Definition 1:[Reference: Metric spaces, Micheal O Searcoid]
Suppose $V$ is a linear space over $\mathbb R$ or $\mathbb C $. Suppose $||·||$ is a real function defined on $V$ such that, for each $x, y ∈ V$ and each scalar $α$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||αx|| = |α| ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
Definition 2:[Reference: Introduction to topology and modern analysis, George F. Simmons]
Suppose $V$ is a linear space over $\mathbb R$ or $\mathbb C $. Suppose $||·||$ is a real function defined on $V$ such that, for each $x, y ∈ V$, we have
• $||x|| ≥ 0$ with equality if, and only if, $x = 0$;
• $||-x|| = ||x||$; and
• (triangle inequality) $||x + y|| ≤ ||x|| + ||y||$.
Then $||·||$ is called a norm on $V.$
How does the definition 1 and definition 2 are equivalent?
Suppose $||·||$ satisfies the three conditions in the definition 1. It obviously satisfies the three conditions of definition 2. Only change in the definition is the second condition only.
Suppose $||·||$ satisfies the three conditions in the definition 2. Only change in the definition is the second condition only.
each scalar $α $ scalar and $x\in V$ , $||\alpha x||=|\alpha|||x||$. Suppose $\alpha \in \mathbb R$,
Case 1:- $\alpha=0$, then condition (2) of definition (1) satisfies trivially.
I don't know how to proceed further.
They are not. For example $\|x\|=0$ for $x=0$ and $\|x\|=1$ otherwise obeys definition 2, but not definition 1.