Unbounded symmetric convex sets

Question

If $X$ is a normed space, can one find a open, unbounded, symmetric, convex set $C$ that is bounded in every direction? That is, for every $x \neq 0$ there exists $\lambda_x>0$ such that for any $|\lambda|>\lambda_x$, $\lambda x\notin C$?

Answer 1

Let $X = \ell^1$, i.e. the set of infinite sequences $(x_n)_{n \in \mathbb N}$ with the norm $\|x\| = \sum_n |x_n|$. Take $C$ to be all sequences for which $\sup_n |x_n| < 1$.

• $C$ is open because if $x = (x_n) \in C$, then $\sup_n |x_n| = s < 1$. For any $y = (y_n) \in X$ such that $\|y\| < \frac {1-s} 2$, we have: $|y_n| \le \sum_n |y_n| < \frac {1-s} 2$, hence $|x_n + y_n| \le |x_n| + |y_n| < s + \frac {1-s} 2 = \frac {1+s} 2$ and therefore $\sup_n |x_n + y_n| \le \frac {1+s} 2 < 1$, i.e. $B(x,\frac {1-s} 2) \subset C$.

• $C$ is unbounded: it contains all sequences first $n$ terms of which equal to $\frac 1 2$ and the rest are zeros, the norms of these sequences are $\frac n 2$, i.e. are unbounded.

• $C$ is clearly symmetric.

• $C$ is convex: if $x = (x_n) \in C$, $y = (y_n) \in C$ and $\sup_n |x_n| = u < 1$, $\sup_n |y_n| = v < 1$, then for any $\lambda \in (0,1)$ and $n \in \mathbb N$ we have: $|\lambda x_n + (1-\lambda) y_n| \le \lambda u + (1-\lambda) v \le \max(u, v)$, this implies $\sup_n |\lambda x_n + (1-\lambda) y_n| \le \max(u, v) < 1$ and $\lambda x + (1-\lambda) y \in C$.

• $C$ is bounded in every direction. Let $x = (x_n)$ and $x \ne 0$. Then $x_k \ne 0$ for some $k \in \mathbb N$ and we can take $\lambda_x = \frac 1 {|x_k|}$ because for any $\lambda$ such that $|\lambda| > \lambda_x$ we will have for the $k$-th term of $\lambda x$: $|\lambda x_k| = |\lambda| |x_k| > \frac 1 {|x_k|} |x_k| = 1$, therefore $\sup_n |\lambda x_n| > 1$ and $\lambda x \notin C$.