# Want show $\sup_n |T_n(x)|=\infty$ for some $x\in \ell_p(\mathbb N)$

Let $p\in (1,\infty)$. For each $n\geq 1$, let $T_n:\ell_p(\mathbb N)\to \mathbb C$ be the continuous linear functional given by $T_n(x)=\sum_{i=n}^{2n}x_i$, where $x=(x_k)_{k\geq 1}\in \ell_p(\mathbb N)$.

The claims to be proven are:

i) $\sup_n |T_n(x)|<\infty$ for all $x\in \ell_1(\mathbb N)\subset \ell_p(\mathbb N)$,

ii) there exists $x\in \ell_p(\mathbb N)$ such that $\sup_n |T_n(x)|=\infty$.

To answer i), we have from the earlier problem, that $||T_n||=1$. Then given $x\in \ell_1(\mathbb N)$, one has $\sup_n |T_n(x)|\leq ||x||_1<\infty$. I hope that the reason is correct.

Now, to prove ii), I do not know how. My thought says, if only it is correct, since $(\ell_p(\mathbb N))^*$ is the dual space of $\ell_p(\mathbb N)$, we may find some $i$-th coordinate functional in order to reach the conclusion.

Let $p\in (1, \infty)$ and let $q = p/(p-1)$ be its conjugate exponent. Let us define, for every $n$, $$y^{(n)} = (y^{(n)}_k)_{k\geq 1}, \qquad y^{(n)}_k := \begin{cases} 1, &\text{if}\ n\leq k \leq 2n,\\ 0, &\text{otherwise}, \end{cases}$$ so that $T_n(x) = \sum_{k=1}^\infty x_k y^{(n)}_k$. By the Riesz representation theorem, we have that $$\|T_n\|_* = \|y^{(n)}\|_q = (n+1)^{1/q},$$ so that $$\sup_n \|T_n\|_* = +\infty.$$ By the Banach-Steinhaus theorem we can conclude that there exists a dense set $D \subset \ell^p$ such that $$\sup_n |T_n(x)| = +\infty \qquad \forall x\in D.$$
• I have a question, why is $\sup_n \|T_n\|_* = +\infty$, when $\{(n+1)^{1/q}:n\geq 1\}$ has no upper bound, which means that it has no supremum. To make it more general, if one set is non-empty and has no upper bound, would the supremum of it be infinity? Dec 8 '17 at 18:42
• By definition, it is customary to set $\sup A = +\infty$ when a set $A\subset \mathbb{R}$ is not bounded from above. Dec 8 '17 at 18:47
• I suppose that your answer is made for a general case. Some parts of your answer (both theorems) is out of my knowledge, but I really appreciate your help! I tried an example, similar to your first edit, namely the sequence $((\ln k)/k)_k$, and it worked. Dec 10 '17 at 1:29