Let $f:\mathbb{R}^n \rightarrow \mathbb{R} \cup \{ \infty \}$ be a convex function. The convex conjugate of $f$, which we call $f^*$ is defined as
$f^*(y)=\sup \, \left \{ \langle y,x \rangle -f(x) \, \, \vert \, \, x \in \mathbb{R}^n \right \}$
I'm trying to find the convex conjugate of $f(x)=|\langle a,x \rangle|$, where $| \cdot |$ is the absolute value function.
It is easy to find that for $f(x)=\langle a, x \rangle$ we have $f^*(y)=0$ if $y=a$ and $f^*(y) = \infty$ else.
Also for $f(x)=| x |$ we find that $f^*(y)=0$ if $|y|\leq 1$ and $f^*(y)=\infty$ else.
I have trouble though for the composition of both. I appreciate any hints!
Best Answer
In general, there is no good way to find the conjugate of a composition. But you can use the following fact:
Fact. Let $M$ be a linear subspace of $\mathbb R^n$, and let $g:M\to\mathbb R\cup\{+\infty\} $ be a convex function. If $\pi_M$ is the orthogonal projection onto $M$, then $$ (g\circ \pi_M)^*(y) = \begin{cases} g^*(y),\quad &y\in M \\ +\infty,\quad &y\notin M \end{cases} \tag1 $$
Proof. Let $M^\perp$ be the orthogonal complement of $M$. Every vector $x$ decomposes as $x_M+x_{M^\perp}$ where $x_M\in M$ and $x_{M^\perp}\in {M^\perp}$. Therefore, $$(g\circ \pi_M)^*(y) =\sup_{x_M, \ x_{M^\perp}} \left(\langle y, x_M\rangle+ \langle y, x_{M^\perp}\rangle - f(x_M) \right) \tag2 $$ Observe that $x_{M^\perp}$ appears only in the term $\langle y, x_{M^\perp}\rangle$. If $y\in M$, then $\langle y, x_{M^\perp}\rangle=0$ for all $x_{M^\perp}$ and therefore (2) becomes the formula for $g^*$. If $y\notin M$, then $\sup_{ x_{M^\perp}} \langle y, x_{M^\perp}\rangle =+\infty$. $\Box$
Combining the above fact with your knowledge of one-dimensional conjugate $$|cx|^*(y)=\begin{cases} 0 , \quad &|y|\le |c| \\ +\infty , \quad &\text{otherwise} \end{cases} \tag3$$ you can conclude with $$|\langle a,x\rangle|^*(y)=\begin{cases} 0 , \quad &y=ta, \ -1\le t\le 1 \\ +\infty , \quad &\text{otherwise} \end{cases} \tag4$$