Prove some properties of an alternative definition of the norm of a vector

by RoRi

April 28, 2016

Consider an alternative definition for the norm of a vector $A = (a_1, \ldots, a_n)$ given by

$\lVert A \rVert = \sum_{k=1}^n |a_k|.$

Prove that this definition satisfies all of the properties of Theorems 12.4 and 12.5 (pages 453-454 of Apostol).
Consider this definition in $\mathbb{R}^2$ and draw the set of all points $(x,y)$ which have norm 1.
If we defined
$\lVert A \rVert = \left| \sum_{k=1}^n a_k \right|$

which of the properties of Theorems 12.4 and 12.5 would hold?

Proof. For Theorem 12.4(a) we have
$\lVert A \rVert > 0$

if $A \neq O$ since all of the terms in the sum are greater than or equal to 0, with at least one non-zero since $A \neq O$ .

For Theorem 12.4(b) if $A = O$ then $\lVert A \rVert = \sum_{k=1}^n |0| = 0$ .

For Theorem 12.4(c) we have

$\lVert cA \rVert = \sum_{k=1}^n |ca_k| = \sum_{k=1}^n |c| |a_k| = |c| \sum_{k=1}^n |a_k| = |c| \lVert A \rVert.$

For Theorem 12.5 we have

$\begin{align*} \lVert A + B \rVert &= \sum_{k=1}^n |a_k + b_k| \\ &\leq \sum_{k=1}^n (|a_k| + |b_k|) \\ &= \sum_{k=1}^n |a_k| + \sum_{k=1}^n |b_k| \\ &= \lVert A \rVert + \lVert B \rVert. \qquad \blacksquare \end{align*}$
We have the following diagram
Property 12.4(a) fails since if we take $A = (-1,1)$ then $\lVert A \rVert = 0$ , but $A \neq O$ .
Property 12.4(b) holds since if $A = O$ then $\lVert A \rVert =0$ .
Property 12.4(c) holds since

$\begin{align*} \lVert cA \rVert &= \left| \sum_{k= 1}^n ca_k \right| \\[9pt] &= \left| c \sum_{k=1}^n a_k \right| \\[9pt] &= |c| \left| \sum_{k=1}^n a_k \right| \\[9pt] &= |c| \lVert A \rVert. \end{align*}$

Property 12.5 holds since

$\begin{align*} \lVert A +B \rVert &= \left| \sum_{k=1}^n (a_k + b_k) \right| \\[9pt] &= \left| \sum_{k=1}^n a_k + \sum_{k=1}^n b_k \right| \\[9pt] &\leq \left| \sum_{k=1}^n a_k \right| + \left| \sum_{k=1}^n b_k \right| \\[9pt] &= \lVert A \rVert + \lVert B \rVert. \end{align*}$

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