Consider the following sets $$ \mathcal{X}_y\equiv \{x\in \mathbb{R}^d: \quad f(y)*x\leq 0_k\}\quad \text{for each }y\in \mathbb{R}^j\\ \mathcal{Y}\equiv\{y\in \mathbb{R}^j: \quad \exists \quad x\in \mathbb{R}^{d} \quad \text{such that} \quad f(y)*x\leq 0_k\} $$ where $f:\mathbb{R}^j\rightarrow \mathbb{R}^{k\times d}$ is a non-linear function, $k\geq d$, $\mathbb{R}^{k\times d}$ denotes the set of matrices of reals with size $k\times d$, $0_k$ is the $k\times 1$ vector of zeros.
My understanding is that $\mathcal{X}_y$ is closed set because it is a convex polyhedron. Further, $\mathcal{Y}$ is compact when $\mathcal{X}_y$ is bounded for each $y\in \mathbb{R}^j$.
Questions: are there sufficient and necessary conditions on $f$ such that $\mathcal{Y}$ is compact (i.e., if the above is correct, such that $\mathcal{X}_y$ is bounded for each $y\in \mathbb{R}^j$)?