Solution
= Solution
The condition is $\max(|x-1/2|,|y|)\le1/2$. Therefore the ball is the \b[closed square]
$$
\boxed{[0,1]\times[-1/2,1/2].}
$$
Its four <vertices> are $(0,\pm1/2)$ and $(1,\pm1/2)$.
\Image[/past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2008-supremum-balls.png]
{title=Closed supremum-norm ball and its restriction to the L-shaped metric space}