# Proving $\lim_{p\to\infty}$ $\left\lVert x \right\rVert_p = \left\lVert x \right\rVert_\infty$

How can one prove, that

$$\lim_{p\to\infty}$$ $$\left\lVert x \right\rVert_P = \left\lVert x \right\rVert_\infty$$ applies to all $$x \in \mathbb{R^n}$$ ?

I know that two norms$$\left\lVert \cdot \right\rVert_a$$ and $$\left\lVert \cdot \right\rVert_b$$ in $$\mathbb{R^n}$$ are equivalent, if there are constants $$c_1,c_2 > 0$$ so that for all $$x \in \mathbb{R^n}$$ there is an inequation chain

$$c_1 \left\lVert x \right\rVert_a \leq \left\lVert x \right\rVert_b \leq c_2 \left\lVert x \right\rVert_a$$

I think I have to use the inequation above somehow to prove the former, yet I don't know how

We have \begin{align} \|x\|_p =& \left[|x_1|^p+\ldots+|x_i|^p+\ldots+|x_n|^p\right]^{\frac{1}{p}} \\ =& \left[\displaystyle\max_{1\leq k\leq n}{|x_k|^p} \left[ \left( \frac{|x_1|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}} \right)^p +\ldots+ \left( \frac{|x_i|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}} \right)^p +\ldots+ \left( \frac{|x_n|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}} \right)^p \right] \right]^{\frac{1}{p}} \\ =& \displaystyle\max_{1\leq k\leq n}{|x_k|} \left[ \left( \frac{|x_1|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}} \right)^p +\ldots+ \left( \frac{|x_i|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}} \right)^p +\ldots+ \left( \frac{|x_n|}{\displaystyle\max_{1\leq k\leq n}{|x_k|}} \right)^p \right]^{\frac{1}{p}} \end{align} Using a shorter notation $$\|x\|_\infty=\displaystyle\max_{1\leq k\leq n}{|x_k|}$$, $$\|x\|_p= \|x\|_\infty \left[ \left( \frac{|x_1|}{\|x\|_\infty} \right)^p +\ldots+ \left( \frac{|x_i|}{\|x\|_\infty} \right)^p +\ldots+ \left( \frac{|x_n|}{\|x\|_\infty} \right)^p \right]^{\frac{1}{p}}$$ We have two implications.
First. $$0\leq \left(\frac{|x_1|}{\|x\|_\infty}\right)^p\leq 1, \ldots, 0\leq \left(\frac{|x_i|}{\|x\|_\infty}\right)^p\leq 1, \ldots 0\leq \left(\frac{|x_n|}{\|x\|_\infty}\right)^p\leq 1,$$ implies $$\|x\|_p\leq \|x\|_\infty\sqrt[p]{n} \quad (\ast)$$ Second. $$1\leq \left[ \left( \frac{|x_1|}{\|x\|_\infty} \right)^p +\ldots+ \left( \frac{|x_i|}{\|x\|_\infty} \right)^p +\ldots+ \left( \frac{|x_n|}{\|x\|_\infty} \right)^p \right]^{\frac{1}{p}}$$ implies $$\|x\|_\infty\leq \|x\|_{p}\quad (\ast\ast)$$ Putting the Inequalities $$(\ast)$$ and $$(\ast\ast)$$ together we have $$\|x\|_\infty\leq \|x\|_{p}\leq \|x\|_\infty\cdot \sqrt[p]{n}$$ Once $$\lim_{p\to\infty}\sqrt[p]{n}=1$$ for all $$n\in\mathbb{N}-\{0\}$$ we have $$\lim_{p\to \infty} \|x\|_{p}=\|x\|_\infty$$