Norm on Vector Space is Quasinorm

Theorem

Let $\struct {R, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_R$.

Let $V$ be a vector space over $R$.

Let $\norm \cdot$ be a norm on $V$.

Then $\norm \cdot$ is a quasinorm on $V$.

Proof

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Proof of $(\text Q 1)$

This is precisely $(\text N 2)$ in the definition of a norm.

$\Box$

Proof of $(\text Q 2)$

Since $\norm \cdot$ is a norm on $V$, we have:

$\norm {x + y} \le \norm x + \norm y$

for all $x, y \in V$.

In particular, there exists an $M \ge 1$ with:

$\norm {x + y} \le M \paren {\norm x + \norm y}$

So $(\text Q 2)$ holds.

$\blacksquare$

Sources

• 2014: Loukas Grafakos: Classical Fourier Analysis (3rd ed.) ... (previous) ... (next): Appendix G: Basic Functional Analysis