# p-Norm with p $\to$ infinity

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?

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.

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.