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For each $n\geq 1$, let $T_n:\ell_p(\mathbb N)\to \mathbb C$ be the linear functional given by $T_n(x)=\sum_{i=n}^{2n}x_i$, where $x=(x_k)_{k\geq 1}\in \ell_p(\mathbb N)$. Let $p$ be any number in $[1,\infty]$. My problem is to determine $\left \| T_n \right \|$ depending on $n$ and $p$.

My idea is to find the least $c>0$ such that $\left \| T_nx\right \|\leq c \left \| x \right \|$ for all $x\in \ell_p(\mathbb N)$. I am stuck in this part. The hint I got is to write $T_n(x)=\sum_{k=1}^{\infty}x_ky_k^{(n)}$ for a suitable $y^{(n)}=(y_k^{(n)})_{k\geq 1}$. In this case I suppose that $y^{(i)}=(1,1,\dots)$ if $n\leq i\leq 2n$ and $y^{(i)}=(0,0,\dots)$ if otherwise.

Hopeless
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2 Answers2

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Actually it needs much more careful calculation what I did before (revised form): Applying Hölder's inequality we get: $$|T_n(x)|\leq\sum_{i=n}^{2n}|x_i|\leq \bigg(\sum_{i=n}^{2n}1^q\bigg)^{1/q}\bigg(\sum_{i=n}^{2n}|x_i|^p\bigg)^{1/p}$$ $$=\bigg(\sum_{i=n}^{2n}1^q\bigg)^{1/q}\bigg(\sum_{i=n}^{2n}|x_i|^p\bigg)^{1/p}\leq \bigg(n+1\bigg)^{1/q}\bigg(\sum_{i=1}^{\infty}|x_i|^p\bigg)^{1/p}$$ $$\leq\bigg(n+1\bigg)^{1/q}\|x\|_{\ell_p},$$where $\frac{1}{p}+\frac{1}{p}=1$. Then we have $$|T_n(x)|\leq \bigg(n+1\bigg)^{1/q}\|x\|_{\ell_p}\|x\|_{\ell_{p}}\Rightarrow \|T_n\|\leq \bigg(n+1\bigg)^{1/q}.$$ Here $q=1-1/p$ and $\|T_n\|=\displaystyle\sup_{x\in\ell_p} |T_{n}(x|$. For the oher direction, let $x=(x_1,x_2,\cdots,)$ be in $\ell_p$ with $x_i=1$ for $n\leq i\leq 2n$ so that $$\|x\|_{\ell_p}=\bigg(n+1\bigg)^{1/p},$$ which is the sum of 1's from $n$ to $2n$, and $T_{n}(x)=n+1$. Now, since $T_n$ is a bounded operator, it follows from the fact $|T_{n}(x|\leq\|T_n\|\|x\|_{\ell_p}$ that $$n(n+1\leq\|T_n\|\bigg(n+1\bigg)^{1/p}$$ and hence we find $$\bigg(n+1\bigg)^{1-1/p}\leq\|T_n\|$$. Therefore, $$\|T_n\|=\bigg(n+1\bigg)^{1/q},$$ since $1/q=1-1/p$.

daulomb
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  • Thanks for your answer, especially the Hölder's inequality that helped me out. Since the sum gives $2n - (n-1)=n+1$, the factor should have been $(n+1)^{1/q}$. If you have time, could you please enlighten me how to prove the reverse inequality, that is, $|T_n|\geq (n+1)^{1/q}$? – Hopeless Dec 07 '17 at 16:43
  • you should take a specific $x\in\ell_p$ and use the fact that $|T_n(x)|\leq |T_n||x|_{\ell_p}$. – daulomb Dec 07 '17 at 16:47
  • When you say about picking a sequence, I would take the zero ones, $x:=(0,0,0,\dots)$, because then it would give $T_n(x)=0$. Is that all? – Hopeless Dec 07 '17 at 16:50
  • I mean a vector $x$ other zero, which will give you what you want to show. – daulomb Dec 07 '17 at 16:54
  • Are you thinking about the sequence that "normalizes", that is, the sum of each absolute value of terms gives $1$? I'm out of ideas about picking a specific one. – Hopeless Dec 07 '17 at 17:10
  • it is a stupid mistake :) I dont remember why I wrote $2n$ for sum, it would be sum of natural numbers from $n$ to $2n$. By the way I doubt about your factor – daulomb Dec 07 '17 at 17:20
  • @Hopeless: I completed the solution of your problem you can take a look. – daulomb Dec 07 '17 at 22:35
  • It is not true that $\sum_{i=n}^{2n}1=(1/2)n(3n+1)$. Try check for example $n=2,3$. The sum should be $n+1$. Nevertheless, I'll accept the answer. Thanks for the help! – Hopeless Dec 08 '17 at 15:44
  • @Hopeless: what about $5$ and @10$?Really I so tired if your claim is true put that factor but basically the solution will be so just compute that stupid sum and sorry this – daulomb Dec 08 '17 at 15:45
  • I am not sure what you mean. If $n=5$, then $\sum_{i=5}^{10}1=6$, while your formula gives $40$. No worries, I found a way through your strategy. Thanks, anyway :) – Hopeless Dec 08 '17 at 15:49
  • T thought like this: I computed numbers $1,,,2,,\cdots.n-1,,n,\cdots, 2n$ and subtract the sum from$1$ to $n-1$ and then take the difference. But it seems I made a mistake in calculation. – daulomb Dec 08 '17 at 15:55
  • the sum of difference should be $\frac{3n(n+1)}{2}$. Do you agree? – daulomb Dec 08 '17 at 16:01
  • The sum $\sum_{i=n}^{2n}1$ can be obtained through $$2n=\sum_{i=1}^{2n}1=\sum_{i=1}^{n-1}1+\sum_{i=n}^{2n}1=(n-1)+\sum_{i=n}^{2n}1.$$ Soon, I'm going to create a new question, that is related to this, OP, problem. – Hopeless Dec 08 '17 at 16:06
  • I think we together eventually have discovered the correct sum:) now it works for your sample values $n=2,,3$. Be careful the sum from $1$ to $n-1$ is not $n-1$, it is $\frac{n(n-1)}{2}$. – daulomb Dec 08 '17 at 16:08
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    In fact, the sum $\sum_{i=1}^{n-1}1$ gives $n-1$, while the sum $\sum_{i=1}^{n-1}i$ gives $(1/2)n(n-1)$ :) – Hopeless Dec 08 '17 at 16:14
  • @Hopeless: well done, your nickname hopeless and from now on my real name creless:) you were on the right track so you deserve an upvote:) – daulomb Dec 08 '17 at 16:18
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I think you're on the right track.

The lower bound on the norm can be found from an example vector such as $x = (x_k)$ with $x_k = 1$ for $k$ in $S_n = \{n, \dots, 2n\}$ and zero otherwise.

The upper bound can be found by considering the operator $T_n|_{S_n}$ on the subspace of $\ell_p(\mathbb{N})$ of sequences with support on $S_n$ and applying the generalized mean inequality. (The projection onto this subspace has norm one.)

These two bounds coincide, giving the operator norm for $T_n$ on $\ell_p$.

halfflat
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