#### Proof

Since *E* does not satisfy the Schur property, we can find
$\begin{array}{}\{{x}_{n}{\}}_{n=1}^{\mathrm{\infty}}\subseteq {B}_{E}\end{array}$
and *ε* > 0 so that
${x}_{n}\underset{n\to \mathrm{\infty}}{\overset{\omega}{\to}}0$
but ∥*x*_{n}∥_{E} ≥ *ε* for every *n* in ℕ. Let us denote the set of all weakly null sequences in *E* by
$\begin{array}{}{E}_{\omega}^{\mathbb{N}}\end{array}$. Then, by the uniform boundedness principle, every element in
$\begin{array}{}{E}_{\omega}^{\mathbb{N}}\end{array}$
is bounded (in norm) and therefore, (
$\begin{array}{}{E}_{\omega}^{\mathbb{N}}\end{array}$,∥ ⋅ ∥_{∞}) is a normed space.

Let us also denote
$$\mathfrak{S}=\left\{\{{y}_{n}{\}}_{n=1}^{\mathrm{\infty}}\in {E}_{\omega}^{\mathbb{N}}:{y}_{n}\underset{n\to \mathrm{\infty}}{\overset{\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\parallel \cdot {\parallel}_{\mathrm{\infty}}}{\to}}0\right\}.\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}/$$

Next, we enumerate the set of prime numbers as {*p*_{k}: *k* ≥ 1} in the usual way (in increasing order), and consider
$$\begin{array}{}F:\mathbb{N}\mathrm{\setminus}\{1\}\u27f6\mathbb{N}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mapsto \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}F(m)=l\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{with}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{p}_{l}={\displaystyle min\{p\phantom{\rule{thinmathspace}{0ex}}\text{prime}:p|m\}.}\end{array}$$

Notice that, in particular, *F*(*p*_{n}) = *n* and hence *F* is surjective. Define now
$$\begin{array}{}T:({\ell}_{\mathrm{\infty}},\parallel \cdot {\parallel}_{\mathrm{\infty}})\u27f6({E}_{\omega}^{\mathbb{N}},\parallel \cdot {\parallel}_{\mathrm{\infty}})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\{{a}_{n}{\}}_{n=1}^{\mathrm{\infty}}\mapsto \phantom{\rule{1em}{0ex}}\{T(\{{a}_{n}{\}}_{n=1}^{\mathrm{\infty}})(k){\}}_{k=1}^{\mathrm{\infty}},\end{array}$$
where
$$\begin{array}{}T(\{{a}_{n}{\}}_{n=1}^{\mathrm{\infty}})(k)={a}_{F(k+1)}{x}_{k},\end{array}$$
and
$\begin{array}{}\{{x}_{k}{\}}_{k=1}^{\mathrm{\infty}}\end{array}$
is the sequence we considered at the beginning of this proof.

*T* is obviously linear. Also, for every natural number *n*,
$$\begin{array}{}\epsilon |{a}_{n}|=\epsilon |{a}_{F({p}_{n})}|\le |{a}_{F({p}_{n})}|\parallel {x}_{{p}_{n}-1}{\parallel}_{E}=\parallel T(\{{a}_{k}{\}}_{k=1}^{\mathrm{\infty}})({p}_{k}-1){\parallel}_{E}\le \parallel T(\{{a}_{k}{\}}_{k=1}^{\mathrm{\infty}}){\parallel}_{\mathrm{\infty}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=sup\{\parallel T(\{{a}_{k}{\}}_{k=1}^{\mathrm{\infty}})(l){\parallel}_{E}:l\in \mathbb{N}\}\le max\{|{a}_{F(l+1)}|\parallel {x}_{l}{\parallel}_{E}:l\in \mathbb{N}\}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le max\{|{a}_{l}|:l\in \mathbb{N}\}=\parallel \{{a}_{k}{\}}_{k=1}^{\mathrm{\infty}}{\parallel}_{\mathrm{\infty}}.\end{array}$$

Hence,
$$\begin{array}{}\epsilon \parallel \{{a}_{n}{\}}_{n=1}^{\mathrm{\infty}}{\parallel}_{\mathrm{\infty}}\le \parallel T(\{{a}_{n}{\}}_{n=1}^{\mathrm{\infty}}){\parallel}_{\mathrm{\infty}}\le \parallel \{{a}_{n}{\}}_{n=1}^{\mathrm{\infty}}{\parallel}_{\mathrm{\infty}}\end{array}$$
and therefore *T* is an isomorphism.

To finish, let us show that if
$\begin{array}{}\{{y}_{k}{\}}_{k=1}^{\mathrm{\infty}}\in \overline{T({\ell}_{\mathrm{\infty}})},\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\{{y}_{k}{\}}_{k=1}^{\mathrm{\infty}}\in \mathfrak{S}.\end{array}$

Indeed, consider {*a*^{(n)}} ⊆ *ℓ*^{∞} so that
$${\left\{{a}_{F(k+1)}^{(n)}{x}_{k}\right\}}_{k=1}^{\mathrm{\infty}}\underset{n\to \mathrm{\infty}}{\overset{\parallel \cdot {\parallel}_{\mathrm{\infty}}}{\to}}\{{y}_{k}{\}}_{k=1}^{\mathrm{\infty}}.$$

Then, given
$\begin{array}{}\stackrel{~}{\epsilon}>0\end{array}$, there exists *N* ∈ ℕ so that, if *n* > *m* ≥ *N* and *k* in ℕ,
$$\begin{array}{}\stackrel{~}{\epsilon}>|{a}_{F(k+1)}^{(n)}-{a}_{F(k+1)}^{(m)}|\parallel {x}_{k}{\parallel}_{E}\ge \epsilon |{a}_{F(k+1)}^{(n)}-{a}_{F(k+1)}^{(m)}|,\end{array}$$
so that
$$\begin{array}{}{\displaystyle |{a}_{F(k+1)}^{(n)}-{a}_{F(k+1)}^{(m)}|<\frac{\stackrel{~}{\epsilon}}{\epsilon},}\end{array}$$
for every *k* ∈ ℕ. Therefore,
$\begin{array}{}\{{a}^{(n)}{\}}_{n=1}^{\mathrm{\infty}}\end{array}$
is Cauchy in (*ℓ*_{∞},∥ ⋅ ∥_{∞}) and then we can find
$\begin{array}{}a=\{{a}_{k}{\}}_{k=1}^{\mathrm{\infty}}\in {\ell}_{\mathrm{\infty}}\end{array}$
with
${a}^{(n)}\underset{n\to \mathrm{\infty}}{\overset{\parallel \cdot {\parallel}_{\mathrm{\infty}}}{\to}}a.$

We claim that
$\begin{array}{}\{{y}_{k}{\}}_{k=1}^{\mathrm{\infty}}=\{{a}_{F(k+1)}{x}_{k}{\}}_{k=1}^{\mathrm{\infty}}.\end{array}$
Indeed, if *r* > 0 we can find
$\begin{array}{}\stackrel{~}{N}\end{array}$
∈ ℕ so that, for every *k* ∈ ℕ and *n* ≥
$\begin{array}{}\stackrel{~}{N}\end{array}$,
$$\begin{array}{}|{a}_{k}^{(n)}-{a}_{k}|<\frac{r}{2M},\\ \parallel {a}_{F(k+1)}^{(n)}{x}_{k}-{y}_{k}{\parallel}_{E}<\frac{r}{2},\end{array}$$
where *M* = sup{∥*x*_{l}}∥_{E} : *l* ∈ ℕ}. Then,
$$\begin{array}{}{\displaystyle \parallel {a}_{F(k+1)}{x}_{k}-{y}_{k}{\parallel}_{E}\le \parallel {a}_{F(k+1)}^{(n)}{x}_{k}-{a}_{F(k+1)}{x}_{k}{\parallel}_{E}+\parallel {a}_{F(k+1)}^{(n)}{x}_{k}-{y}_{k}{\parallel}_{E}<\frac{r}{2}+\frac{r}{2}=r.}\end{array}$$

Make *r* → 0^{+} to obtain the claim.

Finally, it is obvious that
${a}_{F(k+1)}{x}_{k}\underset{k\to \mathrm{\infty}}{\overset{\omega}{\to}}0.$
Also, if *a* ≠ 0, we can find *n* ∈ ℕ so that *a*_{n} ≠ 0 and therefore, for every *s* ∈ ℕ,
$$\begin{array}{}\parallel {a}_{F({p}_{n}^{s})}{x}_{{p}_{n}^{s}-1}{\parallel}_{E}\ge |{a}_{n}|\epsilon ,\end{array}$$
which implies that
$\begin{array}{}\{{y}_{k}{\}}_{k=1}^{\mathrm{\infty}}=\{{a}_{F(k+1)}{x}_{k}{\}}_{k=1}^{\mathrm{\infty}}\in \mathfrak{S}.\end{array}$

We then conclude that
$\begin{array}{}\overline{T({\ell}_{\mathrm{\infty}})}\mathrm{\setminus}\{0\}\subseteq \mathfrak{S},\end{array}$
so yielding the spaceability of 𝔖. □

Taking a look into the proof, we can have the following:

#### Proof

In Theorem 1.4, the authors considered the function
$$\begin{array}{}f(x)={\displaystyle \sum _{n=1}^{\mathrm{\infty}}{f}_{n}(\varphi (x)){e}_{n},}\end{array}$$
where {*e*_{n}} is a sequence that prevents *E* from being a Schur space (in particular, it is weakly null but uniformly bounded below in norm) and *f*_{n}, *ϕ* are auxiliary functions that do not depend on the sequence {*e*_{n}}. Using the fact that {*e*_{n}} does not satisfy the Schur property, the authors managed to prove that *f* ∈ 𝒜_{u}(
$\begin{array}{}{\overline{B}}_{X}\end{array}$,*E*_{ω}) ∖ 𝒜_{u}(
$\begin{array}{}{\overline{B}}_{X}\end{array}$,*E*).

In Theorem 2.1, we considered an isomorphim *T* : (*ℓ*_{∞},∥ ⋅ ∥_{∞}) →(𝔖 ∪{0},∥ ⋅ ∥_{∞}). To prove this corollary, we can just consider
$$\begin{array}{}{\stackrel{~}{T}}_{1}:\phantom{\rule{1em}{0ex}}{\ell}_{\mathrm{\infty}}\phantom{\rule{1em}{0ex}}\u27f6{\mathcal{A}}_{u}({\overline{B}}_{X},{E}_{\omega})\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\{{x}_{n}{\}}_{n=1}^{\mathrm{\infty}}\mapsto \sum _{n=1}^{\mathrm{\infty}}{f}_{n}(\varphi (x)){x}_{n}.\end{array}$$

It is easy to see that
$\begin{array}{}{\stackrel{~}{T}}_{1}\end{array}$
is well-defined and that, following the proof of Theorem 1.4, for every
$\begin{array}{}\{{x}_{n}{\}}_{n=1}^{\mathrm{\infty}}\in {\ell}_{\mathrm{\infty}}\mathrm{\setminus}\{0\}\end{array}$
we have that
$\begin{array}{}{\stackrel{~}{T}}_{1}(\{{x}_{n}{\}}_{n=1}^{\mathrm{\infty}})\notin {\mathcal{A}}_{u}({\overline{B}}_{X},E).\end{array}$
Using the fact that *f*_{n} and *ϕ* are independent of the sequence not satisfying the Schur property, it is easy to see that
$\begin{array}{}{\stackrel{~}{T}}_{1}\end{array}$
is injective. □

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