Proof Correct? Show that $\sum_{n \in \mathbb N}s_{n}t_{n}<\infty \Rightarrow s \in \ell^{1}$

Question

Let $(s_{n})_{n}\subseteq \mathbb R$ and $c_{0}$ the space of null sequences. Show that $\sum_{n \in \mathbb N}s_{n}t_{n}<\infty \operatorname{for all }t\in c_{0} \Rightarrow s \in \ell^{1}$.

My idea: We know that $(c_{0})^{*}\simeq \ell^{1}$, so we get a map:

$J:\ell^{1}\to(c_{0})^{*}, (Js)(t)=\sum\limits_{n \in \mathbb N}s_{n}t_{n}$ that is an isometric isomorphism.

And we know for our particular sequence $(s_{n})_{n}=:s$ that $\sum_{n \in \mathbb N}s_{n}t_{n}<\infty$ for any $t\in c_{0}$ so similarly we get a functional:

$l_{s}:c_{0}\to\mathbb K, t \mapsto \sum_{n \in \mathbb N}s_{n}t_{n}$ by the isometric isometry of $J$ there must exist a $k\in \ell^{1}$ so that:

$\sum_{n \in \mathbb N}k_{n}t_{n}=(Jk)(t)=l_{s}(t)=\sum_{n \in \mathbb N}s_{n}t_{n}$. I am unsure whether I can now say that $k=s$? If I can, this would then imply $s \in \ell^{1}$

Any ideas?

Answer 1

From what you wrote isn't clear that $\phi : t \mapsto \sum_{k=1}^\infty s_kt_k$ is in $(c_0)^*$, i.e. a bounded linear functional on $c_0$.

To obtain this, notice that for every $n \in \mathbb{N}$ we have that $\phi_n :t \mapsto \sum_{k=1}^n s_kt_k$ is an element of $(c_0)^*$ and for each particular $t \in c_0$ we have $\phi_n(t) \to \phi(t)$. In particular $(\phi_n(t))_n$ is bounded so the uniform boundedness principle implies that $(\phi_n)_n$ is bounded in operator norm. Now for any $t \in c_0$ we have

$$|\phi(t)| = \lim_{n\to\infty} |\phi_n(t)| \le \limsup_{n\to\infty} (|\phi_n\|\|t\|_\infty) = \left(\limsup_{n\to\infty}\|\phi_n\|\right)\|t\|_\infty$$

Therefore $\phi \in (c_0)^*$ so there exists $r = (r_k)_k \in \ell^1$ such that $\phi(t)= \sum_{k=1}^\infty r_kt_k$. In particular, plugging in the $n$-th canonical vector $t = e_n$ gives $r_n = \phi(e_n) = s_n$.

We conclude $s = r \in \ell^1$.

Answer 2

I think that you can not use the map $J$ to prove your result because the surjectivity of your map it is equivalent to prove your result, infact:

If it is true your result then if you consider $\phi\in c_0^*$ then you can define the following sequence

$s:=\{s_n:=\phi(\delta_n)\}_n$ and you have that for each $t\in c_0$

$\sum s_nt_n=\sum \phi(\delta_n)t_n=\sum\phi(t_n\delta_n)=$

$=\phi(\sum t_n\delta_n)=\phi(t)<\infty$

so by hypothesis you have that $s\in l^1$ and so

$J(s)=\phi$

By other side you have that if $J$ is surjective then your result is true and the prove is the answer of mecanodroid.

I think that if you want prove your result you must choose the sequence $t_n:=sign(s_n)$ and so

$\sum s_nt_n=\sum |s_n|<\infty$

The problem is that sequence t is not in $c_0$ so you must built a succession of sequence $\{t_m\}_m\subseteq c_0$ that converges to $t$.

For example you can define $t_m(n):=sign(s_n)$ if $n<m$ and $t_m(n):=\frac{sign(s_n)}{n}$ if $n\geq m$ so you have that $t_m\in c_0$ for all $m$ and

$\infty> \sum s_nt_m(n)=\sum_{k<m}|s_k|+\sum_{k\geq m} \frac{|s_k|}{k}$

Answer 3

Replacing $s_n$ by $\vert s_n\vert$, we may assume $s_n\geq 0$.

Let us argue by contradiction. We assume that the serie $\sum s_n$ diverges, and we will construct a sequence $t_n$, converging to $0$ such that $\sum s_n . t_n= \infty$.

Indeed, we can find an increasing sequence $n_k$ of integers so that $\sum _{n_k+1}^{n_{n+k} }s_n \geq k^2$ (as $\sum _{n_k+1}^{\infty }s_n =\infty $).

Now, for $n_k+1\leq n\leq n_{k+1}$; let $t_n = {1\over k}$, we see that $\sum s_n.t_n \geq \sum k =\infty$.