Conjugate of a function 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 $f^*(y) = \sup (y^Tx - f(x))$ 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 $dom f^*$
 A: 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 $\operatorname{dom}f^* = \mathcal{B}_\infty$.
