Recession cone of a set in $l_2$

Question

Let $\mathbb{C}=\{(x_n)_{n\geq1}\in l_2/|x_n|\leq n, n=1,2,3...\}$.

Prove that this set is convex, closed and not bounded but $recc(A)=\{(0,0,...,0,...)\}$.

I am using the first definition of recession cone found here Recession cone

And $l_2$ is this space: $l_2=\{(x_n)_{n\geq1}\in \mathbb{R^\mathbb{N}}:\sum_{n=1}^\infty|x_n|^2<\infty\}$

Convexity is easy to prove, but I'm stuck in the rest.

Answer 1

The set is closed since the mapping $x\mapsto x_n$ is continuous on $l_2$. The set is unbounded as $ne_n\in C$.

Assume $d\in recc(A)$. Then $0+\lambda d \in C$ for all $\lambda\ge0$, i.e., $\lambda |d_n|\le n$ for all $\lambda\ge0$. This implies $d_n=0$ for all $n$, and $recc(C)=\{0\}$.