Determine $\left \| T_n \right \|$ depending on $n$ and $p$ 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.
 A: 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 $1/q=1-1/p$.
A: 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 $\ell_p$.
