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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!


marked as duplicate by Nate Eldredge, Chris Janjigian, user61527, Dan Rust, M Turgeon Apr 22 '14 at 2:41

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    $\begingroup$ Duplicate of math.stackexchange.com/q/242779/264 $\endgroup$ – Zev Chonoles Mar 10 '13 at 5:55
  • $\begingroup$ Not true if you read the questions carefully. Your link is for Lp space. My question is for lp space which is two different things. $\endgroup$ – Leo Spencer Mar 10 '13 at 6:05
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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. $\endgroup$ – Zev Chonoles Mar 10 '13 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. $\endgroup$ – Leo Spencer Mar 10 '13 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 '16 at 5:42

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 '15 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 '16 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 '17 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 '17 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 '17 at 13:02

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