A sequence of operators converging to zero on a Banach space

Question

I would like to check if my reasoning in the following problem is correct (one-way, that is), and if there are any other ways to see the following result:

Let $\{ x_n \}$ be a sequence in $l^p$ (so $\{x_n\}$ is a sequence whose elements are sequences) . Then, $\displaystyle\sup_n ||x_n|| < \infty$ and $x_n(j) \to 0$ for all $j \geq 1$, if and only if for all $y \in l^q$, $\displaystyle\sum_{j=1}^\infty x_n(j)y(j) \to 0$, where $p+q=pq$, $1< p,q < \infty$.

Attempt:

I attempted to the "only if" part by trying to use a special value of $y$, namely $\{ x_n^{\frac{p}{q}}\}$, which gives: $$ \displaystyle\sum_{j=1}^\infty x_n(j)y(j) \to 0 \implies \displaystyle\sum_{j=1}^\infty \big(x_n(j)\big)^{1+p/q} \to 0 \implies \displaystyle\sum_{j=1}^\infty \big(x_n(j)\big)^p \to 0 \implies ||x_n|| \to 0 $$

Hence, as $||x_n||$ is a convergent sequence, $||x_n||$ is bounded. Furthermore, suppose that $j>0$ and $\epsilon>0$ fixed. Then, for this $\epsilon$, there exists $N \in \mathbb{N}$ such that $n > N \implies ||x_n|| < \epsilon$. But then, $||x_n|| < \epsilon \implies x_n(j) < \epsilon$. Hence,$x_n(j)\to 0$ for all $j$.

I know I have done something contentious above, but I can't figure out what it is (something to do with absolute value, maybe). I cannot get a headway on the reverse part either.

Answer 1

$\Rightarrow)$ Let $y\in \ell^q$. Then the map

$$z\mapsto \sum_jy(j)z(j)=\int_\mathbb{N}yz\ d\mu$$

(where $\mu$ is the counting measure) is well defined (Hölder's inequality) and continuous, by the dominated convergence theorem. In particular, if $x_n\to 0$ in $\ell^p$ (which is implied by your hypothesis), then $\int yx_n\ d\mu\to \int 0\ d\mu=0$.

$\Leftarrow)$ What you've written in your answer looks fine to me, since $|x(j)|\le \|x\|_p$.

Answer 2

1....If $x_n(i)$ does not converge to $0$ for some $i,$ let $y=(y(j))_{j\in N}$ where $y(j)=0$ when $j\ne i$ and $y(i)=1.$ Then $(\sum_j x_n(j)y(j))_n=(x_n(i))_n$ doesn't converge to $0.$

2....Since $l^q =(l_p)^*$ it is convenient to write $\sum_jx_n(j)y(j)=y(x)$ for $x\in l_p$ and $y\in l_q$. Note that $|y(x)|\leq \|y\|_q\cdot \|x\|_p.$

Suppose $\|x_n\|_p\leq M<\infty$ for all $n,$ and $x_n(j)\to 0$ as $n\to \infty$ for each $n.$ Given $e>0,$ and given $y\in l^q,$ take $m\in N$ large enough that $[\sum_{j>m}|y_j|^q]^{1/q}<\frac {e}{2M}.$ And for each $i\leq m$ take $n_i$ large enough that $n\geq n_i\implies |x_n(i)y(i)|<\frac {e}{2m}.$ Let $n'=\max \{n_i:i\leq m\}.$

Now for each $n$ let let $x'_n=(x'_n(j))_j$ where $x'_n(j)=0$ for $j\leq m$ and $x'_n(j)=x_n(j)$ for $j>m.$ And let $y'=(y'(j))_j$ where $y'(j)=0$ for $j\leq m,$ and $y'(j)=y(j)$ for $j>m.$ Note that $\|x'_n\|_p\leq \|x_n\|_p\leq M$ and that $\|y'\|_q<\frac {e}{2M}.$

Then $$ n\geq n' \implies |y(x_n)|\leq \sum_{i=1}^m|y(i)x_n(i)|+|\sum_{j>m}y(j)x'_n(j)|<$$ $$< m\frac {e}{2m}+|\sum_{j>m}y(j)x'_n(j)|=$$ $$=\frac {e}{2}+|y'(x'_n)|\leq$$ $$\leq \frac{e}{2}+\|y'\|_q \|x'_n\|_p<$$ $$<\frac {e}{2}+\frac {e}{2M}M=e.$$

Remark. The condition that $(x_n)_n$ is a bounded sequence is necessary. For example if $x_n(j)=0$ for $j\ne n$ and $x_n(n)=n$, then for each $j$ we have $x_n(j)\to 0 $ as $n\to \infty.$ But when $y(j)=1/j$ for each $j,$ we have $y(x_n)=1$ for every $n.$