(From Lax's functional analysis)
Let p denote a positive homogeneous subadditive function defined on a linear space $X$ over the reals.
(1)The set of points $x$ satisfying $p(x) < 1$ is a convex set S of X, and $0$ is an interior point of it (i.e. for any y $\in$ X, there is an $\theta$, $0 + t y\in S $ for all real $t$, $|t|<\theta$);
(2)The set of points $x$ satisfying $p(x) \le 1$ is a convex set of X.
PS: It's easy to prove the convexity. However, I have some problems to prove the second statement in (1), and I am wondering why (2) cannot get the conclusion that $0$ is an interior point.
Best Answer
Suppose $p(y) >0$ and $p(-y) >0$. Then $p(0+ty) \leq p(0)+p(ty)=0+|t| p(\pm y) <1$ if $|t| <\frac 1 {|p(y)|}$ and $|t| <\frac 1 {|p(-y)|}$. So $0$ is an interior point. I will leave the cases $p(y)=0$ and $p(-y)=0$ to you. ($0$ is also an interior point of $\{x: p(x) \leq 1\}$)