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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

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    $\begingroup$ Is $A$ a matrix or a vector ? The last formula is false in the matrix case and have no sense for a vector. $\endgroup$
    – nicomezi
    Oct 25, 2019 at 3:38

1 Answer 1

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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 $\varliminf_{p \to \infty}\Vert x \Vert_p \ge \Vert x \Vert_\infty$.

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