Determining when $A(e_n) = \alpha_n\sum_{i=n}^{2n}e_i$ is a bounded linear function on $l^1(\mathbb{N})$

Question

Consider the space $S = l^1(\mathbb{N})$ over complex coefficients and define $A$ by $A(e_n) = \alpha_n\sum_{i=1}^{2n}e_n$ for the standard basis $\left(e_n\right)_{n\in\mathbb{N}}$ of $S$. I am trying to determine when $A$ is a bounded linear function, but currently I am stuck at showing that $A$'s codomain really is $l^1(\mathbb{N})$, which in turn seems to be equivalent for $A$ to be continuous. I mean the following: $u \equiv \sum_{i=1}^\infty u_ie_i, u_i\in\mathbb{C}$ such that $\sum_{i=1}^\infty |u_i| < \infty$. Now,

$$A\left(u\right) = A\left(\sum_{i=1}^\infty u_ie_i\right)$$

and I would like to be able to conclude that

$$A\left(u\right) = \sum_{i=1}^\infty A(u_ie_i) = \sum_{i=1}^\infty u_i\alpha_i\sum_{j=i}^{2i}e_i$$

But I don't really see an immediate way to conclude this. Being able to pull out the limit is equivalent to

$$||\lim_{N\to\infty}\sum_{i=1}^NA(u_ie_i) - A(u)|| = \lim_{N\to\infty}||\sum_{i=1}^NA(u_ie_i) - A(u)|| = 0$$

but the problem is that we don't a priori know that $A(u)$ is, and hence concluding the convergence feels sketchy.

How should I proceed with this proof?

Answer 1

Let $\mathbf{e}_k=\mathbb{1}_{k}$, that is, $\mathbf{e}_k(m)=1$ iff $m=k$. Any $\mathbf{x}\in\ell_1(\mathbb{N})$ can be written as $$\mathbf{x}=\sum_{n\in\mathbb{N}}\mathbf{x}(n)\mathbf{e}_n$$

Notice that $\|A\mathbf{e}_n\|_1=2n|\alpha_n|$. Hence, if $C:=\sup_nn|\alpha_n|<\infty$ we have that $$\|A\mathbf{x}\|_1\leq\lim_N\sum^N_{k=1}|\mathbf{x}(k)|\|A\mathbf{e}_k\|_1\leq C\lim_N\sum^N_{k=1}|\mathbf{x}(k)|=C\|\mathbf{x}\|_1$$ for all $\mathbf{x}\in\ell_1(\mathbb{N})$.

Conversely, $A\in L(\ell_1,\ell_1)$ then $$\|A\mathbf{e}_k\|=2n|\alpha_n|\leq \|A\|\|\mathbf{e}_n\|_1=\|A\|$$ for all $n$. Hence $\sup_n|\alpha_n|<\infty$.

Answer 2

$\rhd$ The problem is that $(e_n)_{n \in \mathbb{N}}$ is not a bisis of $\ell^1( \mathbb{N})$, therefore your $A$ is not well defined. For example, $u= \left(\dfrac{1}{2^n}\right)_{n \in \mathbb{N}}$ is not in Span($(e_n)_{n \in \mathbb{N}}$). It is then possible to build such an operator $A$ that is not bounded for any choice of $(\alpha_n)_{n \in \mathbb{N}}$.

For example, with $f_k = \left(\dfrac{1}{k^n}\right)_{n \in \mathbb{N}}$, one can show that as $(e_n)_{n \in \mathbb{N}} \cup (f_k)_{k \geq 2}$ is an independant family of $\ell^1( \mathbb{N})$, and then any operator $A$ such that :

• $A(e_n)= \alpha_n \sum\limits_{i=n}^{2n}e_i$ for all $n \geq 2$ ;
• $A(f_k)= k.f_k$ for all $k \geq 2$

won't be a bounded operator.

$\rhd$ Therefore, to proceed, you have to assume that $A(\sum_{i=1}^{+\infty} u_ie_i)) = \sum_{i=1}^{+\infty} u_iA(e_i)$. With this, you should be able to proceed and show that : $$A \text{ is bounded }\;\; \Longleftrightarrow \;\; \exists C\geq 0 : \forall n \in \mathbb{N}, |\alpha_n|\leq \dfrac{C}{n+1}.$$