Let $Y\subset \mathbb R^p$ be a convex and
$Y_\infty=\{d\in \mathbb R^p : \exists y s.t. y+\alpha d \in Y \forall \alpha >0\}$
How I can show Y is bounded iff $Y_\infty =\{0\} $ ?
And if $Y=\{(y_1,y_2)\in\mathbb R^2 : y_2>=y_1^2\}$ , what is $Y_\infty$ ??
Here I show that $Y_\infty =\{(0,t):t\geq0\}$.
First note that $Y$ is the epigraph of the function $x\mapsto x^2$, and we can write $$ Y = \{ (x, x^2 + r) : x,r\in\mathbb{R},\,r\geq 0\}. $$ Since $(0,0)\in Y$, it is clear that $(0,t)\in Y$ for all $t\geq0$ and thus $\{(0,t):t\geq0\}\subseteq Y_\infty$.
Now let $(s,t)\in\mathbb{R}^2$ such that $(s,t)\not\in \{(0,t):t\geq0\}$. We will show that $(s,t)\not\in Y_\infty$. There are three cases:
First consider case (1) and suppose that $s=0$ and $t<0$. Let $(x,x^2+r)\in Y$ be arbitrary. Since $t<0$, there exists an $\alpha>0$ large enough so that $x^2+r<-\alpha t$ and thus $(x,x^2+r)+\alpha(0,t)\not\in Y$. Hence $(0,t)\not\in Y_\infty$.
Now consider case (2). Without loss of generality we may suppose that $s=1$ (otherwise we may divide by $s$). As before, let $(x,x^2+r)\in Y$ be arbitrary. We may choose $\alpha>0$ large enough so that $$ r <\alpha(\alpha +2x-t) $$ and consider the point $(x,x^2+r) + \alpha(1,t) = (x+\alpha,x^2+r+\alpha t)$. Note that $$ (x+\alpha)^2= x^2 + 2\alpha x + \alpha^2> x^2+r+\alpha t $$ for our choice of $\alpha$, and thus $(x,x^2+r) + \alpha(1,t)\not\in Y$. It follows that $(1,t)\not\in Y_\infty$.
Case (3) is analogous, except wlog $s=-1$.