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

So the objective function can be written in the form $$ 1/2f(p)^2- f(p) \left(\frac{p}{f(p)}\cdot g(p)\right) $$

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

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

I don't understand the result, especially why is it natural to introduce $\pi^\star$ ?

We can easily get convinced in dim 1 = n.

Thanks for your enlightenment

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With $f(p) = p$ you obtain $\pi^*=1$ and $p^* = g(1)$. So, the claim is that: $$\text{argmin}_{p} 1/2p^2 - p g(p) = g(1).$$ This is not necessarily true, for example when $g(p) = -(p-1)$: $$\text{argmin}_{p} 1/2p^2 + p(p-1) = \frac{1}{3} \neq 0 = g(1).$$

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  • $\begingroup$ no the objective function contains $p \cdot g(p)$ and it is homogeneous of degree 1 in $p$ so your function is not a valid candidate $\endgroup$ – Smilia Jan 4 '17 at 20:46
  • $\begingroup$ @Smilia I changed the objective function in my answer. What exactly should be homogeneous of degree 1? I think $f$ is. $\endgroup$ – LinAlg Jan 4 '17 at 21:38
  • $\begingroup$ oh that's right thanks for this point, I edtited my post to consider the case where $f(p)$ is in fact a norm ... it should be more interesting $\endgroup$ – Smilia Jan 5 '17 at 8:09
  • $\begingroup$ @Smilia that does not affect the answer much. Just take $f(p)=|p|$, then $\pi^* = -1$ (for my choice of $g$), but $g(-1) \neq 1/3$. $\endgroup$ – LinAlg Jan 7 '17 at 10:22

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