Infinity norm proof
my attempt
$$\lim_{p\to\infty}\|A\|_{p} = \lim_{p\to\infty} (|A_1|^p + \cdots + |A_n|^p)^{1/p}$$
$$= \lim_{p\to\infty} \left( \sum_{i=1}^{n} |x_i|^p \right)^{1/p}$$
I know the infinity norm is max row but I'm unsure how to get to that.
Turn above to
$$\|A\|_{\infty} = \max_j \sum_{i=1}^{\infty}|a_{ij}|$$
Not sure which properties to use to solve this I'm very confused. Any help is appreciated
$\endgroup$ 11 Answer
$\begingroup$It seems like you are confusing about operator norm and vector norm. To clarify, the p-norm of a matrix/operator is defined to be $\Vert A \Vert_{p,\text{op}}= \sup_{\Vert x \Vert_{p,\text{vec}}=1} \Vert A x \Vert_{p,\text{vec}} $.
And the limiting result holds for vector norm $\Vert \cdot \Vert_p = \Vert \cdot \Vert_{p,\text{vec}} $.
Proof:
$\Vert x \Vert_p = (\sum_i \vert x_i \vert^p)^{\frac{1}{p}} \le (\sum_i \max_i \vert x_i\vert^p )^{\frac{1}{p}} = n^{\frac{1}{p}} \max_i \vert x_i\vert \to \max_i \vert x_i \vert = \Vert x \Vert_\infty$ since $n$ is a fixed number, so $\varlimsup_{p \to \infty} \Vert x \Vert_p \le \Vert x \Vert_\infty$.
On the other hand, $\Vert x \Vert_p = (\sum_i \vert x_i \vert^p)^{\frac{1}{p}} \ge (\max_i \vert x_i \vert^p)^{\frac{1}{p}} = \max_i \vert x_i \vert = \Vert x \Vert_\infty$ So $\varliminf_{p \to \infty}\Vert x \Vert_p \ge \Vert x \Vert_\infty$.
$\endgroup$
