Proving certain properties involving sublinear functional

Question

I'm trying to prove this statement.

For nonnegative sublinear functional $p : X \rightarrow R$ and $\forall c > 0$ , the set $E = \{ x : p(x) \leq c \}$ is convex and absorbing set. Also If $p$ is a seminorm, then E is also balanced.

Here are definitions I use.

$p$ is sublinear functional if $\forall x,y \in X, \forall \alpha > 0$ these conditions hold
$p(x+y) \leq p(x) + p(y)$
$p(\alpha x) = \alpha p(x)$

$p$ is seminorm if $\forall x,y \in X, \forall \alpha \in R$ these conditions hold
$p(x+y) \leq p(x) + p(y)$
$p(\alpha x) = |\alpha| p(x)$
$p(x) \geq 0$ (this condition could be skipped)

$E$ is a convex set, if $\forall t \in [0,1]$ and $\forall x_1, x_2 \in X,$ $(1-t)x_1 + tx_2 \in E$
$E$ is absorbing set , if $\forall x \in X$ $\exists \alpha > 0$ s.t. $\alpha x \in E$
$E$ is balanced set, if $\forall x \in E$ and $\forall \alpha \in R : |\alpha| \leq 1,$ $\alpha x \in E$

I was able to prove that $E$ is a convex set, but failed to prove other conditions. For proving convexity here's what I've done.

$\forall x_1, x_2 \in E,$ $\forall t \in [0,1]$ I need to prove that $(1-t)x_1 + tx_2 \in E$

$p((1-t)x_1 + tx_2) \leq p((1-t)x_1) + p(tx_2) = (1-t)p(x_1) + tp(x_2) \leq (1-t)c + tc = c$

So $p((1-t)x_1 + tx_2) \leq c$, which means that $(1-t)x_1 + tx_2 \in E$

Answer 1

Let $x\in X$. If $p(x)=0$ you have $x\in E$, if it is not zero you have $p\left(\frac{c}{p(x)}\,x\right)=\frac{c}{p(x)}p(x)=c$ and $\frac c{p(x)}x\in E$ so $E$ is absorbing.

If $p$ is a semi-norm and $x\in E$ you have for $|\alpha|≤1$ that $p(\alpha x)=|\alpha|p(x)≤|\alpha| c≤c$ so $\alpha x\in E$ and $E$ is balanced.