Determine $\sup \{xy−x^2/2\mid x\in [-1,1]\}$

Question

I am trying to find the supremum of the following set $\{xy−x^2/2\mid x\in [-1,1]\}$, where $y$ is a real number. I am not sure if this is correct, but I managed to find that $$ \sup \{ xy-x^2/2\mid x\in [-1,1]\}=\begin{cases} y^2/2 & \text{ if } y\in [-1,1] \\ y-1/2 & \text{ if } y>1 \\ -y-1/2 & \text{ if } y<-1 \end{cases} $$

Answer 1

You are computing the convex conjugate of $(f+g)(x)$ where $f(x) = 0.5 x^2$ and $g(x) = \delta(x, [-1,1])$ (taking the value 0 if $x$ is in the interval and $\infty$ otherwise. Note that $f^*(y) = 0.5 y^2$ and $g^*(y) = |y|$. $$\sup_x \{ xy - (f+g)(x)\} = (f+g)^*(y)= \inf_z \{ f^*(z)+g^*(y-z)\} $$ $$ = \inf_z \{ 0.5 z^2+ |y-z|\} $$ This is an unconstrained convex optimization problem. The derivative is $z-1$ if $y > z$, and $z+1$ if $y < z$, and the gradient is $[z-1,z+1]$ if $y=z$. We have that 0 is in the gradient if ($z=1$ and $y>z$) or ($z \in [-1,1]$ and $y=z$) or ($z=-1$ and $y<z$). This gives the three cases in your question.