Let $S$ be a closed proper noncompact convex set of $\mathbb{R}^n$ and $\epsilon>0$. Let us note $S_{\epsilon} := S+ B(0,\epsilon)$ the thickening of $S$. Do we have that $S_{\infty}=(S_{\epsilon})_{\infty}$ ? (Here $\infty$ is the recession cone.)
From the definition, we directly get $$S_{\infty}\subset(S_{\epsilon})_{\infty}$$ but the other inclusion is not obvious. Nonetheless, intuitively, add a uniform thickening should not add extra asymptotic directions. Thanks for any help.
Proof by contradiction. Assume there exists $d \in (S_{\epsilon})_{\infty}$ but $d \notin S_{\infty}$. So there is a $\lambda > 0$ and $s \in S$ such that $ s + \lambda d \notin S $.Due to convexity of $S$ we even can say more: $$ s + t d \notin S \quad \quad \forall t \in [ \lambda, + \infty )$$ Now define $f: [0 , +\infty] \to \Bbb R$ with $f(t) = d(s+td, S)$. Observe that $f$ is a convex function achieving its minimum at $t=0$ that implies $$\lim_{ t \to +\infty} f(t) = + \infty. \quad \text{"not hard to prove "}$$
Then the contradiction is obvious.