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If we are in a sequence space, then the $ l^{p} $-norm of the sequence $ \mathbf{x} = (x_{i})_{i \in \mathbb{N}} $ is $ \displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p} $.

The $ l^{\infty} $-norm of $ \mathbf{x} $ is $ \displaystyle \sup_{i \in \mathbb{N}} |x_{i}| $.

Prove that the limit of the $ l^{p} $-norms is the $ l^{\infty} $-norm.

I saw an answer for $ L^{p} $-spaces, but I need one for $ l^{p} $-spaces. Besides, I didn’t really understand the $ L^{p} $-answer either.

Thanks for your help!

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    $\begingroup$ Duplicate of math.stackexchange.com/q/242779/264 $\endgroup$ Mar 10, 2013 at 5:55
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    $\begingroup$ $\ell^p$ spaces are $L^p$ spaces; as Wikipedia explains, These are special cases of L^p spaces for the counting measure on the set of natural numbers. $\endgroup$ Mar 10, 2013 at 6:07
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    $\begingroup$ I realize that but as I said in the op, I did not completely understand the solution I found which was the one you linked. Also, I feel that since I desire a solution simply for lp spaces, that there is a different solution without considering Lp spaces. And yes you are right I should not have said two different things as that is not completely accurate. $\endgroup$ Mar 10, 2013 at 6:15
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    $\begingroup$ @ZevChonoles The question you linked to asks for finite measure space, yet this question asks for a measure that is not finite. $\endgroup$
    – ZQ Wan
    Jan 25, 2016 at 5:42
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    $\begingroup$ @ZevChonoles I am not sure if this should be marked as a duplicate - the other question specifically asks about a finite measure space, and the accepted answer makes use of this fact. As such, that answer does not answer this question asked by the OP. Nevertheless, it is worth mentioning that the answer to the ‘duplicate’ question can be modified for $\sigma$-finite measure spaces, so can be used as a basis for an answer to the OP’s question. $\endgroup$
    – John Don
    Mar 17, 2019 at 9:10

1 Answer 1

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Let me state the result properly:

Let $x=(x_n)_{n \in \mathbb{N}} \in \ell^q$ for some $q \geq 1$. Then $$\|x\|_{\infty} = \lim_{p \to \infty} \|x\|_p. \tag{1}$$

Note that $(1)$ fails, in general, not hold if $x=(x_n)_{n \in \mathbb{N}} \notin \ell^q$ for all $q \geq 1$ (consider for instance $x_n := 1$ for all $n \in \mathbb{N}$.)

Proof of the result: Since $$|x_k| \leq \left(\sum_{j=1}^{\infty} |x_j|^p \right)^{\frac{1}{p}}=\|x\|_p$$ for all $k \in \mathbb{N}$, $p \geq 1$, we have $\|x\|_{\infty} \leq \|x\|_p$. Thus, in particular $$\|x\|_{\infty} \leq \liminf_{p \to \infty} \|x\|_p. \tag{1}$$

On the other hand, we know that $$\|x\|_p = \left( \sum_{j=1}^{\infty} |x_j|^{p-q} \cdot |x_j|^q \right)^{\frac{1}{p}} \leq \|x\|_{\infty}^{\frac{p-q}{p}} \cdot \left( \sum_{j=1}^{\infty} |x_j|^q \right)^{\frac{1}{p}} = \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}$$ for all $q<p$ where we used $|x_j| \leq \|x\|_{\infty}$ for all $j \in \mathbb{N}$. Therefore, we arrive at

$$ \limsup_{p \to \infty} \|x\|_p \leq \limsup_{p \to \infty} \left( \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}\right) = \|x\|_{\infty} \cdot 1. \tag{2}$$

Hence, $$\limsup_{p \to \infty} \|x\|_p \leq \|x\|_{\infty} \leq \liminf_{p \to \infty} \|x\|_p.$$ This shows that $\lim_{p \to \infty} \|x\|_p$ exists and equals $\|x\|_{\infty}$.

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    $\begingroup$ @Gustavo Since this question is not directly related to my answer and of independent interest, I recommend opening a new question. $\endgroup$
    – saz
    Sep 10, 2015 at 18:09
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    $\begingroup$ I have a simple proof for $\|x\|_{\infty}\geq\lim_{p\rightarrow\infty}\|x\|_{p}$ : By $x\in l^{p}$, $\sum_{j=1}^{\infty}|x_{j}|^{p}<\infty$. And $\left(\sum_{j=1}^{\infty}|x_{j}|^{p}\right)/\|x\|_{\infty}^{p}=\sum_{j=1}^{\infty}\left(\frac{|x|_{j}}{\|x\|_{\infty}}\right)^{p}$ is decrease as $p\rightarrow\infty$. So $\exists C$ s.t. $\sum_{j=1}^{\infty}\left(\frac{|x|_{j}}{\|x\|_{\infty}}\right)^{p}<C$ for large $p$. Then $\left(\sum_{j=1}^{\infty}\left(\frac{|x|_{j}}{\|x\|_{\infty}}\right)^{p}\right)^{1/p}<C^{1/p}\rightarrow1$ as $p\rightarrow\infty$. $\endgroup$
    – kayak
    Dec 30, 2016 at 5:14
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    $\begingroup$ @R.T. I didn't claim that this inequality holds; I just used that $$|x_j|^{p-q} \leq \|x\|_{\infty}^{p-q}$$ implies that $$\sum_j |x_j|^{p-q} |x_j|^q \leq \|x\|_{\infty}^{p-q} \sum_j |x_j|^q.$$ $\endgroup$
    – saz
    May 16, 2017 at 5:02
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    $\begingroup$ @R.T. Why do you think so? Surely we agree that $$|x_j|^{p-q} |x_j|^q \leq \|x\|_{\infty}^{p-q} |x_j|^q$$ for each $j$. Summing both sides over $j \geq 1$ yields $$\sum_j |x_j|^{p-q} |x_j|^q \leq \|x\|_{\infty}^{p-q} \sum_j |x_j|^q.$$ $\endgroup$
    – saz
    May 16, 2017 at 17:38
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    $\begingroup$ @catch22 The statement of the OP is somewhat inaccurate. Properly stated the result reads as follows: If $x=(x_n)_{n \in \mathbb{N}} \in \ell^{\infty} \cap \ell^{p_0}$ for some $p_0 \geq 1$, then $$\|x\|_{\infty} = \lim_{p \to \infty} \|x\|_p. \tag{1}$$ Note that $x \in \ell^{p_0}$ implies $x \in \ell^q$ for all $q \in [p_0,\infty)$; in the proof we can WLOG assume that $q=p_0$. Let me finally remark that if $x \notin \ell^p$ for all $p>0$, then $(1)$ does, in general, not hold true (consider for instance $x_n := 1$ for all $n$). $\endgroup$
    – saz
    Oct 29, 2017 at 13:02

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