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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

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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 $$ \lim_{p\to \infty} \|x\|_{p}=\|x\|_\infty $$

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