Prove 1-Norm is a Norm
I am just curious how you would simply prove that a 1-norm is a norm. Step-by-step would be very helpful. Proofs are not my strong point. Thank you!
$\endgroup$ 11 Answer
$\begingroup$Let $V$ be a vector space with basis $(v_k)_{k=0}^{n-1}$. Let $x = \sum_{k=0}^{n-1} a_k v_k$. Define the $1$-norm to be $$ \lVert x \rVert_1 = \sum_{k=0}^{n-1} |a_k| $$
Identity of indiscernible: If $x = 0$, then $a_k = 0$ for all $k$, so $\lVert x \rVert_1 = 0$. Conversely, if $\lVert x \rVert_1 = 0$, by the nonnegativity of absolute value, $|a_k| = 0$ for all $k$. Hence $a_k = 0$ and $x = 0$.
Triangle inequality: Let $y = \sum_{k=0}^{n-1} b_k v_k$. Then $$ \lVert x + y \rVert_1 = \lVert \sum_{k=0}^{n-1} (a_k + b_k) v_k \rVert_1 = \sum_{k=0}^{n-1} |a_k + b_k| \leq \sum_{k=0}^{n-1} |a_k| + \sum_{k=0}^{n-1} |b_k| = \lVert x \rVert_1 + \lVert y \rVert_1 $$
Homogeneity: Let $\lambda$ be a scalar. $$ \lVert \lambda x \rVert_1 = \sum_{k=0}^{n-1} |\lambda a_k| = |\lambda| \sum_{k=0}^{n-1} |a_k| = |\lambda| \lVert x\rVert_1 $$
$\endgroup$ 1
