Subadditivity of Invariant Metric on Vector Space
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Theorem
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $d$ be an invariant metric on $X$.
Then:
- $\map d {n x, {\mathbf 0}_X} \le n \map d {x, {\mathbf 0}_X}$
for each $n \in \N$ and $x \in X$.
Proof
We have:
\(\ds \map d {n x, {\mathbf 0}_X}\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \map d {k x, \paren {k - 1} x}\) | Metric Space Axiom $(\text M 2)$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{k \mathop = 1}^n \map d {k x - \paren {k - 1} x, \paren {k - 1} x - \paren {k - 1} x}\) | Definition of Invariant Metric on Vector Space | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \map d {x, {\mathbf 0}_X}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \map d {x, {\mathbf 0}_X}\) |
$\blacksquare$
Sources
- 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.28$: Theorem