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!

Haskell Curry
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Leo Spencer
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    Duplicate of http://math.stackexchange.com/q/242779/264 – Zev Chonoles Mar 10 '13 at 05:55
  • 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. – Leo Spencer Mar 10 '13 at 06:05
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    $\ell^p$ spaces **are** $L^p$ spaces; [as Wikipedia explains](http://en.wikipedia.org/wiki/Sequence_space), `These are special cases of L^p spaces for the counting measure on the set of natural numbers.` – Zev Chonoles Mar 10 '13 at 06:07
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    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. – Leo Spencer Mar 10 '13 at 06:15
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    @ZevChonoles The question you linked to asks for finite measure space, yet this question asks for a measure that is not finite. – ZQ Wan Jan 25 '16 at 05:42
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    @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. – John Don Mar 17 '19 at 09:10

1 Answers1


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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  • Sorry if this is a dumb question. There is a sense which this limit $\displaystyle \lim_{p \to \infty} \lVert x \rVert_p = \lVert x \rVert_{\infty}$ is uniform in $x$? – Gustavo Sep 10 '15 at 03:19
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    @Gustavo Since this question is not directly related to my answer and of independent interest, I recommend opening a new question. – saz Sep 10 '15 at 18:09
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    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} – kayak Dec 30 '16 at 05:14
  • So we conclude $\|x\|_{\infty}\geq\lim_{p\rightarrow\infty}\|x\|_{p}$. – kayak Dec 30 '16 at 05:14
  • @saz I agree that $|x_j|\le ||x||_\infty$, but how do you conclude $(\sum_{j=1}^{\infty} |x_j|^{p-q})^{1/p}\le ||x||_\infty^{(p-q)/p}$? In that case, you contradicted with your previous conclusion that $||x||_\infty \le ||x||_p$. –  May 16 '17 at 03:07
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    @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 $$\sum_j |x_j|^{p-q} |x_j|^q \leq \|x\|_{\infty}^{p-q} \sum_j |x_j|^q.$$ – saz May 16 '17 at 05:02
  • @saz Thank you, but for your second inequality in the comment above to hold, shouldn't you have $\sum_j |x_j|^{p-q} \le \|x\|^{p-q}_{\infty}$? But how can you get this from the first inequality? –  May 16 '17 at 15:43
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    @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 $$\sum_j |x_j|^{p-q} |x_j|^q \leq \|x\|_{\infty}^{p-q} \sum_j |x_j|^q.$$ – saz May 16 '17 at 17:38
  • Thank you! That makes sense! @saz –  May 16 '17 at 18:47
  • What if $\sum_{j=1}^{\infty} |x_j|^q$ is infinite? – catch22 Oct 29 '17 at 12:45
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    @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 $n$). – saz Oct 29 '17 at 13:02
  • @saz: Thanks. Maybe $x \in \ell^{\infty}$ is enough though? Since $\big( \sum |x_i|^q \big)^\frac1p \leq (n \cdot ||x||_{\infty}^q)^\frac1p \rightarrow 1$ as $p \rightarrow \infty$ – catch22 Oct 29 '17 at 13:10
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    @catch22 No, $x \in \ell^{\infty}$ is not enough... just consider $x_n := 1$ for all $n$ – saz Oct 29 '17 at 13:26
  • sorry if this is a stupid question, but why do you need to use lim inf and lim sup? wouldn't each step still be true if you just took $lim_{p \to \infty}$? – user428487 Mar 26 '18 at 11:54
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    @user428487 A priori I don't know whether $\lim_{p \to \infty}$ exists. – saz Mar 26 '18 at 12:00
  • @saz Hi. Is there a reason why you didn't factor $\sup x_i$ out of $\sum_i (|x_i|^p)^{1/p}$ and take the limit as $p\rightarrow \infty?$ – Teodorism Sep 19 '19 at 22:07
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    @Teodorism What do you mean by factor? If we estimate $$\sum_{i=1}^{\infty} |x_i|^p $$ by $$\sup_i |x_i| \sum_{i=1}^{\infty}$$ we don't gain anything since the latter is infinite (because we are summing over infinitely many $i$). – saz Sep 20 '19 at 18:06
  • @saz This is what I mean: https://math.stackexchange.com/q/3362824/371990 – Teodorism Sep 20 '19 at 18:10
  • @saz Hi again. When you’re factoring out the $\|x\|_\infty^{1-q/p}$ how do you ensure norm infinity exists? Or how do you know it doesn’t affect the convergence of the p-sum? – Teodorism Sep 24 '19 at 08:17
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    @Teodorism As I stated at the beginning of my answer, we need to assume that $x \in \ell^q$ for some $q \geq 1$. From $\sum_{i \geq 1} |x_i|^q < \infty$it follows, in particular, $\lim_{i \to \infty} x_i=0$, and so $\|x\|_{\infty} = \sup_i |x_i|<\infty$. – saz Sep 24 '19 at 08:22
  • @saz Even if the sup exists we don’t know it’s attained though? So the factoring it out may change the convergence of the p sum? – Teodorism Sep 24 '19 at 08:26
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    @Teodorism What do you mean by "change the convergence of the $p$ sum"? It doesn't play a role whether the supremum is attained or not; i'm just using the trivial estimate $|x_j| \leq \|x\|_{\infty}$. – saz Sep 24 '19 at 08:30