Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $1<p<\infty$ and $\{x_n\}\subset l^p$, then $\sum_{j=1}^\infty x_n(j)y(j)\to 0$ for every $y\in l^q$, $\frac{1}{p}+\frac{1}{q}=1$, iff $\sup_n||x_n||_P<\infty$ and $x_n(j)\to 0$ for every $j\geq 1$.

I've proved $\sup_n||x_n||_P<\infty$ and $x_n(j)\to 0$ easily. But for converse I stuck. please help me.

I can not put any comment. so I have to write here. Thanks to triple_sec.

share|improve this question
    
I guess you mean $y\in\ell^q$, instead of $y\in\ell^p$. –  triple_sec May 5 at 9:18
add comment

1 Answer 1

EDIT: The reverse direction does not make use of the “almost magical power” (as Folland, 1999, p. 163, put it) of the uniform boundedness principle, so it is accordingly somewhat boring and tedious. The idea is as follows: For a given $y\in \ell^q$, the key observation is that we can separate the domain of the summation of $\sum_{j=1}^{\infty}|x_n(j)y(j)|$ into two parts for a given $n\in\mathbb N$: (1) If $J$ is large enough, then the partial sum $\sum_{j=J+1}^{\infty}|y(j)|^q$ will be small enough and $\sum_{j=J+1}^{\infty}|x_n(j)|^p$ will also remain uniformly bounded, thanks to the assumption that $\sup_{m\in\mathbb N}\|x_m\|_p<\infty$. (2) As for the first $J$ terms, we can exploit the fact that the $x_n(j)$ individually converge to zero as $n\to\infty$, and, since there are only finitely many sequences $(x_n(1),\ldots,x_n(J))_{n\in\mathbb N}$ left, we can make the convergence uniform.

Formally, suppose that $S\equiv\sup_{n\in\mathbb N}\|x_n\|_p<\infty$ and $x_n(j)\to0$ as $n\to\infty$ for any integer $j\geq 1$. Fix any $\varepsilon>0$ and take any $y\in\ell^q$. This means that $$\|y\|_q^q=\sum_{j=1}^{\infty}|y(j)|^q<\infty.$$ In particular, this implies that there exists some $J\in\mathbb N$ such that $$\sum_{j=J+1}^{\infty}|y(j)|^q<\left(\frac{\varepsilon}{2(S+1)}\right)^q.$$ (The $+1$ in the denominator controls for the case in which $S=0$.) Since $\lim_{n\to\infty}x_n(j)=0$ for any fixed integer $j\geq1$, it is the case that for any $j\in\{1,\ldots,J\}$, there exists some $N_j\in\mathbb N$ such that if $n>N_j$ and $n\in\mathbb N$, then $$|x_n(j)|<\frac{\varepsilon}{2(J^{1/p})(\|y\|_q+1)}.$$ Now, if $n\in\mathbb N$ and $n>\max\{N_1,\ldots,N_J\}$, then Hölder's inequality and the previously established facts imply the following: \begin{align*} \sum_{j=1}^{\infty}|x_n(j)||y(j)|=&\,\sum_{j=1}^{J}|x_n(j)||y(j)|+\sum_{j=J+1}^{\infty}|x_n(j)||y(j)|\\ \leq&\left(\sum_{j=1}^{J}|x_n(j)|^p\right)^{1/p}\left(\sum_{j=1}^{J}|y(j)|^q\right)^{1/q}+\left(\sum_{j=J+1}^{\infty}|x_n(j)|^p\right)^{1/p}\left(\sum_{j=J+1}^{\infty}|y(j)|^q\right)^{1/q}\\ \leq&\left(\sum_{j=1}^{J}|x_n(j)|^p\right)^{1/p}\left(\sum_{j=1}^{\infty}|y(j)|^q\right)^{1/q}+\left(\sum_{j=1}^{\infty}|x_n(j)|^p\right)^{1/p}\left(\sum_{j=J+1}^{\infty}|y(j)|^q\right)^{1/q}\\ =&\left(\sum_{j=1}^{J}|x_n(j)|^p\right)^{1/p}\|y\|_q+\left(\sum_{j=1}^{\infty}|x_n(j)|^p\right)^{1/p}\left(\sum_{j=J+1}^{\infty}|y(j)|^q\right)^{1/q}\\ \leq&\left(\sum_{j=1}^{J}\frac{\varepsilon^p}{2^pJ(\|y\|_q+1)^p}\right)^{1/p}\|y\|_q+\left(\sum_{j=1}^{\infty}|x_n(j)|^p\right)^{1/p}\left(\sum_{j=J+1}^{\infty}|y(j)|^q\right)^{1/q}\\ =&\,\frac{\varepsilon}{2}\frac{\|y\|_q}{\|y\|_q+1}+\|x_n\|_p\times\left(\sum_{j=J+1}^{\infty}|y(j)|^q\right)^{1/q}\\ <&\,\frac{\varepsilon}{2}+\|x_n\|_p\times\left(\sum_{j=J+1}^{\infty}|y(j)|^q\right)^{1/q}\\ \leq&\,\frac{\varepsilon}{2}+S\times\frac{\varepsilon}{2(S+1)}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*}

This chain of inequalities establishes that $\lim_{n\to\infty}\sum_{j=1}^{\infty}|x_n(j)||y(j)|=0$, from which it readily follows that $\lim_{n\to\infty}\sum_{j=1}^{\infty}x_n(j)y(j)=0$ as well.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.