Suppose that we have a sequence $(a_k)_k$ s.t. $a_k \to 0$(not necessarily in $\ell^p(1 \le p < \infty)$). Then we can define a bounded linear operator $T : \ell^p \to \ell^p, (x_k)_k \mapsto (a_k x_k)_k$, I want to show that $T(B_1)$ is a closed set, where $B_1$ is the closed unit ball in $\ell^p$.
Some observations are the following :
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$T$ is compact, because we can approximate it by finite rank operator, just taking the first n terms of $(a_k)_k$, let the other terms be zero.
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If $p > 1$, then I can use some machinery to show $T(B_1)$ is closed, becasue in this case $\ell^p$ is reflexive, so I can use weak compactness of the unit ball to find the desired preimage.
My question is, how can I prove this for $p = 1$ ? Or probably this is not true, then I'd like to see a conterexample. And I was stuck to find a suitable preimage, if we assume that $T\vec{x}_k \to \vec{y}$, one of the candidate should be $(y_k/a_k)_k$, but I can't even show this is an element in $\ell^p$.
Best Answer
In order that $T$ maps $\ell^{p}$ into itself, we need boundedness of $(a_k)$. But $a_k \to 0$ is not necessary for showing that $T(B_1)$ is closed.
Suppose $\sum_k |x_k^{(n)}| \leq 1$ for all $n$ and $T(x^{(n)})=(a_k x_k^{n})\to (y_k)$ in $\ell^{1}$. Let $x_k=\frac {y_k}{a_k}$ if $a_k \neq 0$ and $x_k=0$ if $a_k=0$. Then $a_kx_k=y_k$ for all $k$. If we show that $\sum|x_k| \leq 1$ we are done since $(y_k)=T(x_k)$. By Fatou's Lemma we have $\sum_{a_k \neq 0} |\frac {y_k}{a_k}| \leq \lim \inf \sum_{a_k \neq 0} |\frac {a_kx_k^{n}}{a_k}|\leq \lim \inf \sum | x_k^{n}| \leq 1$. Thus, $\sum |x_k| \leq 1$.