Yes, this is homework.
Given $f(x) = 1^{T}(x)_+$, where $(x)_+ := \max\{0,x\}$, what is $f^*$?
I know that the conjugate of a function $f$ is $$ f^*(y) = \sup \left( y^T x - f(x) \right) $$ but I do not know how to show the conjugate of $f(x) = 1^{T}(x)_+$. I am looking for the steps to determine the conjugate and the $\operatorname{dom} f^*$.
Let us first examine the case where $x\in\mathbb{R}$, $f(x) = \max(0,x)$.
Then, the conjugate is
$$ \begin{align} f^*(y) &= \sup_{x\in\mathbb{R}} \{yx-\max(0,x)\}\\ &= \max \left\{ \sup_{x\geq 0}(yx-x), \sup_{x<0}(yx+x)\right\}\\ &= \max \left\{ \sup_{x\geq 0}x(y-1), \sup_{x<0}x(y+1)\right\}\\ &= \max \left\{ \begin{cases}0,&\text{if } y\leq 1\\ +\infty,&\text{otherwise } \end{cases}, \begin{cases}0,&\text{if } y\geq -1\\ +\infty,&\text{otherwise } \end{cases} \right\}\\ &=\begin{cases} 0,&\text{if } y \in [-1,1]\\ +\infty,&\text{otherwise } \end{cases} \end{align} $$
Clearly $\operatorname{dom} f^* = [-1,1]$. Note here that $f^*$ is the indicator function of $[-1,1]$, therefore $f$ is the support function of this interval.
Now for the case where $x\in\mathbb{R}^n$ we have
$$ f(x) = 1'[x]_+, $$
where $[x]_+=(\max(0,x_1), \max(0,x_2), \ldots, \max(0,x_n))$. Then $f^*$ can be written as
$$ \begin{align} f^*(y) &= \sup_{x\in\mathbb{R}^n} \{y'x - f(x)\}\\ &= \sup_{x\in\mathbb{R}^n} \left\{\sum_{i=1}^{n}y_i x_i - \sum_{i=1}^{n}\max(0,x_i)\right\}\\ &= \sup_{x\in\mathbb{R}^n} \left\{\sum_{i=1}^{n}y_i x_i - \max(0,x_i)\right\} \end{align} $$
The sum is separable, therefore we can easily break down the big $\sup$ into a sum of $\sup$'s and use the previous result for the case $x\in\mathbb{R}$.
$$ \begin{align} f^*(y) &= \sum_{i=1}^{n}\sup_{x_i\in\mathbb{R}}\{y_ix_i-\max(0,x_i)\}\\ &= \sum_{i=1}^{n}\delta(x_i\mid [-1,1])\\ &= \delta(x\mid \mathcal{B}_{\infty}), \end{align} $$
and of course $\operatorname{dom}f^* = \mathcal{B}_\infty$.
