I have to show that:

for all vectors $v\in \Bbb R^n$: $\lim_{p\to \infty}||v||_p = \max_{1\le i \le n}|v_i|$

with the $||\cdot ||_p$ norm defined as $$ ||\cdot ||_p: (v_1, \dots ,v_n) \to (\sum^n_{i=i} |v_i|^p)^{1/p} $$

I think I once read something about mixing the root and the same power with the power going to infinity but i can't really remember anything concrete. Any Ideas?

Thanks in advance


2 Answers 2


Hint: For the upper bound, observe that

$$ \left(\sum_{i=1}^n |v_i|^p\right)^{1/p}\leq\left(\sum_{i=1}^n \max|v_i|^p\right)^{1/p}=n^{1/p}\max|v_i|. $$

For the lower bound, observe that

$$ \left(\sum_{i=1}^n |v_i|^p\right)^{1/p}\geq\left( \max|v_i|^p\right)^{1/p}=\max|v_i|. $$

Now, take limits.

  • 1
    $\begingroup$ I've been looking on several "duplicate" type questions hoping for a nice clean simple proof. Thanks! $\endgroup$
    – Serafina
    Aug 15, 2023 at 2:46

As a norm is completely described by its unit ball, let us see the way unit balls of $||.||_p$ converge.

See (classical) pictures below of the unit balls of $||.||_1 (square), ||.||_2 (circle), ||.||_3$ and $||.||_9$ in $\mathbb{R}^2$. These balls are getting more and more "square" as $p$ increases, the limit square being described by equation $\max(|x|,|y|)=1$, providing a geometric intuition about the way the limit is obtained.

enter image description here

see https://en.wikipedia.org/wiki/Lp_space.

  • 2
    $\begingroup$ for a 1000 dimensional unit sphere the distance from every point on the sphere is 1 from the origin, while for the 1000 dimensional unit box, the corner of the box is 1000 units from the origin. In higher dimensions the unit box gets very pointy, in that sense. $\endgroup$ Jun 30, 2022 at 2:50

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