Why $p(x-y) \geq p(y)-p(x)$,implies $p(x-y) \geq |p(y)-p(x) |$ for the seminorm $p$

Question

In this lectureFucntional Analisys II at page 2 the author has

Definition 1.1. Let $X$ be a vector space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. A function $p: X \rightarrow \mathbb{R}$ is called a seminorm if

(i) $p(x+y) \leq p(x)+p(y)$ for $x, y \in X$ (sub-additivity),

(ii) $p(\lambda x)=|\lambda| \cdot p(x)$ for $\lambda \in \mathbb{K}$ and $x \in X$ (positive homogeneity).

Lemma 1.2. Let $p$ be a seminorm on a vector space $X$. Then

(a) $p(0)=0$,

(b) $p(x) \geq 0$,

(c) $|p(x)-p(y)| \leq p(x-y)$,

Proof. By homogeneity, $p(0)=p(0 \cdot x)=0 \cdot p(x)=0$, which is (a). By sub-additivity, $p(x) \leq p(x-y)+p(y)$, so $p(x-y) \geq p(x)-p(y)$. Substituting $x$ for $y$ we get $p(x-y)=|-1| \cdot p(y-x) \geq p(y)-p(x)$, so (c) follows. In particular, $p(x) \geq|p(x)-p(0)| \geq 0$, which is (b).

My question is why $p(x-y) \geq p(y)-p(x)$,implies $p(x-y) \geq |p(y)-p(x) |$ ?

Answer 1

By (ii) $P(-x)=p(x)$. So $p(x-y) \geq p(y)-p(x)$ gives $p(x-y)=p(y-x) \geq p(x)-p(y) =-(p(y)-p(x))$. combining these two inequalities we get $p(x-y) \geq |p(y)-p(x)|$.

($t \geq s$ and $t \geq -s$ imply $t \geq |s|$ since $|s|$ is either $s$ or $-s$).