Triangle Inequality for Summation over Finite Set
Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f : S \to \mathbb A$ be a mapping.
Let $\size {\, \cdot\,}$ denote the standard absolute value.
Let $\size f$ be the absoute value of $f$.
Then we have the inequality of summations on finite sets:
- $\ds \size {\sum_{s \mathop \in S} \map f s} \le \sum_{s \mathop \in S} \size {\map f s}$
Outline of Proof
Using the definition of summation, we reduce this to Triangle Inequality for Indexed Summations.
Proof
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{<n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following inequality of indexed summations:
- $\ds \size {\sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i} } \le \sum_{i \mathop = 0}^{n - 1} \map {\paren {\size f \circ \sigma} } i$
By Absolute Value of Mapping Composed with Mapping:
- $\size f \circ \sigma = \size {f \circ \sigma}$
The above equality now follows from Triangle Inequality for Indexed Summations.
$\blacksquare$