has multiple solutions, or when argmax has multiple solutions ($g=0$)? Are there any nonlinear smooth 1-homogeneous functions for which argmax exists? – 2017-01-10
  • 0
    ok, what if first we look for conditions when this result is true, don't we ? – 2017-01-14

0 Answers 0