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Is a knowledge fact that in $\mathbb{R}^n$ we have $$ \|\cdot\|_{\infty}=\lim_{p\to \infty}\|\cdot\|_{p} $$ where we define $$ \|x\|_{p}=\left(\sum_{i=1}^{n}x_i^p\right)^{\frac{1}{p}} $$ and $$ \|x\|_{\infty}=\sup_{1\leq i\leq n}|x_i|. $$

Now consider $l_p=\{\underline{x}=(x_1,\ldots,x_n,\ldots), x_i\in \mathbb{R}; ~~\sum_{n=1}^{n}x_i^p <\infty \}$ and $l_{\infty}=\{ \underline{x}=(x_1,\ldots,x_n,\ldots), x_i\in \mathbb{R};~\sup_{i}|x_i|<\infty \}$

Consider in $l_p$ the norm $\|x\|_{p}=\left(\sum_{i=1}^{\infty}x_i^p\right)^{\frac{1}{p}}$ and in $l_{\infty}$ the norm $\|x\|_{\infty}=\sup_{i\in \mathbb{N}}|x_i|.$

My question: It is true that if $\underline{x}\in l_{\infty}$ then $$ \|\underline{x}\|_{\infty}=\lim_{p\to \infty}\|\underline{x}\|_{p}~\text{?} $$

O Empalador de Cabras
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1 Answers1

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Yes, this holds if $(x_n)\in l^1$. The proof is a bit long, since it involves proving two inequalities. We first prove $\|(x_n)\|_{\infty}\le\lim \limits_{p\to \infty} \|(x_n)\|_p$. For arbitrary $p$ and all $n\in \mathbb{N}$ it's obviously true that $|x_n|^p\le \sum \limits_{n=1}^{\infty}|x_n|^p$, so we obtain $\|(x_n)\|^p_{\infty}\le \sum \limits_{n=1}^{\infty}|x_n|^p$ and this is equivalent to $\|(x_n)\|_{\infty}\le \|(x_n)\|_{p}$. Our claimed inequality follows immediately.

Now, in order to prove $\lim \limits_{p\to \infty} \|(x_n)\|_p \le \|(x_n)\|_{\infty}$, we can assume $p\in \mathbb{N}$, since if $1\le p<q<\infty$, then $\|(x_n)\|_q\le \|(x_n)\|_p$. We first prove the following fact: $$\lim \limits_{N\to \infty} \|(x_n)\|_N=\lim\limits_{N\to\infty}\left(\sum \limits_{n=1}^N|x_n|^N\right)^{\frac{1}{N}}$$ which is of course equivalent to: $$\lim \limits_{N\to \infty} \|(x_n)\|_N-\left(\sum \limits_{n=1}^N|x_n|^N\right)^{\frac{1}{N}}=0$$ To prove this, let $y_N$ be the sequence that agrees with $(x_n)$ for the first $N$ terms and is zero for all other terms. Then we have $$\|y_N\|_N=\left(\sum \limits_{n=1}^N|x_n|^N\right)^{\frac{1}{N}}$$ allowing us to use the reverse triangle inequality:$$\left| \|(x_n)\|_N-\left(\sum \limits_{n=1}^N|x_n|^N\right)^{\frac{1}{N}}\right|\le\|(x_n)-y_N\|_N=\left(\sum \limits_{n=N+1}^{\infty}|x_n|^N\right)^{\frac{1}{N}}\le \sum \limits_{n=N+1}^{\infty}|x_n|<\epsilon$$

for large enough $N$, since $(x_n)\in l^1$.

We now use this fact to prove the claimed inequality. For all $N\in \mathbb{N}$ it is true that: $$\left(\sum\limits_{n=1}^N|x_n|^N\right)^{\frac{1}{N}}\le\left(\sum\limits_{n=1}^N\|(x_n)\|^N_{\infty}\right)^{\frac{1}{N}}=N^{\frac{1}{N}}\|(x_n)\|_{\infty}$$ and from this we finally obtain: $$\lim \limits_{N\to\infty} \|(x_n)\|_N \le \lim \limits_{N\to\infty}N^{\frac{1}{N}}\|(x_n)\|_{\infty}=\|(x_n)\|_{\infty}$$

Vincent Boelens
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