Norm of Summation

Theorem
Let $\struct {X, \norm \cdot}$ be a normed vector space.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.

Then:


 * $\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$ for all $n \in \N$.

Proof
The proof proceeds by induction.

For each $n \in \N$, let $\map P n$ be the proposition:


 * $\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$ for all $n \in \N$.

Basis for the Induction
$\map P 1$ is the case:


 * $\ds \norm {\sum_{k \mathop = 1}^1 x_k} \le \sum_{k \mathop = 1}^1 \norm {x_k}$

This is immediate, since:


 * $\ds \norm {\sum_{k \mathop = 1}^1 x_k} = \norm {x_1}$

and:


 * $\ds \sum_{k \mathop = 1}^1 \norm {x_k} = \norm {x_1}$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.

So this is the induction hypothesis:


 * $\ds \norm {\sum_{k \mathop = 1}^n x_k} \le \sum_{k \mathop = 1}^n \norm {x_k}$

from which it is to be shown that:


 * $\ds \norm {\sum_{k \mathop = 1}^{n + 1} x_k} \le \sum_{k \mathop = 1}^{n + 1} \norm {x_k}$

Induction Step
This is the induction step:

We have:

So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.