Let $$||x||_p=\left( \displaystyle \sum_{i=1}^n|x_i|^p\right)^{\frac{1}{p}}$$ and $$||x||_M=\max\{|x_1|,|x_2|,...,|x_n|\},$$ norms in $\mathbb{R}^n$. Show that $$\lim_{p \to \infty}||x||_p=||x||_M, \ \forall x \in \mathbb{R}^n.$$
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What kind of convergence are you referring to with $\lim_p ||x||_p$ ? – Gabriel Romon Mar 18 '14 at 19:25
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the norm $p$ of a element converg to norm of máximus of the element, when $p$(of norm $p$) convergs to infty. – Jarbas Dantas Silva Mar 18 '14 at 19:32
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Not sure I understand the downvotes. Searching for symbols can be a little difficult. – copper.hat Mar 18 '14 at 19:34
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Given $x = \big( x_1, x_2, \ldots, x_n \big)$, denote $\ell = \max\big\{ \vert x_1 \vert , \ldots, \vert x_n \vert \big\}$.
Case 1. If $\ell = 0$, the proof is trivial.
Case 2, If $\ell > 0$, then
$$\big\| x \big\|_p = \left( \displaystyle \sum_{i=1}^n|x_i|^p\right)^{\frac{1}{p}} \leq \big( n \ell^p \big)^{1/p} = n^{1/p} \cdot \ell \xrightarrow{p\to +\infty} \ell \,\, . $$
This implies that
$$\limsup_{p\to+\infty} \big\| x \big\|_p\leq \ell \,\, .$$
On the other side,
$$ \big\| x \big\|_p = \left( \displaystyle \sum_{i=1}^n|x_i|^p\right)^{\frac{1}{p}} \geq \ell \Longrightarrow \liminf_{p\to+\infty} \big\| x \big\|_p\geq\ell $$
Combine these two inequality, we can get the desired limit.
Q.E.D.
Chival
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