I have the following set:
\begin{align} M = \{ x \in \mathbb{R}^n: x \geq 0, x^{T}y \leq 1, \forall y \text{ with } \lVert y \rVert \leq 1 \} \end{align}
I would like to rewrite this set so as to find out whether it is a polyhedron, defined as the intersection of finitely many halfspaces of the form:
\begin{align} P=\{x \in \mathbb{R}^n:Ax \leq b\} \end{align}
How can I do this so as to have the polyhedron, or how can I argument that it is impossible, and thus the set is not a polyhedron?
What I have: if the condition on $y$ from the set m were $\lVert y \rVert = 1$ instead of the inequality, then the set would represent the intersection of the unit ball $\{ x: \lVert x \rVert \leq 1 \}$ (using the Cauchy inequality to rewrite the $x^T y \leq 1$ condition) and the non-negative outhunt $R^n_{+}$. To my understanding, this would not be a polyhedron according to the above definition. How does the $\lVert y \rVert \leq 1$ instead of $\lVert y \rVert = 1$ condition change things?