Polar cone of a closed convex cone in $R^4$ defined by a convex inequality constraint

Question

Let $c>1$ be a constant. Consider points in four dimension with coordinates $(x,y,z,p)\in R_{\ge 0} \times R_{\ge 0} \times R \times R_{\ge 0}$ and the cone $$K = \{ (x,y,z,p)\in R_{\ge 0} \times R_{\ge 0} \times R \times R_{\ge 0} : \Big( (\sqrt{x^2+y^2}-z)_+^2+p^2\Big)^{1/2} \le c y \}.$$ Above $t_+=\max(0,t)$ is the positive part of a real $t$. This cone is closed and convex, because the left-hand side of the inequality is a convex function of $(x,y,z,p)$. I am wondering if there are general methods, or just a particular trick for this specific problem, that would give an explicit expression of the polar cone: $$K_* = \Big\{ (x_*,y_*,z_*,p_*)\in R^4: \max_{(x,y,z,p)\in K} \Big(x x_*+y y_*+z z_*+p p_* \Big)\le 0 \Big\}.$$ Remark (notation): $x,y,z,p$ are all scalars, and $x x_*+y y_*+z z_*+p p_*$ is the usual inner product in $R^4$ between $(x,y,z,p)$ and $(x_*,y_*,z_*,p_*)$.

Answer 1

Define a function $g:\mathbb{R}^4 \to \mathbb{R}\cup\{+\infty\}$,

$$ g(x,y,z,p)=\begin{cases}\sqrt{\max\{\sqrt{x^2+y^2}-z,0\}^2+p^2}-cy&\text{if }(x,y,z,p)\in [0,+\infty)\times[0,+\infty)\times\mathbb{R}\times[0,+\infty),\\+\infty,&\text{else.}\end{cases} $$

Then $\{(x,y,z,p)\in\mathbb{R}^4:g(x,y,z,p)<0\}\neq \emptyset$ and $K=\{(x,y,z,p)\in\mathbb{R}^4:g(x,y,z,p)\leq 0\}$.

This means we are in the setting of Lemma 27.20 of

Heinz H. Bauschke, Patrick L. Combettes: Convex Analysis and Monotone Operator Theory In Hilbert Spaces, 2nd edition, 2017. CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC. Springer, Cham. ISBN 978-3-319-48310-8/hbk; 978-3-319-48311-5/ebook.

Since you are looking for the polar cone, you have to compute the normal cone $N_K(0,0,0,0)$ to $K$ at $(0,0,0,0)$, see Definition 6.38, and since $g(0,0,0,0)$, you'll have

$$N_K(0,0,0,0)=N_{\mathrm{dom}(g)}(0,0,0,0)\cup \mathrm{cone}(\partial g(0,0,0,0)).$$

The domain of $g$ is $\mathrm{dom}(g)=[0,+\infty)\times[0,+\infty)\times\mathbb{R}\times[0,+\infty)$, so we should have

$$N_{\mathrm{dom}(g)}(0,0,0,0)=(-\infty,0]\times(-\infty,0]\times\{0\}\times (-\infty,0].$$

It remains to compute the subdifferential of $g$ at $(0,0,0,0)$. This should be possible in a finite amount of time, using Proposition 16.42 about subdifferentials of pointwise sums and Corollary 16.72 about the subdifferential of a composition.