Why is the support function of a convex cone the indicator function of its polar cone? Let $K ⊆ \mathbb{R}^{d}$ be a non-empty, closed, convex cone. Consider the support function $\sigma_K(x):= \sup_{y \in K} \langle x, y \rangle$. This function describes the (signed) distances of supporting hyperplanes of $K$ from the origin. Now let $p \notin K^{*}$, where $K^{*}$ denotes the polar cone of $K$. Then, there exists $y \in K$ s.t. $\langle p, y \rangle > 0$. Since $λy$ belongs to $K$ for all $λ ≥ 0$ this implies that $\sigma_K(p) = +∞$.
But I mean surely if we look at the supporting hyperplanes of $K$ in $0$ that are not in $K^{*}$, then their distance to the origin is $0$.
What is the error in reasoning?
 A: \begin{align}
\newcommand{\R}{{\mathfrak{R}}}
\end{align}
It might help to consider the indicator function.
Also we will use the convention that column vectors are elements of $\R^d$ and row vectors are elements of the dual space $(\R^d)^*$ (even though $(\R^d)^*\cong \R^d$).
When $K\subseteq\R^d$ is a nonempty, closed and convex cone, then the support function $\sigma_K:(\R^d)^*\to\R\cup\{+\infty\}$ coincides with the conjugate function of the indicator of $K$ given by $\delta_K:\R^d\to\R\cup\{+\infty\}$
$$
\delta_K(x) = \begin{cases} 0,&\text{if $x\in K$} \\ +\infty,&\text{o.w.}\end{cases}
 = \delta_{K^\circ}
$$
That is, $\sigma_K(\bullet) = \delta_K^*(\bullet) = \sup_x\{\bullet\cdot x - \delta_K(x)\} = \sup_{x\in K} \{\bullet \cdot x\}$.
We can also see that $\sigma_K = \delta_{K^\circ}$ where $K^\circ = \{y:y\cdot x \leq 0,;\forall x\in K\}$ is the polar cone of $K$.
When $f$ is convex and smooth, then for any $y\in(\R^d)^*$, $f^*(y)$ can be interpreted as the signed distance between zero and the intercept of a supporting hyperplane of $f$ with slope $y$.
We have this interpretation since in this special case, the slope of $f$ is monotone and hence there is a 1:1 map slopes $y$ and points $x$ (see here).
However, when $f$ is not convex, we don't have this relationship, so the $\sup$ is taken to make this relationship well-defined.
We're in the latter case when $f=\delta_K$, which is nonconvex, and the only supporting hyperplane of $\delta_K$ is $y=0$ when $x\in K$ (if $x\notin K$, then there are no supporting hyperplanes).
It helps to draw this out in 1d, for example with $K=\{x\in\R:x\geq0\}$ (see below).
This means that for any $y\in(\R^d)^*$, we can only interpret $\sigma_K(y)$ as the signed distance to zero of a supporting hyperplane for $\delta_K$ when $y=0$ and $x\in K$.
Otherwise, there are no supporting hyperplanes, so the signed distance should be $+\infty$.


