# Convex Conjugate of Absolute Norm

Let $f:\mathbb{R}\rightarrow[-\infty,\infty]$ be a continuous function. The convex conjugate of $f$ is:

$$f^*(p) := \sup_{x\in\mathbb{R}}\{px-f(x)\}~.$$

Furthermore, let us define the subderivative $\partial f(a)$ of $f$ at $a$:

$$\partial f(a) := \{y\in[-\infty,\infty]: f(x)-f(a)\ge y(x-a)\}~.$$

I found out that for $f(x)=|x|$:

$$f^*(p) = \begin{cases} 0 & \text{for } |p|\le 1\\ \infty &\text{else}.\end{cases}$$

How can we show that?

• What is your interest in the set $\partial f(a)$? – MathOverview Jan 21 '12 at 18:34
• I thought (and was wrong) that the subderivative is relevant for the proof. For cases where $f$ is differentiable, one can use that $\frac{\partial}{\partial x}(px-f(x)) =0$, thus $p=f'(x)$ – B0rk4 Jan 21 '12 at 18:41

If $|p|\leq 1$ then $p-1\leq 0$ so $\sup_{x\geq 0}\{px-f(x)\}=\sup_{x\geq 0}\{px-x\}=\sup_{x\geq 0}\{(p-1)x\}$ and $(p-1)x\leq 0$ so $\sup_{x\geq 0}\{px-f(x)\}=0$, and since $p+1\geq 0$ we have $\sup_{x\leq 0}\{px-f(x)\}=\sup_{x\leq 0}(p+1)x=0$ so $f^*(p)=0$.
If $p>1$ then $\sup_{x\geq 0}\{px-f(x)\}=\sup_{x\geq 0}\{px-x\}=+\infty$ since $p+1\geq 0$.
If $p<-1$ then by a similar argument $\sup_{x\leq 0}\{px-f(x)\}=+\infty$.