Every order interval in $l^1$ is norm compact

Question

Let $l^1$ denote the space of sequences $(x_n)\subset \mathbb{R}$ with $\Vert (x_n)\Vert_1:=\sum_{n\geq 1} |x_n|<\infty$. We say that $(x_n^1)\leq (x_n^2)$ whenever $x_n^1\leq x_n^2$ for every $n\in\mathbb{N}$. It is well-known that $(l^1,\Vert\cdot\Vert_1)$ is a Banach space.

Given $(x_n^1),(x_n^2)\in l^1$ with $(x_n^1)\leq (x_n^2)$ we define the order interval

$$[(x_1^n),(x_2^n)]:=\{ (y_n)\in l^1\colon (x_n^1)\leq (y_n)\leq (x_n^2)\}.$$

I suspect that this set is norm compact.

Any hint to prove that?

Answer 1

We may consider the set $S=[0, (x_n)]$ only without loss of generality.

Let $(y_n^k) \in S$. We want to show that it has a convergent subsequence in $S$.

A short but not elementary proof: The set $S$ is compact in the product topology. Thus, there is a point wise convergent subsequence $(y_n^{k_m})$ with a limit $(y_n)\in S$. By dominated convergence theorem, it converges in $\ell^1$.

Answer 2

Call $C$ this set. Closeness is quite easy to see, since $C=\bigcap_{n\in\mathbb N}C_n$ where $C_n=\left\{\left(y_j\right)\in\ell^1\mid x_n^1\leqslant y_n\leqslant x_n^2\right\}=L_n^{-1}\left(\left[x_n^1,x_n^2\right]\right)$ and $L_n\colon \ell^1 \to \mathbb R$ is the continuous linear functional defined by $L_n\left( \left(y_j\right) \right)=y_n$.

Now, we have to see whether $C$ is precompact, that is, if for all positive $\varepsilon$, there exists a finite set $F\subset C$ such that for all $ \left(y_j\right)\in C$, there exists $\left(y'_j\right)\in F$ such that $\left\lVert \left(y_j\right)-\left(y'_j\right)\right\rVert_1\lt \varepsilon$. It suffices to treat the case where $x_n^1=-x_n^2$ for all $n$ because $$C\subset \widetilde{C}:=\left\{\left(y_j\right)\mid \forall n, - \max\left\{\left\lvert x_n^1\right\rvert,\left\lvert x_n^2\right\rvert\right\}\leqslant y_n\leqslant\max\left\{\left\lvert x_n^1\right\rvert,\left\lvert x_n^2\right\rvert\right\} \right\}.$$ To this aim, fix $\varepsilon\gt 0$ and let $R_n:=\max\left\{\left\lvert x_n^1\right\rvert,\left\lvert x_n^2\right\rvert\right\}$. Let $N$ be such that $\sum_{n=N+1}^{\infty}R_n\lt \varepsilon/2$. Then use precompactness of $\prod_{n=1}^N\left[-R_n,R_n\right]$.