show that $x$ is in $\ell^1$ if $\sum x_ny_n$ is convergent for all $y=(y_n)\in{c_0}$

Question

If $x=(x_n)$ is a sequence of complex numbers such that the series $\sum x_ny_n$ is convergent for all $y=(y_n)\in{c_0}$. Then prove that $x\in{\ell^1}$. Can anyone tell me what is the meaning of $c_0$ in this question and how to proceed on proving it.

Answer 1

Suppose that $x=(x_n)_n\notin \ell^1$, that is $\sum_{n=1}^\infty|x_n|=+\infty$. Let $S_n=\sum_{k=1}^n|x_k|$. We will define the sequence $(y_n)_n$ by setting $y_n=0$ if $x_n=0$ and $$ y_n=\frac{1}{x_n}\log\left(\frac{S_n}{S_{n-1}}\right)\qquad\hbox{if $x_n\ne0$} $$ (with $S_0=1$.) Now, clearly, if $x_n\ne 0$, we have $$ |y_n|=\frac{1}{|x_n|}\log\left(1+\frac{|x_n|}{S_n}\right)\leq \frac{1}{S_n}. $$ This allows us to conclude that $\lim_{n\to\infty}y_n=0$, ${\it i.e.}$ $(y_n)_{n}\in c_0$. On the the other hand, if $x_{n_0}\ne 0$ then for $n>n_0$ we have $$0\leq x_ny_n=\log(S_n)-\log(S_{n-1})$$ and this proves that $\sum x_ny_n$ is divergent.

We have proved that if $x=(x_n)_n\notin \ell^1$ then there is a sequence $y=(y_n)_{n}\in c_0$, such that $\sum x_ny_n$ is divergent. This proves the desired result by contraposition.

Answer 2

I would upvote Omran Kouba's answer if I could, since it is probably the one you were looking for. Here is an alternative direct proof, if one is willing to use more tools from functional analysis.

Equip $c_0$ with the sup norm $\|y\|_\infty = \sup |y_n|$. This a Banach space [exercise].

Now let us take your sequence $y=(y_n)$ and consider, for every $k$, the linear functional $$ L_k:x\longmapsto \sum_{n=0}^k x_ny_n. $$ On $c_0$, its norm is [exercise] $$\|L_k\|=\sum_{n=0}^k|y_n|.$$

By assumption $\sup_k |L_kx|<\infty$ for every $x\in c_0$. Therefore, by the Uniform Boundedness Principle, $\sum_{n=0}^\infty|y_n|=\sup_k \|L_k\|<\infty$.

Answer 3

Suppose that $$ \sum_{k=1}^\infty|x_k|=\infty\tag1 $$ The divergence in $(1)$ implies that there is a sequence $\{n_m\}_{m=1}^\infty$ so that $n_1=1$ and $$ \sum_{n_m\le k\lt n_{m+1}}\!\!\!\!|x_k|\ge m\tag2 $$ Let the sequence $\{y_k\}_{k=1}^\infty$ be given by $$ y_k=\left\{\begin{array}{cl} \frac1m\frac{|x_k|}{x_k}&\text{if $n_m\le k\lt n_{m+1}$ and $x_k\ne0$}\\ \frac1m&\text{if $n_m\le k\lt n_{m+1}$ and $x_k=0$} \end{array}\right.\tag3 $$ Then, $\lim\limits_{k\to\infty}y_k=0$, and $$ \begin{align} \sum_{k=1}^\infty y_kx_k &=\sum_{m=1}^\infty\sum_{n_m\le k\lt n_{m+1}}x_ky_k\tag{4a}\\ &=\sum_{m=1}^\infty\frac1m\sum_{n_m\le k\lt n_{m+1}}\!\!\!\!|x_k|\tag{4b}\\ &\ge\sum_{m=1}^\infty\frac1m\cdot m\tag{4c} \end{align} $$ and $\text{(4c)}$ diverges.

Therefore, if $\sum\limits_{k=1}^\infty|x_k|$ diverges, then there is a sequence $y\in c_0$ so that $\sum\limits_{k=1}^\infty y_kx_k$ also diverges.