Suppose that $X=\mathbb{R}^n$ and that $K=\mathbb{R}_{+}^{n}$.

Question

Suppose that $X=\mathbb{R}^n$ and that $K=\mathbb{R}_{+}^{n}$. using the definition of the Fenchel conjugate verify that $\iota_{K}^{*}=\iota_{-k}$ where $\iota_{K}$ is the indicator function.

my first question is am I correct in assuming that that $-k=\mathbb{R}_{-}^{n}$ and if so, does this mean the fenchel conjugate of the indicator function takes in positive values and outputs negative values?

Second I am hoping someone can help along with the following proof:

Given that $\iota_{K}(x)=0$ if $x\in k$ and $\iota_{K}(x)=+\infty$ if $x\not\in k$ we have two cases to consider.

case 1, $ \;\;x\not\in k$ \begin{align} \iota_{K}^{*}(v)&=\sup(\langle x,v\rangle-\iota_{K}(x))\\ &=\sup(\langle x,v\rangle-\infty)\\ &=-\infty \end{align} which not not a supremum.

case 2, $ \;\;x\in k$ \begin{align} \iota_{K}^{*}(v)&=\sup(\langle x,v\rangle-\iota_{K}(x))\\ &=\sup(\langle x,v\rangle-0)\\ &=xv. \end{align} We now must consider the cases where $v<0,v=0$ and $v>0$

and this is pretty much as far as I've got. I know that when $v=0$ then $xv=0$. However, I could use some help with the other two, as well as anywhere else I may have gone wrong. Thank you.

Answer 1

I'll just write down the solution and hope that you catch some tricks from the manipulations :)

Define $\mathbb{R}^n_+ := \{x \in \mathbb{R}^n | x \ge 0\}$ (the nonnegative $n$th orthant), and $\mathbb{R}^n_- := \{x \in \mathbb{R}^n | x \le 0\}$ (the nonpositive $n$th orthant). From these definitions, it's clear that $-\mathbb{R}^n_+ = \mathbb{R}^n_-$. Now, your objective is to show that $i_{\mathbb{R}_+^n}^* = i_{-\mathbb{R}_+^n}$.

Indeed, for any $x \in \mathbb{R}^n$, we have \begin{equation} \begin{split} i_{\mathbb{R}_+^n}^*(x) &:= \sup_{z \in \mathbb{R}^n}\langle x, z\rangle - i_{\mathbb{R}_+^n}(z) = \sup_{z \ge 0}\langle x, z\rangle \begin{cases} \ge \sup_{t \ge 0}tx_i = +\infty, &\mbox{ if some }x_i > 0,\\=0, &\mbox{ otherwise}\end{cases}\\ &= \begin{cases}0, &\mbox{ if }x \le 0,\\+\infty, &\mbox{ otherwise}\end{cases}\\ &= i_{\mathbb{R}_-^n}(x) = i_{-\mathbb{R}_+^n}(x). \end{split} \end{equation}