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
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$\begingroup$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.
$\endgroup$ 0 $\begingroup$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.
see .
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