For any $x ∈ \mathbb{R}^n$ and $p ≥ 1$, define $$||x||_p = \left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p},||x||_∞ = \underset{1≤i≤n}{\max}|x_i| $$ Show that $$ \underset{p→∞}{\lim} ||x||_p = ||x||_∞$$
I understand this in theory, that if $x_i < x_j$ then as they are both raised to the power of ∞ then $x_i$ will become arbitrarily small relative to $x_j$, and so their sum to the power of $\frac1∞$ is $\approx x_j$. But I don't know how to go about proving this formally.
HINT: Note that $$\max_i \left\vert x_i \right\vert^p \leq \sum_{i=1}^n \left\vert x_i \right\vert^p \leq n \max_i \left\vert x_i \right\vert^p$$
Case 1: Assume without loss of generality, that $x_1$ is your maximum and assume that $|x_1| > |x_i|\ \forall i=2,..,n$. Then we get $$\left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p} = \sqrt[p]{|x_1|^p\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_n|^p}{|x_1|^p}) } \to \sqrt[p]{|x_1|^p} =||x||_∞ \ \mathrm{for }\ p\to\infty , $$ because $$\frac{|x_i|^p}{|x_1|^p} \to 0 \ \mathrm{for }\ p\to\infty\forall i=2,..,n,$$this follows from the convergence of the geometric series.
Case 2: Assume w.o. loss of generality that $x_1$ is your maximum and that $|x_1| \geq |x_i|\ \forall i=2,..,n$. Equality be the case $q$ times. Then: $$\left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p} = \sqrt[p]{|x_1|^p\cdot q\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_{n-q}|^p}{|x_1|^p}) } = q^{\frac{1}{p}}\sqrt[p]{|x_1|^p\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_{n-q}|^p}{|x_1|^p}) } \to \sqrt[p]{|x_1|^p} =||x||_∞ \ \mathrm{for }\ p\to\infty, $$because $q^{\frac{1}{p}}\to1 \mathrm{for }\ p\to\infty.$
This completes the proof.
