I know the definition of the direction of set. but the question is a bit confusing for me . how to extend one point to the set?

Question

Let S$\subseteq$$R^N$ be a closed convex set and $x^{\star}$ $\in$ S.Suppose d$\ne$0,d$\in$S, and $x^{\star}$+$\lambda$d$\in$S for all $\lambda \ge 0$.Show that d is a direction of S.

This question [2.33] is from the chapter 2 of NONLINEAR PROGRAMMING Theory and Algorithms.

Answer 1

Let $y\in S,$ $\lambda > 0$ and $\alpha\in (0,1).$ Since $$x^{\star} + \frac{\lambda}{1-\alpha}d\in S$$ and $S$ is convex, $$\alpha y + (1-\alpha)\left(x^{\star} + \frac{\lambda}{1-\alpha}d\right)\in S.$$ The left side simplifies to $\alpha y + \lambda d + (1-\alpha)x^{\star}.$ Now take the limit as $\alpha \rightarrow 1$ and use the fact that $S$ is closed.

Answer 2

Suppose d isn't a direction of S, then $\exists x' \in S$, and $x'+\lambda d\notin S$. Hence $\exists p\ne 0$, $p^T(x'+\lambda d)>\alpha$, $p^Tx\le \alpha$. Since $x^{\ast}+\lambda d\in S$ for all $\lambda \ge 0$, then $p^T(x^{\ast}+\lambda d)\le \alpha$, since $p^Tx^{\ast}\le \alpha $, thus $p^Td\le 0$(otherwise when $\lambda$ is large, $p^T(x^{\ast}+\lambda d)> \alpha$). Since $x'\in S$, hence $p^Tx'\le \alpha$, $p^T(x'+\lambda d)\le p^Tx'\le \alpha$, there is a contradiction to our suppose, i.e. d is a direction of S.