$\\{x_{n}\\}_{n\in\omega}$ is also $F$). If $\chi=(\chi_{i})\in E^{\prime}$
and, in the case (i), $G$ is the finite support of $\chi$, then

$\Big{\langle}\chi,y_{n}-\sum_{j\leq n}x_{j}\Big{\rangle}=\sum_{i\in F\cup
G}\Big{\langle}\chi_{i},P_{i}(y_{n})-\sum_{j\leq n}x_{j}^{i}\Big{\rangle}\to
0\quad\mbox{ as }\;n\to\infty.$

Hence the sequence $\big{\\{}y_{n}-\sum_{j\leq n}x_{j}\big{\\}}$ is weakly
null. Thus $E$ has the property $p$-$(u)$.

###### Proposition 12.9

Let $p\in[1,\infty)$, and let $(E,\tau)$ be a locally convex space. If
$\mathcal{T}$ is a locally convex topology on $E$ compatible with $\tau$, then
$(E,\tau)$ has the property $p$-$(u)$ if and only if $(E,\mathcal{T})$ has the
property $p$-$(u)$.

###### Proof 79

The assertion immediately follows from the fact that the notions of being
weakly null, weakly Cauchy and weakly $p$-summable depend only on the duality
$(E,E^{\prime})$.

We finish this section with the several open problems.

###### Problem 12.10

For every distinct pairs $(p,q)$ and $(p^{\prime},q^{\prime})$ with $p\leq q$
and $p^{\prime}\leq q^{\prime}$, find (Banach, Fréchet, barrelled, etc.)
locally convex spaces $E$ with the property $V_{(p,q)}$ but without the
property $V_{(p^{\prime},q^{\prime})}^{\prime}$ (analogously, for the
properties $EV_{(p,q)}$, $V_{(p,q)}^{\ast}$, $EV_{(p,q)}^{\ast}$ and
$p$-$(u)$).

###### Problem 12.11

Characterize Tychonoff spaces $X$ for which the space $C_{k}(X)$ has one of
the property $V_{(p,q)}$ ($EV_{(p,q)}$, $V_{(p,q)}^{\ast}$,
$EV_{(p,q)}^{\ast}$ or $p$-$(u)$).

###### Problem 12.12

Characterize Tychonoff spaces $X$ for which the free lcs $L(X)$ over $X$ has
one of the property $V_{(p,q)}$ ($EV_{(p,q)}$, $V_{(p,q)}^{\ast}$,
$EV_{(p,q)}^{\ast}$ or $p$-$(u)$).

###### Problem 12.13

Characterize Tychonoff spaces $X$ for which the space $C_{p}(X)$ has the
property $p$-$(u)$.

##  13 Weakly sequentially $p$-(pre)compact and $p$-convergent operators

In what follows we shall use intensively the following $p$-versions of weakly
compact-type properties, which extend the corresponding notions in the case of
Banach spaces defined in [12] and [42].

###### Definition 13.1

Let $p\in[1,\infty]$. A subset $A$ of a separated tvs $E$ is called

  1. $\bullet$

(relatively) weakly sequentially $p$-compact if every sequence in $A$ has a
weakly $p$-convergent subsequence with limit in $A$ (resp., in $E$);

  2. $\bullet$

weakly sequentially $p$-precompact if every sequence from $A$ has a weakly
$p$-Cauchy subsequence.

Note that if $E$ is a Banach space, weakly sequentially $p$-precompact sets of
$E$ are known as conditionally weakly $p$-compact sets, see Definition 2.5 of
[16].

###### Lemma 13.2

Let $p\in[1,\infty]$, $E$ and $L$ be locally convex spaces, and let $T:E\to L$
be a weak-weak continuous linear map. If $A$ is a $($relatively$)$ weakly
sequentially $p$-compact set (resp., a weakly sequentially $p$-precompact set)
in $E$, then the image $T(A)$ is a $($relatively$)$ weakly sequentially
$p$-compact (resp., weakly sequentially $p$-precompact) set in $L$.

###### Proof 80

We consider only the case when $A$ is a (relatively) weakly sequentially
$p$-compact set, the case when $A$ is weakly sequentially $p$-precompact can
be considered analogously. Let $\\{T(a_{n})\big{\\}}_{n\in\omega}$ be a
sequence in $T(A)$. Since $A$ is (relatively) weakly sequentially $p$-compact,
the sequence $\\{a_{n}\\}$ has a subsequence
$\big{\\{}a_{n_{k}}\big{\\}}_{k\in\omega}$ which weakly $p$-converges to some
$a\in A$ (resp., $a\in E$). Let now $\chi\in L^{\prime}$. Since $T$ is weakly
continuous, Theorem 8.10.3 of [67] implies that $T^{\ast}(\chi)\in E^{\prime}$
and we have

$\big{(}\langle\chi,T(a_{n_{k}})-T(a)\rangle\big{)}_{k\in\omega}=\big{(}\langle
T^{\ast}(\chi),a_{n_{k}}-a\rangle\big{)}_{k\in\omega}\in\ell_{p}\;(\mbox{or
$\in c_{0}$ if $p=\infty$}).$

Therefore $T(a_{n_{k}})$ weakly $p$-converges to $T(a)$. Thus $T(A)$ is a
$($relatively$)$ weakly sequentially $p$-compact set in $L$.

Below we extend the notion of $p$-convergent operators between Banach spaces
defined by Castillo and Sánchez [12] to the general case of all separated
topological vector spaces. Weakly $p$-precompact operators between Banach
spaces were defined by Ghenciu [42].

###### Definition 13.3

Let $p\in[1,\infty]$. A linear map $T:E\to L$ between separated topological
vector spaces $E$ and $L$ is said to be

  1. $\bullet$

$p$-convergent if it sends weakly $p$-summable sequences in $E$ into null-
sequences in $L$;

  2. $\bullet$

weakly sequentially $p$-compact if for some $U\in\mathcal{N}_{0}(E)$, the set
$T(U)$ is a relatively weakly sequentially $p$-compact subset of $L$;

  3. $\bullet$

weakly sequentially $p$-precompact if for some $U\in\mathcal{N}_{0}(E)$, the
set $T(U)$ is weakly sequentially $p$-precompact in $L$;

  4. $\bullet$

unconditionall converging if $T$ sends $wuC$ series into unconditionally
convergent series.

To characterize $p$-convergent operators we need the next lemma.

###### Lemma 13.4

Let $p\in[1,\infty]$, $E$ and $L$ be locally convex spaces, and let $T:E\to L$
be a $p$-convergent linear map. If $\\{x_{n}\\}_{n\in\omega}\subseteq E$ is
weakly $p$-Cauchy, then the sequence $\big{\\{}T(x_{n})\big{\\}}_{n\in\omega}$
is Cauchy in $L$. Consequently, if $L$ is sequentially complete, then
$\big{\\{}T(x_{n})\big{\\}}_{n\in\omega}$ converges in $L$.

###### Proof 81

Take a pair of strictly increasing sequences $(k_{n}),(j_{n})$ in $\omega$.
Then $\\{x_{k_{n}}-x_{j_{n}}\\}$ is weakly $p$-summable. Since $T$ is
$p$-convergent, we obtain $T(x_{k_{n}})-T(x_{j_{n}})\to 0$ in $L$, and hence
$\big{\\{}T(x_{n})\big{\\}}_{n\in\omega}$ is Cauchy in $L$.

For $1\leq p<\infty$ and operators between Banach spaces the next proposition
is Theorem 2.6 of [16].

###### Proposition 13.5

Let $p\in[1,\infty]$, and let $T$ be a weak-weak sequentially continuous
linear map from a locally convex space $E$ to a sequentially complete locally
convex space $L$. Then the following assertions are equivalent:

  1. (i)

$T$ is $p$-convergent;

  2. (ii)

$T(A)$ is relatively sequentially compact in $L$ for each weakly sequentially
$p$-precompact subset $A$ of $E$;

  3. (iii)

$T(A)$ is sequentially precompact in $L$ for any weakly sequentially
$p$-precompact subset $A$ of $E$.

###### Proof 82

(i)$\Rightarrow$(ii) Let $A$ be a weakly sequentially $p$-precompact subset of
$E$, and let $S=\\{a_{n}\\}_{n\in\omega}$ be a sequence in $A$. Then $S$ has a
weakly $p$-Cauchy subsequence $\\{a_{n_{k}}\\}_{k\in\omega}$. Then, by Lemma
13.4, the sequence $\big{\\{}T(a_{n_{k}})\big{\\}}_{k\in\omega}$ converges to
some element $x\in L$. Thus $T(A)$ is relatively sequentially compact in $L$.

(ii)$\Rightarrow$(iii) is trivial.

(iii)$\Rightarrow$(i) Let $\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-summable
sequence in $E$. Then $\\{x_{n}\\}_{n\in\omega}$ is weakly $p$-Cauchy in $E$,
and hence $\\{T(x_{n})\\}_{n\in\omega}$ is sequentially precompact in $L$.
Since $L$ is sequentially complete, Lemma 3.5 implies that
$\\{T(x_{n})\\}_{n\in\omega}$ is relatively sequentially compact in $L$.
Therefore, any subsequence $S$ of $\\{T(x_{n})\\}_{n\in\omega}$ has a
subsequence $\\{T(x_{n_{k}})\\}_{k\in\omega}$ which converges to some element
$z\in L$. On the other hand, since $T$ is weak-weak sequentially continuous
and $x_{n}\to 0$ in the weak topology, we obtain that $T(x_{n})\to 0$ in the
weak topology of $L$. Therefore $z=0$. Since $S$ was arbitrary, it follows
that $T(x_{n})\to 0$ in the space $L$. Thus $T$ is $p$-convergent.

The problem to find pairs $(E,L)$ of locally convex spaces for which
$\mathcal{L}(E,L)$ contains only $p$-convergent operators is of independent
interest. Below we consider partial cases when this problem has a positive
solution.

###### Proposition 13.6

Let $E$ and $L$ be locally convex spaces.

  1. (i)

If $p\in[1,\infty]$ and $E$ or $L$ has the $p$-Schur property, then each
operator $T\in\mathcal{L}(E,L)$ is $p$-convergent.

  2. (ii)

Assume that $L$ has the Schur property $($for example, $L$ carries its weak
topology$)$. Then for every $p\in[1,\infty]$, each operator
$T\in\mathcal{L}(E,L)$ is $p$-convergent.

###### Proof 83

(i) If $E$ has the $p$-Schur property, the assertion is trivial. If $L$ has
the $p$-Schur property, the assertion follows from (iii) of Lemma 4.6 which
states that continuous images of weakly $p$-summable sequences are weakly
$p$-summable.

(ii) If $L$ has the Schur property, it has the $p$-Schur property for every
$p\in[1,\infty]$. Now (i) applies.

A sufficient condition of being a $p$-convergent operator is given in the next
assertion.

###### Proposition 13.7

Let $p\in[1,\infty)$, $E$ and $L$ be locally convex spaces, and let $T:E\to L$
be a weak-weak sequentially continuous linear map. Assume that one of the
following conditions holds:

  1. (i)

for any operator $S:\ell_{p^{\ast}}^{0}\to E$ $($or $S:c_{0}^{0}\to E$ if
$p=1$$)$, the linear map $T\circ S$ is sequentially precompact;

  2. (ii)

$E$ is sequentially complete and for any operator $S:\ell_{p^{\ast}}\to E$
$($or $S:c_{0}\to E$ if $p=1$$)$, the linear map $T\circ S$ is sequentially
precompact.

Then $T$ is $p$-convergent.

###### Proof 84

Let $T\circ S$ be sequentially precompact for any operator $S$ from (i) or
(ii). To show that $T$ is $p$-convergent, fix an arbitrary weakly $p$-summable
sequence $\\{x_{n}\\}_{n\in\omega}$ in $E$. We have to prove that $T(x_{n})\to
0$ in $L$. Suppose for a contradiction that $T(x_{n})\not\to 0$ in $L$. Then
we can find a neighborhood $U$ of zero in $L$ and a subsequence
$\\{x_{n_{i}}\\}_{i\in\omega}$ of $\\{x_{n}\\}$ such that $T(x_{n_{i}})\not\in
U$ for every $i\in\omega$. Since $\\{x_{n_{i}}\\}_{i\in\omega}$ is also weakly
$p$-summable, without loss of generality we can assume that $x_{n}\not\in U$
for every $n\in\omega$.

By Proposition 4.14, there is $S\in\mathcal{L}(\ell_{p^{\ast}}^{0},E)$ (resp.,
$S\in\mathcal{L}(\ell_{p^{\ast}},E)$ or $S\in\mathcal{L}(c_{0}^{0},E)$,
$S\in\mathcal{L}(c_{0},E)$ if $p=1$) such that $S(e_{n}^{\ast})=x_{n}$ for
every $n\in\omega$. Since $T\circ S$ is sequentially precompact from a normed
space and $\\{e_{n}^{\ast}\\}_{n\in\omega}$ is bounded, the sequence
$\\{TS(e_{n}^{\ast})\\}_{n\in\omega}$ has a Cauchy subsequence
$\\{TS(e_{n_{k}}^{\ast})\\}_{k\in\omega}$. Denote by $D$ the co-extension of
$TS$ into the completion $\overline{L}$ of the space $L$. Since the sequence
$\\{TS(e_{n_{k}}^{\ast})\\}_{k\in\omega}$ is also Cauchy in the complete space
$\overline{L}$, it converges to some point $\overline{x}\in\overline{L}$.
Observe that $\\{e_{n}^{\ast}\\}_{n\in\omega}$ is a weakly null-sequence in
$\ell_{p^{\ast}}$ (or in $c_{0}$ if $p=1$), and hence the weak-weak sequential
continuity of $D$ implies that $D(e_{n}^{\ast})=TS(e_{n}^{\ast})\to 0$ in the
weak topology of $\overline{L}$. Since also
$TS(e_{n_{k}}^{\ast})\to\overline{x}$ in the weak topology on $\overline{L}$,
we obtain that $\overline{x}=0$, and hence
$T(x_{n_{k}})=TS(e_{n_{k}}^{\ast})\to 0$ in the original topology of $L$. But
this contradicts the choice of $\\{x_{n}\\}$.

For $p\in[1,\infty]$, the $p$-version of the weak sequential completeness of
Banach spaces is introduced in Definition 2.17 of [61]. Below we extend this
notion to the class of all locally convex spaces.

###### Definition 13.8

Let $p\in[1,\infty]$. A locally convex space $E$ is called weakly sequentially
$p$-complete if every weakly $p$-Cauchy sequence is weakly $p$-convergent.

So weakly sequentially $\infty$-complete spaces are exactly weakly
sequentially complete. In particular, $\ell_{1}$ is weakly sequentially
$\infty$-complete. Even a more general assertion holds true.

###### Proposition 13.9

If a Banach space $E$ has the Schur property, then $E$ is weakly sequentially
$p$-complete for every $p\in[1,\infty]$.

###### Proof 85

Let $S=\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-Cauchy sequence in $E$. Then
$S$ is weakly Cauchy and hence, by the Schur property, $S$ is Cauchy in $E$.
Since $E$ is complete, there is $x\in E$ such that $\|x_{n}-x\|\to 0$. Choose
a strictly increasing sequence $(n_{k})$ in $\omega$ such that

$\|x_{n_{k}}-x\|<\tfrac{1}{2^{k}}\;\;\mbox{ for every }\;k\in\omega.$

Then $\\{x_{n_{k}}\\}_{k\in\omega}$ is weakly $p$-converges to $x$ since (we
consider the case $p<\infty$)

$\sum_{k\in\omega}|\langle\chi,x_{n_{k}}-x\rangle|^{p}\leq\sum_{k\in\omega}\|\chi\|\cdot\tfrac{1}{2^{k}}<\infty$

for every $\chi\in E^{\prime}$. Thus, by Lemma 4.5, $S$ weakly $p$-converges
to $x$.

In Corollary 13.13 below we shall show that for every $1<p<\infty$, the space
$\ell_{p}$ is weakly sequentially $p^{\ast}$-complete.

The next assertion will play a crucial role to characterize $p$-convergent
operators.

###### Proposition 13.10

Let $p\in[1,\infty)$, and let $E$ be a locally convex space.

  1. (i)

Each operator $S:\ell_{p^{\ast}}^{0}\to E$ $($or $S:c_{0}^{0}\to E$ if
$p=1$$)$ is weakly sequentially $p$-precompact.

  2. (ii)

If $p>1$, then for each operator $S:\ell_{p^{\ast}}\to E$, the set
$S(B_{\ell_{p^{\ast}}})$ is weakly sequentially $p$-compact in $E$; in
particular, $S$ is weakly sequentially $p$-compact.

  3. (iii)

If $p=1$, then every operator $S:c_{0}\to E$ is weakly sequentially
$1$-precompact. If additionally $E$ is weakly sequentially $1$-complete, then
for each operator $S:c_{0}\to E$, the set $S(B_{c_{0}})$ is a weakly
sequentially $1$-compact subset of $E$; in particular, $S$ is weakly
sequentially $p$-compact.

###### Proof 86

To avoid an unnecessary complication with notations and taking into account
that $p<\infty$, below we set $\ell_{1^{\ast}}^{0}:=c_{0}^{0}$ and
$\ell_{1^{\ast}}:=c_{0}$. We show that the set $S(B_{\ell_{p^{\ast}}^{0}})$
(resp., $S(B_{\ell_{p^{\ast}}})$) is weakly sequentially $p$-precompact
(resp., weakly sequentially $p$-compact) in $E$. To this end, let
$\\{x_{n}\\}_{n\in\omega}$ be an arbitrary sequence in
$B_{\ell_{p^{\ast}}^{0}}$ (resp., in $B_{\ell_{p^{\ast}}}$). To prove the
proposition we have to find a weakly $p$-Cauchy (resp., weakly $p$-convergent
in $S(B_{\ell_{p^{\ast}}})$) subsequence of $\\{S(x_{n})\\}_{n\in\omega}$. We
shall consider $\ell_{p^{\ast}}^{0}$ as a dense subspace of $\ell_{p^{\ast}}$.
We distinguish between the next three possible cases.

Case 1. The sequence $\\{x_{n}\\}_{n\in\omega}$ contains a subsequence
$\\{x_{n_{k}}\\}_{k\in\omega}$ which is Cauchy in the norm topology. Passing
to a subsequence if needed, we can assume that $x_{n}$ norm converges to some
element $x\in B_{\ell_{p^{\ast}}}$ and moreover, that
$\|x_{n}-x\|_{\ell_{p^{\ast}}}<\tfrac{1}{2^{n}}$ for every $n\in\omega$. Then
for every $n\in\omega$ and each $\chi\in E^{\prime}$, we have

$\big{|}\langle\chi,S(x_{n})-S(x)\rangle\big{|}=\big{|}\langle
S^{\ast}(\chi),x_{n}-x\rangle\big{|}\leq\|S^{\ast}(\chi)\|\cdot\tfrac{1}{2^{n}}.$

Let $(k_{n}),(j_{n})\subseteq\omega$ be two strictly increasing sequences.
Then for every $\chi\in E^{\prime}$, we obtain

$\big{|}\langle\chi,S(x_{k_{n}})-S(x_{j_{n}})\rangle\big{|}\leq\big{|}\langle\chi,S(x_{k_{n}})-S(x)\rangle\big{|}+\big{|}\langle\chi,S(x_{j_{n}})-S(x)\rangle\big{|}\leq\|S^{\ast}(\chi)\|\cdot(\tfrac{1}{2^{k_{n}}}+\tfrac{1}{2^{j_{n}}})$

and hence $\\{S(x_{n})\\}_{n\in\omega}$ is weakly $p$-Cauchy (resp.,
$S(x_{n})$ weakly $p$-converges to $S(x)\in S(B_{\ell_{p^{\ast}}})$) and we
are done.

Case 2. The sequence $\\{x_{n}\\}_{n\in\omega}$ contains a subsequence
$\\{x_{n_{k}}\\}_{k\in\omega}$ weakly convergent in $B_{\ell_{p^{\ast}}}$
which does not contain a norm-Cauchy subsequence.

In this case, passing to a subsequence if needed, without loss of generality
we can assume that $\\{x_{n}\\}_{n\in\omega}$ weakly converges to some element
$x\in B_{\ell_{p^{\ast}}}$. Since it does not contain a norm-Cauchy
subsequence, there is $\varepsilon>0$ such that
$\|x_{n}-x\|_{\ell_{p^{\ast}}}\geq\varepsilon>0$ for every $n\in\omega$. Then,
by Proposition 2.1.3 of [1] (see also Theorem 1.3.2 of [1]), there are a basic
subsequence $\\{x_{n_{k}}-x\\}_{k\in\omega}\subseteq\ell_{p^{\ast}}$ of
$\\{x_{n}-x\\}_{n\in\omega}$ and a linear topological isomorphism
$R:\overline{\mathsf{span}}\big{(}\\{x_{n_{k}}-x\\}_{k\in\omega}\big{)}\to\ell_{p^{\ast}}$
such that (recall that $(e_{k}^{\ast})$ is the standard unit basis of
$\ell_{p^{\ast}}$)

$R(x_{n_{k}}-x)=a_{k}e^{\ast}_{k}\quad\mbox{ for every $k\in\omega$, where
}\;a_{k}=\|x_{n_{k}}-x\|_{\ell_{p^{\ast}}},$ (13.1)

and such that
$V:=\overline{\mathsf{span}}\big{(}\\{x_{n_{k}}-x\\}_{k\in\omega}\big{)}$ is
complemented in $\ell_{p^{\ast}}$. Note that

$\varepsilon\leq a_{k}=\|x_{n_{k}}-x\|_{\ell_{p^{\ast}}}\leq 2\quad\mbox{ for
every $k\in\omega$},$

and therefore the map $Q:\ell_{p^{\ast}}\to\ell_{p^{\ast}}$ defined by
$Q(\xi_{k}):=(a_{k}\xi_{k})$ is a topological linear isomorphism. Hence,
replacing $R$ by $Q^{-1}\circ R$, without loss of generality we can assume
that $a_{k}=1$ for every $k\in\omega$.

Subcase 2.1. Let $S:\ell_{p^{\ast}}^{0}\to E$ be an operator.

Since the sequence $\\{x_{n}\\}$ was arbitrary, to show that $S$ is weakly
sequentially $p$-precompact it is sufficient to prove that the subsequence
$\\{S(x_{n_{k}})\\}_{k\in\omega}$ is weakly $p$-Cauchy in $E$. To this end, it
suffices to show that
$\big{\\{}S\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\big{\\}}_{k\in\omega}$
is weakly $p$-summable for every strictly increasing sequence $(k_{j})$ in
$\omega$.

Let
$H_{(k_{j})}:=\mathsf{span}\big{\\{}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{\\}}_{j\in\omega}\subseteq
V$, $R_{(k_{j})}$ be the restriction of $R$ onto $H_{(k_{j})}$, and let
$X_{(k_{j})}:=R_{(k_{j})}\big{(}H_{(k_{j})}\big{)}\subseteq\mathsf{span}\\{e^{\ast}_{k_{j}}\\}_{j\in\omega}$.
Then $R_{(k_{j})}$ is a linear topological isomorphism. Observe also that
$H_{(k_{j})}\subseteq\ell_{p^{\ast}}^{0}$. Let
$\mathsf{id}_{(k_{j})}:H_{(k_{j})}\to\ell_{p^{\ast}}^{0}$ be the identity
embedding. Then we obtain the following two sequences

$\textstyle{X_{(k_{j})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{R_{(k_{j})}^{-1}}$$\textstyle{H_{(k_{j})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathsf{id}_{(k_{j})}}$$\textstyle{\ell_{p^{\ast}}^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{S}$$\textstyle{E}$
and
$\textstyle{E^{\prime}_{\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{S^{\ast}}$$\textstyle{\ell_{p}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathsf{id}_{(k_{j})}^{\ast}}$$\textstyle{\hskip
5.69054pt\big{(}H_{(k_{j})}\big{)}^{\prime}_{\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\big{(}R_{(k_{j})}^{-1}\big{)}^{\ast}}$$\textstyle{\big{(}X_{(k_{j})}\big{)}^{\prime}_{\beta}.}$

Fix now an arbitrary $\chi\in\ E^{\prime}$ and set
$\eta:=\big{(}R_{(k_{j})}^{-1}\big{)}^{\ast}\circ\mathsf{id}_{(k_{j})}^{\ast}\circ
S^{\ast}(\chi)\in\big{(}X_{(k_{j})}\big{)}^{\prime}_{\beta}$. Since
$X_{(k_{j})}$ is a subspace of
$W:=\mathsf{span}\\{e^{\ast}_{k_{j}}\\}_{j\in\omega}$, we can consider $\eta$
as an element $(b_{j})$ of $\ell_{p}=W^{\prime}_{\beta}$. By (13.1), we have
$R_{(k_{j})}\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}=e^{\ast}_{k_{j}}-e^{\ast}_{k_{j+1}}$
and hence

$\displaystyle\langle\chi,S\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\rangle$
$\displaystyle=\langle
S^{\ast}(\chi),\mathsf{id}_{(k_{j})}\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\rangle=\Big{\langle}\big{(}R_{(k_{j})}^{-1}\big{)}^{\ast}\circ\mathsf{id}_{(k_{j})}^{\ast}\circ
S^{\ast}(\chi),e^{\ast}_{k_{j}}-e^{\ast}_{k_{j+1}}\Big{\rangle}$
$\displaystyle=\big{\langle}(b_{j}),e^{\ast}_{k_{j}}-e^{\ast}_{k_{j+1}}\big{\rangle}=b_{j}-b_{j+1},$

and hence

$\sum_{j\in\omega}\big{|}\langle\chi,S\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\rangle\big{|}^{p}=\sum_{j\in\omega}\big{|}b_{j}-b_{j+1}\big{|}^{p}<\infty.$

Thus the sequence
$\big{\\{}S\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\big{\\}}_{k\in\omega}\subseteq
E$ is weakly $p$-summable, as desired.

Subcase 2.2. Let $S:\ell_{p^{\ast}}\to E$ be an operator.

Since the sequence $\\{x_{n}\\}$ was arbitrary, to show that
$S\big{(}B_{\ell_{p^{\ast}}}\big{)}$ is weakly sequentially $p$-compact it is
sufficient to prove that the sequence
$\\{S(x_{n_{k}}-x)\\}_{k\in\omega}\subseteq E$ is weakly $p$-summable, and
hence $S(x_{n_{k}})$ weakly $p$-converges to $S(x)$. To this end, we denote by
$\mathsf{id}_{V}:V\to\ell_{p^{\ast}}$ the identity embedding. Then we obtain
the following sequences

$\textstyle{\ell_{p^{\ast}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{R^{-1}}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathsf{id}_{V}}$$\textstyle{\ell_{p^{\ast}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{S}$$\textstyle{E}$
and
$\textstyle{E^{\prime}_{\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{S^{\ast}}$$\textstyle{\ell_{p}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathsf{id}_{V}^{\ast}}$$\textstyle{V^{\prime}_{\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(R^{-1})^{\ast}}$$\textstyle{\ell_{p}.}$

Take an arbitrary $\chi\in E^{\prime}$. Set
$(b_{k}):=(R^{-1})^{\ast}\circ\mathsf{id}_{V}^{\ast}\circ
S^{\ast}(\chi)\in(\ell_{p^{\ast}})^{\prime}=\ell_{p}$. Then (13.1) implies

$\sum_{k\in\omega}\big{|}\langle\chi,S(x_{n_{k}}-x)\rangle\big{|}^{p}=\sum_{k\in\omega}\Big{|}\big{\langle}S^{\ast}(\chi),\mathsf{id}_{V}\circ
R^{-1}(e^{\ast}_{k})\big{\rangle}\Big{|}^{p}=\sum_{k\in\omega}\Big{|}\Big{\langle}(b_{k}),e^{\ast}_{k}\Big{\rangle}\Big{|}^{p}<\infty.$

Thus the sequence $\\{S(x_{n_{k}}-x)\\}_{k\in\omega}\subseteq E$ is weakly
$p$-summable.

Case 3. The sequence $\\{x_{n}\\}_{n\in\omega}$ does not contain a weakly
convergent subsequence.

Observe first that this case is possible only if $p=1$. (Indeed, if
$1<p<\infty$, then the space $\ell_{p^{\ast}}$ is separable and reflexive, and
hence $B_{\ell_{p^{\ast}}}$ is a metric compact space in the weak topology.
Thus $\\{x_{n}\\}_{n\in\omega}$ contains weakly convergent (in
$\ell_{p^{\ast}}$) subsequences.) So, let $S:c_{0}^{0}\to E$ (or $S:c_{0}\to
E$) be an operator.

By assumption, the norm closure
$\overline{\\{x_{n}\\}_{n\in\omega}}^{\,\|\cdot\|}$ does not contain $0$.
Therefore, by the Kadets–Pełczyński Theorem 1.5.6 of [1], either (1) the weak
closure $K$ of $\\{x_{n}\\}_{n\in\omega}$ is weakly compact in $c_{0}$ and
fails to contain $0$, or (2) $\\{x_{n}\\}_{n\in\omega}$ contains a basic
subsequence. However, the case (1) is impossible. Indeed, since $c_{0}$ is a
separable metrizable spaces, by [66], $K$ is cosmic and hence metrizable.
Therefore, the sequence $\\{x_{n}\\}_{n\in\omega}$ must contain a weakly
convergent subsequence which is impossible by the assumption of Case 3. Hence
passing to a subsequence we can assume that $\\{x_{n}\\}_{n\in\omega}$ is a
basic sequence in $c_{0}$.

Considering $c_{0}$ as a subspace of $\mathbb{F}^{\omega}$, the bounded
sequence $\\{x_{n}\\}_{n\in\omega}$ has a convergent subsequence. Therefore,
without loss of generality we can and will assume that
$\\{x_{n}\\}_{n\in\omega}$ pointwise converges to a point
$z\in\mathbb{F}^{\omega}$. Let $n_{0}=0$. Since $x_{n_{0}}\in c_{0}$, there is
$N_{0}\in\omega$ such that $x_{n_{0}}(i)<1$ for every $i>N_{0}$. As $x_{n}\to
z$ in $\mathbb{F}^{\omega}$, there is $n_{1}>n_{0}$ such that
$|x_{n_{1}}(i)-z(i)|<\tfrac{1}{4}$ for every $i\leq N_{0}$. Since
$x_{n_{1}}\in c_{0}$, there is $N_{1}>N_{0}$ such that
$|x_{n_{1}}(i)|<\tfrac{1}{4}$ for every $i>N_{1}$. Continuing this process we
can find a subsequence $\\{x_{n_{k}}\\}_{n\in\omega}$ of
$\\{x_{n}\\}_{n\in\omega}$ which satisfies the following property: there is a
strictly increasing sequence $(N_{k})\subseteq\omega$ such that for every
$k\in\omega$, the following inequalities hold

$\displaystyle\big{|}x_{n_{k+1}}(i)-z(i)\big{|}<\tfrac{1}{4^{k+1}}\quad\mbox{
for every }\;i\leq N_{k},$ (13.2)
$\displaystyle\big{|}x_{n_{k+1}}(i)\big{|}<\tfrac{1}{4^{k+1}}\quad\mbox{ for
every }\;i>N_{k+1}.$

Then for every $s>k>0$ and each $\eta\in\ell_{1}=(c_{0}^{0})^{\prime}$, (13.2)
implies (recall that all $x_{n}\in B_{c_{0}}$)

$\displaystyle\big{|}\langle\eta,x_{n_{k}}-x_{n_{s}}\rangle\big{|}$
$\displaystyle\leq\sum_{i\leq
N_{k-1}}|\eta(i)|\cdot\tfrac{2}{4^{k}}+\sum_{i=N_{k-1}+1}^{N_{s+1}}|\eta(i)|\cdot
2+\sum_{i>N_{s+1}}|\eta(i)|\cdot\tfrac{2}{4^{k}}$ (13.3)
$\displaystyle\leq\|\eta\|_{\ell_{1}}\cdot\tfrac{2}{4^{k}}+2\sum_{i=N_{k-1}+1}^{N_{s+1}}|\eta(i)|.$

We claim that that the sequence $\\{S(x_{n_{k}})\\}_{n\in\omega}$ is weakly
$1$-Cauchy. To this end, fix an arbitrary strictly increasing sequence
$(k_{j})\subseteq\omega$. For every $\chi\in E^{\prime}$, let
$\eta:=S^{\ast}(\chi)\in\ell_{1}$. Taking into account that $k_{j+1}+1\leq
k_{j+3}-1$ and hence $N_{k_{j+1}+1}<N_{k_{j+3}-1}+1$, (13.3) implies

$\sum_{j\in\omega}\big{|}\big{\langle}\chi,S\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\big{\rangle}\big{|}\leq\sum_{j\in\omega}\tfrac{2}{4^{k_{j}}}\cdot\|\eta\|_{\ell_{1}}+\sum_{j\in\omega}2\sum_{i=N_{k_{j}-1}+1}^{N_{k_{j+1}+1}}|\eta(i)|<\|\eta\|_{\ell_{1}}\cdot\Big{(}6+\sum_{j\in\omega}\tfrac{2}{4^{k_{j}}}\Big{)}<\infty$

Therefore the sequence $\\{S(x_{n_{k}})\\}_{n\in\omega}$ is weakly $1$-Cauchy.
Thus $S\big{(}B_{c_{0}^{0}}\big{)}$ (or $S\big{(}B_{c_{0}}\big{)}$) is weakly
sequentially $1$-precompact.

If additionally $E$ is weakly sequentially $1$-complete and $S:c_{0}\to E$,
the sequence $\\{S(x_{n_{k}})\\}_{n\in\omega}$ weakly $1$-converges to some
point of $E$. Thus $S\big{(}B_{c_{0}}\big{)}$ is weakly sequentially
$1$-compact.

###### Corollary 13.11

Let $p\in[1,\infty)$, and let $\Gamma$ be an infinite set. Then:

  1. (i)

the identity operators
$\mathsf{id}_{\ell_{p^{\ast}}^{0}}:\ell_{p^{\ast}}^{0}(\Gamma)\to\ell_{p^{\ast}}^{0}(\Gamma)$
and
$\mathsf{id}_{\ell_{p^{\ast}}}:\ell_{p^{\ast}}(\Gamma)\to\ell_{p^{\ast}}(\Gamma)$
$($or $\mathsf{id}_{c_{0}^{0}}:c_{0}^{0}(\Gamma)\to c_{0}^{0}(\Gamma)$ and
$\mathsf{id}_{c_{0}}:c_{0}(\Gamma)\to c_{0}(\Gamma)$ if $p=1$$)$ are weakly
sequentially $p$-precompact;

  2. (ii)

(Proposition 1.4 of [12]) if $p>1$, then the identity operator
$\mathsf{id}_{\ell_{p^{\ast}}}:\ell_{p^{\ast}}(\Gamma)\to\ell_{p^{\ast}}(\Gamma)$
is weakly sequentially $p$-compact.

###### Proof 87

(i) follows from Proposition 13.10 in which $E=\ell_{p^{\ast}}^{0}$,
$c_{0}^{0}$, $\ell_{p^{\ast}}^{0}$ or $c_{0}$, respectively, and the fact that
any sequence sits in a direct summand which is isomorphic to $E$.

(ii) follows from (ii) of Proposition 13.10 and the fact from (i).

###### Remark 13.12

The condition $p\not=1$ in (ii) of Corollary 13.11 is essential. Indeed, for
$p=1$ the sequence $\\{e_{0}+\cdots+e_{n}\\}_{n\in\omega}\subseteq B_{c_{0}}$
(which has no weakly convergent subsequences) does not have weakly
$1$-convergent subsequences. Therefore $\mathsf{id}_{c_{0}}$ is not weakly
sequentially $1$-compact.

###### Corollary 13.13

For every $1<p<\infty$, the space $\ell_{p}$ is weakly sequentially
$p^{\ast}$-complete.

###### Proof 88

By (ii) of Proposition 13.10 applied to the identity operator
$\mathsf{id}_{\ell_{p}}:\ell_{p}\to\ell_{p}$, we obtain that $B_{\ell_{p}}$ is
a weakly sequentially $p^{\ast}$-compact subset of $\ell_{p}$. Now, let
$S=\\{x_{n}\\}_{n\in\omega}$ be a weakly $p^{\ast}$-Cauchy sequence in
$\ell_{p}$. Since $S$ is bounded, without loss of generality we can assume
that $S\subseteq B_{\ell_{p}}$. Then $S$ has a subsequence
$\\{x_{n_{k}}\\}_{k\in\omega}$ which weakly $p^{\ast}$-converges to some point
$x\in B_{\ell_{p}}$. Since $S$ is weakly $p^{\ast}$-Cauchy, Lemma 4.5 implies
that also $S$ weakly $p^{\ast}$-converges to $x$. Thus $\ell_{p}$ is weakly
sequentially $p^{\ast}$-complete.

Recall (see Definition 2.1 of [13]) that a Banach space $X$ belongs to the
class $W_{p}$, $p\in[1,\infty]$, if any bounded sequence admits a weakly
$p$-convergent subsequence. We extend this notion as follows.

###### Definition 13.14

Let $p\in[1,\infty]$. A locally convex space $E$ is said to belong to the
class $W_{p}$ if any bounded subset of $E$ is relatively weakly sequentially
$p$-compact.

The next assertion immediately follows from Definitions 13.1 and 13.14.

###### Proposition 13.15

Let $p\in[1,\infty]$, and let a locally convex space $E$ belong to the class
$W_{p}$. If $E$ has the $p$-Schur property, then every bounded subset of $E$
is relatively sequentially compact.

###### Corollary 13.16

Let $1\leq p<\infty$. Then $\ell_{p}\in W_{p}$ if and only if $p\geq 2$.

###### Proof 89

Assume that $1<p<2$. Then $p<p^{\ast}$, and hence, by Proposition 6.5,
$\ell_{p}$ has the $p$-Schur property. Recall also that $\ell_{1}$ has the
Schur property. Since $B_{\ell_{p}}$ is not compact, $\ell_{p}$ does not
belong to $W_{p}$ for every $1\leq p<2$.

Assume that $2\leq p<\infty$. By (ii) of Corollary 13.11, $B_{\ell_{p}}$ is
weakly sequentially $p^{\ast}$-compact. Since $1<p^{\ast}\leq 2$, we have
$p^{\ast}\leq p$. Therefore any weakly $p^{\ast}$-convergent sequence is also
weakly $p$-convergent. Thus $B_{\ell_{p}}$ is weakly sequentially $p$-compact
and hence $\ell_{p}\in W_{p}$.

The following theorem generalizes and extends the corresponding assertion for
Banach spaces, see [12, p. 45] and Proposition 13 of [42].

###### Theorem 13.17

Let $p\in[1,\infty)$. For any locally convex spaces $E$ and $L$ and each weak-
weak sequentially continuous linear map $T:E\to L$, the following assertions
are equivalent:

  1. (i)

$T$ is $p$-convergent;

  2. (ii)

$T\circ S$ is a sequentially precompact operator for any operator
$S:\ell_{p^{\ast}}^{0}\to E$ $($or $S:c_{0}^{0}\to E$ if $p=1$$)$; moreover
the set $T\circ S(B_{\ell_{p^{\ast}}^{0}})$ (or $T\circ S(B_{c_{0}^{0}})$) is
sequentially precompact in $E$.

If in addition $E$ is sequentially complete, then (i) and (ii) are equivalent
to the following

  1. (iii)

$T\circ S$ is a sequentially precompact operator for any operator
$S:\ell_{p^{\ast}}\to E$ $($or $S:c_{0}\to E$ if $p=1$$)$; moreover, the set
$T\circ S(B_{\ell_{p^{\ast}}})$ (or $T\circ S(B_{c_{0}})$) is sequentially
precompact in $E$.

If $1<p<\infty$ and $E$ is sequentially complete, then (i)–(iii) are
equivalent to the following

  1. (iv)

$T\circ S$ is a sequentially compact operator for any operator
$S:\ell_{p^{\ast}}\to E$; moreover, the set $T\circ S(B_{\ell_{p^{\ast}}})$ is
sequentially compact in $E$.

If $p=1$, $E$ and $L$ are sequentially complete, and $E$ is weakly
sequentially $1$-complete, then (i)–(iii) are equivalent to

  1. (v)

$T\circ S$ is a sequentially compact operator for any operator $S:c_{0}\to E$;
moreover, the set $T\circ S(B_{c_{0}})$ is sequentially compact in $E$.

###### Proof 90

(i)$\Rightarrow$(ii) and (i)$\Rightarrow$(iii): Let $\\{x_{n}\\}_{n\in\omega}$
be an arbitrary sequence in $B_{\ell_{p^{\ast}}^{0}}$ (resp., $B_{c_{0}^{0}}$,
$B_{\ell_{p^{\ast}}}$ or $B_{c_{0}}$). By Proposition 13.10, any operator
$S:\ell_{p^{\ast}}^{0}\to E$ (resp., $S:c_{0}^{0}\to E$, $S:\ell_{p^{\ast}}\to
E$ or $S:c_{0}\to E$) is weakly sequentially $p$-precompact. Therefore there
is a subsequence $\\{x_{n_{k}}\\}_{k\in\omega}$ of $\\{x_{n}\\}_{n\in\omega}$
such that for every increasing sequence $(k_{j})$ in $\omega$, the sequence
$\big{\\{}S\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\big{\\}}_{k\in\omega}\subseteq
E$ is weakly $p$-summable. Since $T$ is $p$-convergent it follows that
$TS\big{(}x_{n_{k_{j}}}-x_{n_{k_{j+1}}}\big{)}\to 0$ in the space $L$. As the
sequence $(k_{j})$ was arbitrary, we obtain that the sequence
$\\{TS(x_{n_{k}})\\}$ is Cauchy in $L$. Therefore $TS$ is sequentially
precompact. Since any sequentially precompact subset is bounded, it follows
that the linear map $TS$ is bounded and hence continuous.

(ii)$\Rightarrow$(i) and (iii)$\Rightarrow$(i) follow from Proposition 13.7.

Assume that $1<p<\infty$ and $E$ is sequentially complete.

(i)$\Rightarrow$(iv) Let $\\{x_{n}\\}_{n\in\omega}$ be an arbitrary sequence
in $B_{\ell_{p^{\ast}}}$. Then, by (ii) of Proposition 13.10, there is a
subsequence $\\{x_{n_{k}}\\}_{k\in\omega}$ of $\\{x_{n}\\}_{n\in\omega}$ and
$x\in B_{\ell_{p^{\ast}}}$ such that the sequence
$\big{\\{}S\big{(}x_{n_{k}}-x\big{)}\big{\\}}_{k\in\omega}\subseteq E$ is
weakly $p$-summable. Since $T$ is $p$-convergent it follows that
$TS\big{(}x_{n_{k}}-x\big{)}\to 0$ in $E$, and hence
$TS\big{(}x_{n_{k}}\big{)}\to TS(x)\in TS(B_{\ell_{p^{\ast}}})$. Therefore
$TS(B_{\ell_{p^{\ast}}})$ is a sequentially compact subset of $L$. As above
since any sequentially compact subset is bounded, we obtain that the linear
map $TS$ is bounded and hence continuous. In particular, $TS$ is a
sequentially compact operator.

(iv)$\Rightarrow$(iii) is trivial.

Assume that $p=1$, $E$ and $L$ are sequentially complete, and $E$ is weakly
sequentially $1$-complete.

(i)$\Rightarrow$(v) Let $\\{x_{n}\\}_{n\in\omega}$ be an arbitrary sequence in
$B_{c_{0}}$. Then, by (iii) of Proposition 13.10, there are a subsequence
$\\{x_{n_{k}}\\}_{k\in\omega}$ of $\\{x_{n}\\}_{n\in\omega}$ and $x\in
B_{\ell_{p^{\ast}}}$ such that $S(x_{n_{k}})$ weakly $1$-converges to $S(x)$.
Then the $1$-convergence of $T$ implies $TS\big{(}x_{n_{k}}\big{)}\to TS(x)\in
TS(B_{\ell_{p^{\ast}}})$. Therefore the set $T\circ S(B_{c_{0}})$ is
sequentially compact in $E$, and hence, as above, $TS$ is a sequentially
compact operator.

(v)$\Rightarrow$(iii) is trivial.

Since in angelic spaces the class of relatively sequentially compact sets
coincides with the class of relatively compact sets, Theorem 13.18 implies the
following assertion.

###### Corollary 13.18

Let $p\in[1,\infty)$, $E$ be a sequentially complete locally convex space
which is weakly $1$-complete if $p=1$, and let $L$ be an angelic sequentially
complete locally convex space $($for example $L$ is a strict $(LF)$-space$)$.
Then for each weak-weak sequentially continuous linear map $T:E\to L$, the
following assertions are equivalent:

  1. (i)

$T$ is $p$-convergent;

  2. (ii)

$T\circ S$ is sequentially precompact for any operator
$S:\ell_{p^{\ast}}^{0}\to E$ $($or $S:c_{0}^{0}\to E$ if $p=1$$)$.

  3. (iii)

$T\circ S$ is sequentially precompact for any operator $S:\ell_{p^{\ast}}\to
E$ $($or $S:c_{0}\to E$ if $p=1$$)$.

  4. (iv)

$T\circ S$ is sequentially compact for any operator $S:\ell_{p^{\ast}}\to E$
$($or $S:c_{0}\to E$ if $p=1$$)$.

  5. (v)

$T\circ S$ is compact for any operator $S:\ell_{p^{\ast}}\to E$ $($or
$S:c_{0}\to E$ if $p=1$$)$.

The conclusion of Theorem 13.17 is not true in general for the case $p=\infty$
as the following example shows.

###### Example 13.19

Let $p=\infty$, $E=L=\ell_{1}$, and let $T:E\to L$ and $S:\ell_{1}^{0}\to E$
be the identity maps. Then $T$ is $\infty$-convergent, but $T\circ S$ is not
even weakly sequentially precompact.

###### Proof 91

Every weakly $\infty$-summable sequence is weakly null, and hence the Schur
property of $E$ implies that $T$ is $\infty$-convergent. The Schur property
implies also that the sequence $\\{TS(e_{n})\\}_{n\in\omega}$ does not have a
weakly Cauchy subsequence. Thus $T\circ S$ is not weakly sequentially
precompact.

The condition of being sequentially complete in Theorem 13.17 is also
essential. We denote by
$\mathbf{s}=\\{0\\}\cup\\{\tfrac{1}{n}:n\in\mathbb{N}\\}$ the convergent
sequence with the topology induced from $\mathbb{R}$.

###### Example 13.20

Let $p=1$, $E=L=C_{p}(\mathbf{s})$, $T:E\to L$ be the identity map, and let
$S:c_{0}\to E$ be the identity inclusion defined by
$S(\mathbf{a}):=f_{\mathbf{a}}$, where $\mathbf{a}=(a_{n})_{n\in\omega}\in
c_{0}$ and

$f_{\mathbf{a}}(0)=0\;\mbox{ and
}\;f_{\mathbf{a}}\big{(}\tfrac{1}{n}\big{)}:=a_{n-1}\;\mbox{ for }\;n\geq 1.$

Then $E$ is not sequentially complete, $T$ is $1$-convergent, but $T\circ S$
is not sequentially compact.

###### Proof 92

The space $E=C_{p}(\mathbf{s})$ is not sequentially complete by Theorem 3.15.
The operator $T$ is $1$-convergent by Proposition 13.6. To show that $TS$ is
not sequentially compact, consider the sequence
$\\{\mathbf{a}_{n}\\}_{n\in\omega}$ in $B_{c_{0}}$ where
$\mathbf{a}_{n}:=(\underbrace{1,\dots,1}_{n},0,\dots)$. Then
$TS(\mathbf{a}_{n})=f_{\mathbf{a}_{n}}$ pointwise converges to the function
which is discontinuous at $0$. Therefore the sequence
$\big{\\{}TS(\mathbf{a}_{n})\big{\\}}_{n}$ does not have a convergent
subsequence. Thus $TS(B_{c_{0}})$ is not a relatively sequentially compact
subset of $L$.

Below we show that in general one cannot replace $\ell^{0}_{p^{\ast}}$ in (ii)
of Theorem 13.17 by $\ell_{p^{\ast}}$ without the condition on $E$ being
sequentially complete.

###### Example 13.21

Let $E=L=\ell_{2}^{0}$, $T:E\to L$ be the identity map. Then $E$ is not
sequentially complete, $T$ is not $2$-convergent, but $TS$ is finite-
dimensional for every $S\in\mathcal{L}(\ell_{2},E)$.

###### Proof 93

The normed space $\ell_{p}^{0}$ is not even locally complete because it is not
complete, see Corollary 5.1.9 of [72]. To show that $T$ is not $2$-convergent,
observe that the canonical basis $\\{e_{n}\\}_{n\in\omega}$ of $E$ is weakly
$2$-summable because for every $\chi\in E^{\prime}=\ell_{2}$ we have
$(\langle\chi,e_{n}\rangle)_{n\in\omega}=\chi\in\ell_{2}$. However,
$T(e_{n})=e_{n}\not\to 0$ in $L$. Thus $T$ is not $2$-convergent. The last
assertion follows from the next claim.

Claim 1: For every $p\in[1,\infty)$, each operator
$S\in\mathcal{L}(\ell_{p},\ell_{p}^{0})$ has finite rank.

Indeed, since
$\ell_{p}^{0}=\bigcup_{n\in\omega}\mathsf{span}(e_{0},\dots,e_{n})$, we obtain
$\ell_{p}=\bigcup_{n\in\omega}S^{-1}\big{(}\mathsf{span}(e_{0},\dots,e_{n})\big{)}$.
By the Baire property of $\ell_{p}$, there is $m\in\omega$ such that the
closed subspace $S^{-1}\big{(}\mathsf{span}(e_{0},\dots,e_{m})\big{)}$ of
$\ell_{p}$ has a non-empty interior. Thus
$\ell_{p}=S^{-1}\big{(}\mathsf{span}(e_{0},\dots,e_{m})\big{)}$, and hence
$S(\ell_{p})$ is finite-dimensional.

Now we characterize locally convex spaces with the $p$-Schur property.

###### Proposition 13.22

Let $p\in[1,\infty]$, and let $E$ be a locally convex space. Consider the
following assertions:

  1. (i)

the identity map $\mathsf{id}_{E}:E\to E$ is $p$-convergent, i.e., $E$ has the
$p$-Schur property;

  2. (ii)

every $($relatively$)$ weakly sequentially $p$-compact subset $A$ of $E$ is
$($resp., relatively$)$ sequentially compact in $E$;

  3. (iii)

every weakly sequentially $p$-precompact subset $A$ of $E$ is sequentially
precompact in $E$.

Then (i)$\Leftrightarrow$(ii)$\Rightarrow$(iii). If in addition $E$ is
sequentially complete, then the implication (iii)$\Rightarrow$(i) is also
satisfied.

###### Proof 94

(i)$\Rightarrow$(ii) Assume that $E$ has the $p$-Schur property, and let $A$
be a (relatively) weakly sequentially $p$-compact subset of $E$. Fix an
arbitrary sequence $\\{a_{n}\\}_{n\in\omega}$ in $A$. Then it has a
subsequence $\\{a_{n_{k}}\\}_{k\in\omega}$ weakly $p$-converging to an element
$a\in A$ (resp., $a\in E$). Since $E$ has the $p$-Schur property, we obtain
that $a_{n_{k}}\to a$ in $E$. Thus $A$ is (resp., relatively) sequentially
compact in $E$.

(ii)$\Rightarrow$(i) Assume that every (relatively) weakly sequentially
$p$-compact subset $A$ of $E$ is (resp., relatively) sequentially compact in
$E$. Take an arbitrary weakly $p$-summable sequence
$S=\\{x_{n}\\}_{n\in\omega}$ in $E$. Then $S\cup\\{0\\}$ is a weakly
sequentially $p$-compact set. By assumption, $S$ is (resp., relatively)
sequentially compact in $E$. Now suppose for a contradiction that
$x_{n}\not\to 0$ in $E$. Passing to a subsequence if needed, without loss of
generality we can assume that there is $U\in\mathcal{N}_{0}(E)$ such that
$x_{n}\not\in U$ for every $n\in\omega$. Since $S$ is relatively sequentially
compact in $E$, there is a subsequence $\\{x_{n_{k}}\\}_{k\in\omega}$ of
$\\{x_{n}\\}_{n\in\omega}$ which converges (in $E$) to some element $x$.
Clearly, $x\not=0$ and $x_{n_{k}}\to x$ weakly. On the other hand, since $S$
is weakly null, $\\{x_{n_{k}}\\}_{k\in\omega}$ is also weakly null and hence
it can converge only to zero. Whence $x=0$, a contradiction. Therefore $S$ is
a null-sequence. Thus $E$ has the $p$-Schur property.

(i)$\Rightarrow$(iii) Assume that $E$ has the $p$-Schur property, and let $A$
be a weakly sequentially $p$-precompact subset of $E$. Let
$\\{a_{n}\\}_{n\in\omega}$ be a sequence in $A$. Then it has a weakly
$p$-Cauchy subsequence $\\{a_{n_{k}}\\}_{k\in\omega}$. Therefore for every
sequence $(k_{j})$ in $\omega$, the sequence
$\\{a_{n_{k_{j}}}-a_{n_{k_{j+1}}}\\}_{k\in\omega}$ is weakly $p$-summable. By
the $p$-Schur property, $a_{n_{k_{j}}}-a_{n_{k_{j+1}}}\to 0$ in $E$. But this
means that $\\{a_{n_{k}}\\}_{k\in\omega}$ is a Cauchy sequence in $E$. Thus
$A$ is sequentially precompact.

(iii)$\Rightarrow$(i) Assume that $E$ is sequentially complete and every
weakly sequentially $p$-precompact subset $A$ of $E$ is sequentially
precompact in $E$. Take an arbitrary weakly $p$-summable sequence
$S=\\{x_{n}\\}_{n\in\omega}$ in $E$. Then $S$ is weakly sequentially
$p$-precompact and hence sequentially precompact in $E$. Suppose for a
contradiction that $x_{n}\not\to 0$ in $E$. Then, passing to a subsequence if
needed, we assume that there is an open $U\in\mathcal{N}_{0}(E)$ such that
$x_{n}\not\in U$ for every $n\in\omega$. Since $S$ is sequentially precompact,
there is a Cauchy subsequence $\\{x_{n_{k}}\\}_{k\in\omega}$ of
$\\{x_{n}\\}_{n\in\omega}$. As $E$ is sequentially complete, there is $x\in
E{\setminus}U$ such that $x_{n_{k}}\to x$. Since $S$ is weakly null,
$\\{x_{n_{k}}\\}_{k\in\omega}$ is also weakly null and hence it can converge
only to zero. As $x\not=0$, we obtain a contradiction. Thus $x_{n}\to 0$ in
$E$, and hence $E$ has the $p$-Schur property.

Theorem 13.17 applied to the identity map $T=\mathsf{id}_{E}:E\to E$
immediately implies the following operator characterization of the $p$-Schur
property.

###### Corollary 13.23

Let $p\in[1,\infty)$, and let $E$ be a locally convex space. Then the
following assertions are equivalent:

  1. (i)

the identity operator $\mathsf{id}_{E}:E\to E$ is $p$-convergent, i.e., $E$
has the $p$-Schur property;

  2. (ii)

any operator $S:\ell_{p^{\ast}}^{0}\to E$ $($or $S:c_{0}^{0}\to E$ if $p=1$$)$
is sequentially precompact.

If $E$ is sequentially complete, then (i) and (ii) are equivalent to the
following

  1. (iii)

any operator $S:\ell_{p^{\ast}}\to E$ $($or $S:c_{0}\to E$ if $p=1$$)$ is
sequentially precompact.

If $1<p<\infty$ and $E$ is sequentially complete, then (i)–(iii) are
equivalent to the following

  1. (iv)

any operator $S:\ell_{p^{\ast}}\to E$ is a sequentially compact.

If $p=1$ and $E$ is sequentially complete and weakly sequentially
$1$-complete, then (i)–(iii) are equivalent to

  1. (v)

any operator $S:c_{0}\to E$ is sequentially compact.

###### Remark 13.24

It is natural to ask whether $T^{\ast\ast}$ of a $p$-convergent operator is
also $p$-convergent. A negative answer to this question is given in Theorem
2.7 of [16].

##  14 Pełczyński’s type sets and $p$-convergent operators

Let $E$ and $L$ be locally convex spaces, and let $T:E\to L$ be an operator.
According to Lemma 7.2, if $A$ is a $(p,q)$-$(V^{\ast})$ set then so is its
image $T(A)$, and, by Lemmas 8.2 and 8.6, if $B$ is a $(p,q)$-$(V)$ set or a
weak∗ $(p,q)$-$(V)$ set in $E^{\prime}$, then so is $T^{\ast}(B)$. Taking into
account that $A$ is a bounded subset of $E$ and $B$ is a weak∗ bounded subset
of $E^{\prime}$, these results motivate the following problems:

  1. (1)

Characterize those operators $T$ which map all  bounded sets into
$(p,q)$-$(V^{\ast})$ sets.

  2. (2)

Characterize those operators $T$ such that $T^{\ast}$ maps all  weak∗ bounded
sets or strongly bounded sets into $(p,q)$-$(V)$ sets or weak∗ $(p,q)$-$(V)$
sets.

The purpose of this section is to give complete answers to problems (1) and
(2) for the $(p,\infty)$-case in the general case when $E$ and $L$ are locally
convex spaces. We are interested in this special case also because it is
dually connected with $p$-convergent operators, see in particular Theorems
14.1 and 14.6.

If $1\leq p<\infty$, a characterization of operators $T$ between Banach spaces
for which $T^{\ast}$ is $p$-convergent was obtained by Ghenciu [42]. Below we
generalize this result.

###### Theorem 14.1

Let $p\in[1,\infty]$, and let $T:E\to L$ be an operator between locally convex
spaces $E$ and $L$. Then the following assertions are equivalent:

  1. (i)

for every $B\in\mathsf{Bo}(E)$, the image $T(B)$ is a $p$-$(V^{\ast})$ set in
$L$;

  2. (ii)

$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is $p$-convergent.

If $1\leq p<\infty$, then (i) and (ii) are equivalent to

  1. (iii)

$T^{\ast}\circ S$ is sequentially precompact for any operator
$S:\ell_{p^{\ast}}^{0}\to L^{\prime}_{\beta}$ $($or $S:c_{0}^{0}\to
L^{\prime}_{\beta}$ if $p=1$$)$.

If $1<p<\infty$ and $L^{\prime}_{\beta}$ is sequentially complete, then
(i)-(iii) are equivalent to the following

  1. (iv)

$T^{\ast}\circ S$ is sequentially compact for any operator
$S:\ell_{p^{\ast}}\to L^{\prime}_{\beta}$;

  2. (v)

$T^{\ast}\circ S$ is sequentially precompact for any operator
$S:\ell_{p^{\ast}}\to L^{\prime}_{\beta}$.

If $p=1$, $E^{\prime}_{\beta}$ and $L^{\prime}_{\beta}$ are sequentially
complete and $L^{\prime}_{\beta}$ is weakly sequentially $1$-complete, then
(i)-(iii) are equivalent to the following

  1. (vi)

$T^{\ast}\circ S$ is sequentially compact for any operator $S:c_{0}\to
L^{\prime}_{\beta}$;

  2. (vii)

$T^{\ast}\circ S$ is sequentially precompact for any operator $S:c_{0}\to
L^{\prime}_{\beta}$.

###### Proof 95

(i)$\Rightarrow$(ii) Let $\\{\chi_{n}\\}_{n\in\omega}$ be a weakly
$p$-summable sequence in $L^{\prime}_{\beta}$. We show that
$T^{\ast}(\chi_{n})\to 0$ in $E^{\prime}_{\beta}$. To this end, fix an
arbitrary $B\in\mathsf{Bo}(E)$. Since $T(B)$ is a $p$-$(V^{\ast})$ set in $L$
we have

$\sup_{b\in B}\big{|}\langle T^{\ast}(\chi_{n}),b\rangle\big{|}=\sup_{b\in
B}\big{|}\langle\chi_{n},T(b)\rangle\big{|}\to 0\;\mbox{ as }\;n\to\infty,$

and hence $T^{\ast}(\chi_{n})\in B^{\circ}$ for all sufficiently large
$n\in\omega$. Since $B$ was arbitrary this means that $T^{\ast}(\chi_{n})\to
0$ in $E^{\prime}_{\beta}$, as desired.

(ii)$\Rightarrow$(i) Let $B\in\mathsf{Bo}(E)$. To show that $T(B)$ is a
$p$-$(V^{\ast})$ set in $L$, take any weakly $p$-summable sequence
$\\{\chi_{n}\\}_{n\in\omega}$ in $L^{\prime}_{\beta}$. For every
$\varepsilon>0$, the polar $\varepsilon
B^{\circ}=\big{(}\tfrac{1}{\varepsilon}B\big{)}^{\circ}$ is a neighborhood of
zero in $E^{\prime}_{\beta}$. Since $T^{\ast}$ is $p$-convergent, we have
$T^{\ast}(\chi_{n})\to 0$ in $E^{\prime}_{\beta}$ and hence there is
$N_{\varepsilon}\in\omega$ such that $T^{\ast}(\chi_{n})\in\varepsilon
B^{\circ}$ for all $n\geq N_{\varepsilon}$. Therefore

$\sup_{b\in B}\big{|}\langle\chi_{n},T(b)\rangle\big{|}=\sup_{b\in
B}\big{|}\langle T^{\ast}(\chi_{n}),b\rangle\big{|}\leq\varepsilon\;\mbox{ for
all }\;n\geq N_{\varepsilon}.$

As $\varepsilon$ was arbitrary it follows that $\sup_{b\in
B}\big{|}\langle\chi_{n},T(b)\rangle\big{|}\to 0$. Thus $T(B)$ is a
$p$-$(V^{\ast})$ set.

All other assertions and equivalences follow from Theorem 13.17 applied to
$E_{1}=L^{\prime}_{\beta}$, $L_{1}=E^{\prime}_{\beta}$ and $T_{1}=T^{\ast}$.

Theorem 14.1 applied to the identity map $T=\mathsf{id}_{E}:E\to E$
immediately implies the next corollary in which the equivalence
(i)$\Leftrightarrow$(ii) is proved in Theorem 7.8.

###### Corollary 14.2

Let $p\in[1,\infty]$, and let $E$ be a locally convex space. Then the
following conditions are equivalent

  1. (i)

every bounded subset of $E$ is a $p$-$(V^{\ast})$ set $\big{(}$i.e.,
$\mathsf{Bo}(E)=\mathsf{V}^{\ast}_{p}(E)$$\big{)}$;

  2. (ii)

the identity map $\mathsf{id}_{E^{\prime}}:E^{\prime}_{\beta}\to
E^{\prime}_{\beta}$ is $p$-convergent, i.e., $E^{\prime}_{\beta}$ has the
$p$-Schur property.

If $1\leq p<\infty$, then (i) and (ii) are equivalent to

  1. (iii)

any operator $S:\ell_{p^{\ast}}^{0}\to E^{\prime}_{\beta}$ $($or
$S:c_{0}^{0}\to E^{\prime}_{\beta}$ if $p=1$$)$ is sequentially precompact.

If $1<p<\infty$ and $E^{\prime}_{\beta}$ is sequentially complete, then
(i)-(iii) are equivalent to the following

  1. (iv)

any operator $S:\ell_{p^{\ast}}\to E^{\prime}_{\beta}$ is sequentially
compact;

  2. (v)

any operator $S:\ell_{p^{\ast}}\to E^{\prime}_{\beta}$ is sequentially
precompact.

If $p=1$, $E^{\prime}_{\beta}$ is sequentially complete and weakly
sequentially $1$-complete, then (i)-(iii) are equivalent to the following

  1. (vi)

any operator $S:c_{0}\to E^{\prime}_{\beta}$ is sequentially compact;

  2. (vii)

any operator $S:c_{0}\to E^{\prime}_{\beta}$ is sequentially precompact.

Let $1\leq p\leq q\leq\infty$, and let $E$ be a locally convex space.
According to Lemma 7.2 the family $\mathsf{V}^{\ast}_{(p,q)}(E)$ of all
$(p,q)$-$(V^{\ast})$ sets in $E$ is a saturated bornology containing all
finite subsets of $E$. Therefore one can naturally define:

###### Definition 14.3

Let $1\leq p\leq q\leq\infty$, and let $E$ be a locally convex space. Denote
by $V^{\ast}_{(p,q)}(E^{\prime},E)$ the polar topology on $E^{\prime}$ of
uniform convergence on $(p,q)$-$(V^{\ast})$ sets. Set
$V^{\ast}_{p}(E^{\prime},E)=V^{\ast}_{(p,\infty)}(E^{\prime},E)$ and
$V^{\ast}(E^{\prime},E)=V^{\ast}_{(\infty,\infty)}(E^{\prime},E)$.

Since the family $\mathsf{V}^{\ast}_{(p,q)}(E)$ depends only on the duality,
the topology $V^{\ast}_{(p,q)}(E^{\prime},E)$ is a dual topology. For further
references we select the next simple lemma.

###### Proposition 14.4

Let $1\leq p\leq q\leq\infty$, and let $E$ be a locally convex space. Then:

  1. (i)

$\sigma(E^{\prime},E)\subseteq
V^{\ast}_{(p,q)}(E^{\prime},E)\subseteq\beta(E^{\prime},E)$,

  2. (ii)

$V^{\ast}_{(p,q)}(E^{\prime},E)\subseteq\mu(E^{\prime},E)$ if and only if
every $(p,q)$-$(V^{\ast})$ set $A$ in $E$ is relatively weakly compact.
Consequently, $E$ has the property $V^{\ast}_{(p,q)}$ if and only if the
topology $V^{\ast}_{(p,q)}(E^{\prime},E)$ is compatible with the duality
$(E,E^{\prime})$.

###### Proof 96

(i) is clear, and (ii) follows from the Mackey–Arens theorem and the fact that
$\mathsf{V}^{\ast}_{(p,q)}(E)$ is a saturated bornology.

The next result naturally complements Theorem 14.1.

###### Proposition 14.5

Let $1\leq p\leq q\leq\infty$, and let $T:E\to L$ be an operator between
locally convex spaces $E$ and $L$. Then:

  1. (i)

the adjoint map
$T^{\ast}:\big{(}L^{\prime},V^{\ast}_{(p,q)}(L^{\prime},L)\big{)}\to\big{(}E^{\prime},V^{\ast}_{(p,q)}(E^{\prime},E)\big{)}$
is continuous;

  2. (ii)

the adjoint map
$T^{\ast}:L^{\prime}_{\beta}\to\big{(}E^{\prime},V^{\ast}_{p}(E^{\prime},E)\big{)}$
is $p$-convergent.

###### Proof 97

(i) To show that $T^{\ast}$ is continuous, let $A^{\circ}$ be a standard
$V^{\ast}_{(p,q)}(E^{\prime},E)$-neighborhood of zero, where $A$ is a
$(p,q)$-$(V^{\ast})$ set in $E$. By Lemma 7.2, $T(A)$ is a
$(p,q)$-$(V^{\ast})$ set in $L$. Then for every $\eta\in T(A)^{\circ}$ and
each $a\in A$, we have $|\langle
T^{\ast}(\eta),a\rangle|=|\langle\eta,T(a)\rangle|\leq 1$ and hence
$T^{\ast}\big{(}T(A)^{\circ}\big{)}\subseteq A^{\circ}$. Thus $T^{\ast}$ is
continuous.

(ii) Let $\\{\chi_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in
$L^{\prime}_{\beta}$. To show that $T^{\ast}(\chi_{n})\to 0$ in
$\big{(}E^{\prime},V^{\ast}_{p}(E^{\prime},E)\big{)}$, fix an arbitrary
$B\in\mathsf{V}^{\ast}_{p}(E)$. By Lemma 7.2, $T(B)$ is a $p$-$(V^{\ast})$ set
in $L$ and hence

$\lim_{n\to\infty}\sup_{b\in B}\big{|}\langle
T^{\ast}(\chi_{n}),b\rangle\big{|}=\lim_{n\to\infty}\sup_{b\in
B}\big{|}\langle\chi_{n},T(b)\rangle\big{|}=0.$

Therefore $T^{\ast}(\chi_{n})\in B^{\circ}$ for all sufficiently large
$n\in\omega$. Since $B$ was arbitrary this means that $T^{\ast}(\chi_{n})\to
0$ in $V^{\ast}_{p}(E^{\prime},E)$, as desired.

Below we give sufficient and necessary conditions on operators to be
$p$-convergent.

###### Theorem 14.6

Let $p\in[1,\infty]$, $E$ and $L$ be locally convex spaces, and let
$T\in\mathcal{L}(E,L)$.

  1. (i)

If $T^{\ast}(B)$ is a $p$-$(V)$ set in $E^{\prime}$ for every
$B\in\mathsf{Bo}(L^{\prime}_{\beta})$, then $T$ is $p$-convergent. The
converse assertion is true if $L$ is quasibarrelled.

  2. (ii)

If $T^{\ast}(B)$ is a $p$-$(V)$ set in $E^{\prime}$ for every
$B\in\mathsf{Bo}(L^{\prime}_{w^{\ast}})$, then $T$ is $p$-convergent. The
converse assertion is true if $L$ is barrelled.

###### Proof 98

Assume that $T^{\ast}(B)$ is a $p$-$(V)$ set in $E^{\prime}$ for every
$B\in\mathsf{Bo}(L^{\prime}_{\beta})$ (resp.,
$B\in\mathsf{Bo}(L^{\prime}_{w^{\ast}})$). Fix an arbitrary weakly
$p$-summable sequence $\\{x_{n}\\}_{n\in\omega}$ in $E$. We have to show that
$T(x_{n})\to 0$ in $L$. To this end, fix $U\in\mathcal{N}_{0}^{c}(L)$. Then
the polar $B:=U^{\circ}$ is strongly (hence weak∗) bounded by Theorem 11.3.5
of [67], and hence $T^{\ast}(B)$ is a $p$-$(V)$ set in $E^{\prime}$.
Consequently, by the definition of $p$-$(V)$ sets, we have

$\lim_{n\to\infty}\sup_{\chi\in
B}|\langle\chi,T(x_{n})\rangle|=\lim_{n\to\infty}\sup_{\chi\in B}|\langle
T^{\ast}(\chi),x_{n}\rangle|=0.$

In particular, for all sufficiently large $n\in\omega$, it follows that
$|\langle\chi,T(x_{n})\rangle|\leq 1$ for every $\chi\in B=U^{\circ}$. This
means that $T(x_{n})\in U^{\circ\circ}=U$ for all sufficiently large
$n\in\omega$. Thus $T(x_{n})\to 0$ in $L$ and hence $T$ is $p$-convergent.

Conversely, assume additionally that $L$ is quasibarrelled (resp., barrelled)
and let $T$ be $p$-convergent. Let $B\in\mathsf{Bo}(L^{\prime}_{\beta})$
(resp., $B\in\mathsf{Bo}(L^{\prime}_{w^{\ast}})$). We have to show that

$\lim_{n\to\infty}\sup_{\chi\in B}|\langle
T^{\ast}(\chi),x_{n}\rangle|=\lim_{n\to\infty}\sup_{\chi\in
B}|\langle\chi,T(x_{n})\rangle|=0$

for every weakly $p$-summable sequence $\\{x_{n}\\}_{n\in\omega}$ in $E$. To
this end, fix $\varepsilon>0$. Since $T$ is $p$-convergent, $T(x_{n})\to 0$ in
$L$. As $L$ is quasibarrelled (resp., barrelled) the set $B$ is
equicontinuous. Take $U\in\mathcal{N}_{0}(L)$ such that $B\subseteq
U^{\circ}$. Choose $N_{\varepsilon}\in\omega$ such that
$T(x_{n})\in\varepsilon U$ for every $n\geq N_{\varepsilon}$. Then for every
$n\geq N_{\varepsilon}$, we obtain

$\sup_{\chi\in B}|\langle\chi,T(x_{n})\rangle|\leq\sup\big{\\{}\sup_{\chi\in
B}|\langle\chi,y\rangle|:y\in\varepsilon U\big{\\}}\leq\varepsilon.$

Thus $\sup_{\chi\in B}|\langle T^{\ast}(\chi),x_{n}\rangle|\to 0$, as desired.

###### Corollary 14.7

Let $p\in[1,\infty]$, $E$ be a locally convex space with the property $V_{p}$
(resp., the property $sV_{p}$), and let $L$ be a quasibarrelled space. If an
operator $T:E\to L$ is $p$-convergent, then the adjoint
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ maps bounded sets of
$L^{\prime}_{\beta}$ into relatively weakly (resp., sequentially) compact sets
of $E^{\prime}_{\beta}$. Consequently, if $L$ is a normed space and $T$ is
$p$-convergent, then $T^{\ast}$ is weakly compact (resp., weakly sequentially
compact).

Now we summarize characterizations of the $p$-Schur property of an lcs $E$ and
its strong dual $E^{\prime}_{\beta}$.

###### Corollary 14.8

Let $p\in[1,\infty]$, and let $E$ be a locally convex space. Then:

  1. (i)

if $E$ is quasibarrelled, then
$\mathsf{Bo}(E^{\prime}_{\beta})\subseteq\mathsf{V}_{p}(E^{\prime})$ if and
only if $E$ has the $p$-Schur property;

  2. (ii)

if $E$ is barrelled, then
$\mathsf{Bo}(E^{\prime}_{w^{\ast}})=\mathsf{V}_{p}(E^{\prime})$ if and only
$E$ has the $p$-Schur property;

  3. (iii)

$\mathsf{Bo}(E)=\mathsf{V}_{p}^{\ast}(E)$ if and only if $E^{\prime}_{\beta}$
has the $p$-Schur property.

###### Proof 99

(i) follows from Theorem 14.6 applied to the identity map
$\mathsf{id}_{E}:E\to E$.

(ii) By Theorem 14.6 applied to the identity map $\mathsf{id}_{E}:E\to E$, we
obtain that
$\mathsf{Bo}(E^{\prime}_{w^{\ast}})\subseteq\mathsf{V}_{p}(E^{\prime})$ if and
only $E$ has the $p$-Schur property. It remains to note that
$\mathsf{V}_{p}(E^{\prime})\subseteq\mathsf{Bo}(E^{\prime}_{w^{\ast}})$ always
by (ii) of Lemma 8.2.

(iii) is Theorem 7.8.

Below we give a useful construction of $p$-convergent operators from
$\ell_{1}^{0}$ into locally convex spaces.

###### Proposition 14.9

Let $p\in[1,\infty]$, $\\{x_{n}\\}_{n\in\omega}$ be a bounded sequence in a
locally convex space $(E,\tau)$, and let $T:\ell_{1}^{0}\to E$ be a linear map
defined by

$T(a_{0}e_{0}+\cdots+a_{n}e_{n}):=a_{0}x_{0}+\cdots+a_{n}x_{n}\quad(n\in\omega,\;a_{0},\dots,a_{n}\in\mathbb{F}),$

where $\\{e_{n}\\}$ is the canonical unit basis of $\ell_{1}^{0}$. Then $T$ is
continuous. Moreover, if $E$ is locally complete, then $T$ can be extended to
a continuous operator from $\ell_{1}$ to $E$. In any case, if
$\\{x_{n}\\}_{n\in\omega}$ is a $p$-$(V^{\ast})$ set, then
$T^{\ast}:E^{\prime}_{\beta}\to\ell_{\infty}$ is $p$-convergent.

###### Proof 100

Denote by $A$ the closed absolutely convex hull of $\\{x_{n}\\}_{n\in\omega}$.
Then $A$ is a bounded subset of $E$ (in the cases when
$\\{x_{n}\\}_{n\in\omega}$ is a $p$-$(V^{\ast})$ set, Lemma 7.2 implies that
$A$ is also a $p$-$(V^{\ast})$ set). Consider the normed space
$(E_{A},\|\cdot\|)$ and observe that $\|x_{n}\|\leq 1$ for every $n\in\omega$.
Therefore, if $|a_{0}|+\cdots+|a_{n}|\leq 1$ we obtain

$\big{\|}T(a_{0}e_{0}+\cdots+a_{n}e_{n})\big{\|}=\|a_{0}x_{0}+\cdots+a_{n}x_{n}\|\leq
1$

and hence $T:\ell_{1}^{0}\to(E_{A},\|\cdot\|)$ is continuous. In the case when
$E$ is locally complete, by Proposition 3.6, $(E_{A},\|\cdot\|)$ is a Banach
space and hence $T$ can be extended to a continuous operator from $\ell_{1}$
to $E_{A}$. Since, by Proposition 3.6, the norm topology on $E_{A}$ is finer
than the original topology $\tau{\restriction}_{E_{A}}$ it follows that
$T:\ell_{1}^{0}\to E$ (or $T:\ell_{1}\to E$ if $E$ is locally complete) is
continuous.

Assume now that $\\{x_{n}\\}_{n\in\omega}$ is a $p$-$(V^{\ast})$ set. To show
that the adjoint operator $T^{\ast}$ is $p$-convergent, let
$\\{\chi_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in
$E^{\prime}_{\beta}$. Since $A$ is a $p$-$(V^{\ast})$ set, by definition, we
have $\lim_{n\to\infty}\sup_{x\in A}|\langle\chi_{n},x\rangle|=0$. Therefore

$\|T^{\ast}(\chi_{n})\|_{\ell_{\infty}}=\sup_{k\in\omega}\big{|}\langle
T^{\ast}(\chi_{n}),e_{k}\rangle\big{|}=\sup_{k\in\omega}\big{|}\langle\chi_{n},x_{k}\rangle\big{|}\leq\sup_{x\in
A}\big{|}\langle\chi_{n},x\rangle\big{|}\to 0\;\;\mbox{ as }\;n\to\infty.$

Thus $\big{\\{}T^{\ast}(\chi_{n})\big{\\}}_{n\in\omega}$ is a null sequence in
$\ell_{\infty}$, and hence $T^{\ast}$ is $p$-convergent.

We know that $p$-$(V^{\ast})$ sets are bounded. Below we give an operator
characterization of those spaces $E$ in which the $p$-$(V^{\ast})$ sets have a
stronger topological property than just being bounded as, for example, being
weakly sequentially (pre)compact. The following theorem extends Theorem 15 of
[42].

###### Theorem 14.10

Let $p\in[1,\infty]$, and let $(E,\tau)$ be a locally convex space. Then the
following assertions are equivalent:

  1. (i)

for every normed space $Y$, if $T\in\mathcal{L}(Y,E)$ is such that the adjoint
operator $T^{\ast}:E^{\prime}_{\beta}\to Y^{\prime}_{\beta}$ is
$p$-convergent, then $T$ is weakly sequentially precompact $($resp., weakly
sequentially compact, sequentially precompact or sequentially compact$)$;

  2. (ii)

the same as (i) with $Y=\ell_{1}^{0}$;

  3. (iii)

every $p$-$(V^{\ast})$ subset of $E$ is weakly sequentially precompact
$($resp., relatively weakly sequentially compact, sequentially precompact or
relatively sequentially compact$)$.

If additionally $E$ is locally complete, then (i)-(iii) are equivalent to

  1. (iv)

the same as (i) with $Y=\ell_{1}$.

###### Proof 101

(i)$\Rightarrow$(ii) is clear.

(ii)$\Rightarrow$(iii) Let $A$ be a $p$-$(V^{\ast})$ subset of $E$. To show
that $A$ is weakly sequentially precompact (resp., relatively weakly
sequentially compact, sequentially precompact or relatively sequentially
compact), take an arbitrary sequence $\\{x_{n}\\}_{n\in\omega}$ in $A$ and
define $T:\ell_{1}^{0}\to E$ by

$T(a_{0}e_{0}+\cdots+a_{n}e_{n}):=a_{0}x_{0}+\cdots+a_{n}x_{n}\quad(n\in\omega,\;a_{0},\dots,a_{n}\in\mathbb{F}).$

Then, by Proposition 14.9, $T$ is continuous and such that
$T^{\ast}:E^{\prime}_{\beta}\to\ell_{\infty}$ is $p$-convergent.

By the assumption of (ii), we obtain that $T$ is weakly sequentially
precompact (resp., weakly sequentially compact, sequentially precompact or
sequentially compact). By definition this means that the sequence
$\\{x_{n}\\}=\big{\\{}T(e_{n})\big{\\}}$ has a subsequence which is weakly
Cauchy (resp., weakly convergent in $E$, Cauchy in $E$ or convergent in $E$).
Thus $A$ is a weakly sequentially precompact (resp., relatively weakly
sequentially compact, sequentially precompact or relatively sequentially
compact) subset of $L$, as desired.

(iii)$\Rightarrow$(i) To prove this implication it suffices to show that
$T(B_{Y})$ is weakly sequentially precompact (resp., relatively weakly
sequentially compact, sequentially precompact or relatively sequentially
compact) in $E$ under the assumption that $T^{\ast}$ is $p$-convergent. To
this end, we note that, by the equivalence (i)$\Leftrightarrow$(ii) of Theorem
14.1, the $p$-convergence of $T^{\ast}$ and the boundedness of $B_{Y}$ imply
that the set $T(B_{Y})$ is a $p$-$(V^{\ast})$ set, and hence, by (iii),
$T(B_{Y})$ is weakly sequentially precompact (resp., relatively weakly
sequentially compact, sequentially precompact or relatively sequentially
compact), as desired.

Assume now that $E$ is locally complete. Then the implication
(i)$\Rightarrow$(iv) is evident. To prove the implication
(iv)$\Rightarrow$(ii), let $T\in\mathcal{L}(\ell_{1}^{0},E)$ be such that
$T^{\ast}:E^{\prime}_{\beta}\to\ell_{\infty}$ is $p$-convergent. Then
$\\{T(e_{n})\\}_{n\in\omega}$ is a bounded subset of $E$, and hence, by
Proposition 3.6, the space $E_{B}$ is Banach, where $B$ is the closed
absolutely convex hull of $\\{T(e_{n})\\}_{n\in\omega}$. Since
$T\big{(}B_{\ell_{1}^{0}}\big{)}\subseteq B_{E_{B}}$ the operator
$T:\ell_{1}^{0}\to E_{B}$ is continuous. As $E_{B}$ is complete, $T$ can be
extended to an operator $\bar{T}$ from $\ell_{1}$ to $E_{B}$. Since, by
Proposition 3.6, the topology of $E_{B}$ is stronger than the original
topology on $E_{B}$ it follows that $\bar{T}$ belongs to
$\mathcal{L}(\ell_{1},E)$. Observe that $(\bar{T})^{\ast}=T^{\ast}$ because
$\ell_{1}^{0}$ is dense in $\ell_{1}$, and hence $(\bar{T})^{\ast}$ is
$p$-convergent. Therefore $\bar{T}$ is weakly sequentially precompact (resp.,
weakly sequentially compact, sequentially precompact or sequentially compact),
and hence so is the operator $T$.

For the weakly sequentially precompact case and the weakly sequentially
compact case Theorem 14.10 can be reformulated as an operator characterization
of the property $wsV^{\ast}_{p}$ and the property $sV^{\ast}_{p}$,
respectively.

###### Theorem 14.11

Let $p\in[1,\infty]$, and let $(E,\tau)$ be a locally convex space. Then the
following assertions are equivalent:

  1. (i)

for every normed space $Y$, if $T\in\mathcal{L}(Y,E)$ is such that the adjoint
operator $T^{\ast}:E^{\prime}_{\beta}\to Y^{\prime}_{\beta}$ is
$p$-convergent, then $T$ is weakly sequentially precompact $($resp., weakly
sequentially compact$)$;

  2. (ii)

the same as (i) with $Y=\ell_{1}^{0}$;

  3. (iii)

the space $E$ has the property $wsV^{\ast}_{p}$ $($resp., the property
$sV^{\ast}_{p}$$)$.

If additionally $E$ is locally complete, then (i)-(iii) are equivalent to

  1. (iv)

the same as (i) with $Y=\ell_{1}$.

The following proposition complements Theorem 14.10.

###### Proposition 14.12

Let $p\in[1,\infty]$, and let $E$ be a locally convex space. Then the
following assertions are equivalent:

  1. (i)

each relatively weakly sequentially $p$-compact set in $E$ is a
$p$-$(V^{\ast})$ set;

  2. (ii)

if $\\{x_{n}\\}_{n\in\omega}\subseteq E$ is weakly $p$-summable and
$\\{\chi_{n}\\}_{n\in\omega}$ is weakly $p$-summable in $E^{\prime}_{\beta}$,
then

$\lim_{n\to\infty}\langle\chi_{n},x_{n}\rangle=0.$

###### Proof 102

(i)$\Rightarrow$(ii) The sequence $\\{x_{n}\\}_{n\in\omega}$ is trivially
weakly sequentially $p$-compact. Therefore, by (i), $\\{x_{n}\\}_{n\in\omega}$
is a $p$-$(V^{\ast})$ set. Therefore

$\lim_{n\to\infty}|\langle\chi_{n},x_{n}\rangle|\leq\lim_{n\to\infty}\sup_{k\in\omega}|\langle\chi_{n},x_{k}\rangle|=0.$

(ii)$\Rightarrow$(i) Let $A$ be a relatively weakly sequentially $p$-compact
set in $E$, and suppose for a contradiction that $A$ is not a $p$-$(V^{\ast})$
set. Then there is a weakly $p$-summable sequence
$\\{\chi_{n}\\}_{n\in\omega}$ in $E^{\prime}_{\beta}$ such that
$\lim_{n\to\infty}\sup_{x\in A}|\langle\chi_{n},x\rangle|\not=0$. Passing to a
subsequence if needed, we can assume that there are $a>0$ and a sequence
$\\{x_{n}\\}_{n\in\omega}$ in $A$ such that

$|\langle\chi_{n},x_{n}\rangle|\geq a\quad\mbox{ for every }n\in\omega.$

Since $A$ is relatively weakly sequentially $p$-compact and once more passing
to a subsequence, we can assume that $\\{x_{n}\\}_{n\in\omega}$ weakly
$p$-converges to some point $x\in E$. Therefore $\\{x_{n}-x\\}_{n\in\omega}$
is weakly $p$-summable. As $\\{\chi_{n}\\}_{n\in\omega}$ is weakly
$p$-summable, it follows $\lim_{n\in\omega}\langle\chi_{n},x\rangle=0$.
Therefore, by (ii), we obtain

$a\leq|\langle\chi_{n},x_{n}\rangle|\leq|\langle\chi_{n},x_{n}-x\rangle|+|\langle\chi_{n},x\rangle|\to
0,$

a contradiction.

Let $E$ and $Y$ be locally convex spaces, and let $T:Y\to E$ be an operator.
If $B$ is a $p$-$(V)$ set in $E^{\prime}$, then by (iv) of Lemma 8.2, its
image $T^{\ast}(B)$ is a $p$-$(V)$ set in $Y^{\prime}$. Analogously to
Theorems 14.1 and 14.6 it is natural to describe some classes of operators $T$
for which the image $T^{\ast}(B)$ has some additional (topological) properties
as, for example, being a relatively (weakly) sequentially [pre]compact set.
Below we obtain some necessary and sufficient conditions.

We start from the following construction of $p$-convergent operators into
$\ell_{\infty}$ or $c_{0}$.

###### Lemma 14.13

Let $E$ be a locally convex space, $\\{\chi_{n}\\}_{n\in\omega}\subseteq
E^{\prime}$ be a weak∗ bounded $($resp., weak∗ null$)$ sequence, and let
$S:E\to\ell_{\infty}$ $($resp., $S:E\to c_{0}$$)$ be a linear map defined by

$S(x):=\big{(}\langle\chi_{n},x\rangle\big{)}_{n\in\omega}\quad(x\in E).$

Then:

  1. (i)

$S$ is continuous if and only if $\\{\chi_{n}\\}_{n\in\omega}$ is
equicontinuous.

  2. (ii)

if $p\in[1,\infty]$, then $S$ is $p$-convergent if and only if
$\\{\chi_{n}\\}_{n\in\omega}$ is a $p$-$(V)$ set.

###### Proof 103

(i) Assume that $S$ is continuous. Then for every $\varepsilon>0$, there is
$V\in\mathcal{N}_{0}(E)$ such that $S(V)\subseteq\varepsilon
B_{\ell_{\infty}}$ (resp., $S(V)\subseteq\varepsilon B_{c_{0}}$). This means
that $|\langle\chi_{n},x\rangle|\leq\varepsilon$ for every $n\in\omega$ and
each $x\in V$. Thus $\\{\chi_{n}\\}_{n\in\omega}$ is equicontinuous.
Conversely, assume that $\\{\chi_{n}\\}_{n\in\omega}$ is equicontinuous. Then
there is $U\in\mathcal{N}_{0}(E)$ such that
$\\{\chi_{n}\\}_{n\in\omega}\subseteq U^{\circ}$. Then $S(U)\subseteq
B_{\ell_{\infty}}$ (resp., $S(U)\subseteq B_{c_{0}}$), and hence $S$ is
continuous.

(ii) Assume that $S$ is $p$-convergent. To show that the sequence
$\\{\chi_{n}\\}_{n\in\omega}$ is a $p$-$(V)$ set, fix a weakly $p$-summable
sequence $\\{x_{n}\\}_{n\in\omega}$ in $E$. Since $S$ is $p$-convergent,
$S(x_{n})\to 0$ in $\ell_{\infty}$. Therefore

$\sup_{n\in\omega}\big{|}\langle\chi_{n},x_{k}\rangle\big{|}=\big{\|}S(x_{k})\big{\|}_{\ell_{\infty}}\to
0\quad\mbox{ as }\;\;k\to\infty,$

and hence $\\{\chi_{n}\\}_{n\in\omega}$ is a $p$-$(V)$ set. Conversely, assume
that $\\{\chi_{n}\\}_{n\in\omega}$ is a $p$-$(V)$ set. To show that $S$ is
$p$-convergent, let $\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-summable
sequence in $E$. Then, by the definition of $p$-$(V)$ sets, we have

$\lim_{k\to\infty}\big{\|}S(x_{k})\big{\|}_{\ell_{\infty}}=\lim_{k\to\infty}\sup_{n\in\omega}\big{|}\langle\chi_{n},x_{k}\rangle\big{|}=0$

and hence $S$ is $p$-convergent.

Now we consider an important case when $Y$ is a normed space.

###### Proposition 14.14

Let $p\in[1,\infty]$, $E$ be an $\ell_{\infty}$-$V_{p}$-barrelled space, and
let $T:Y\to E$ be a weakly sequentially $p$-precompact operator from a normed
space $Y$ to $E$. Then for every $p$-$(V)$ set $B$ in $E^{\prime}$, its image
$T^{\ast}(B)$ is relatively (sequentially) compact in the Banach space
$Y^{\prime}_{\beta}$.

###### Proof 104

To prove that $T^{\ast}(B)$ is relatively compact in the Banach space
$Y^{\prime}_{\beta}$, we have to show that for every sequence
$\\{\chi_{n}\\}_{n\in\omega}$ in $B$, $\\{T^{\ast}(\chi_{n})\\}_{n\in\omega}$
has a convergent (in $Y^{\prime}_{\beta}$) subsequence. Since $E$ is
$\ell_{\infty}$-$V_{p}$-barrelled and $\\{\chi_{n}\\}_{n\in\omega}$ is a
$p$-$(V)$ set, the sequence $\\{\chi_{n}\\}_{n\in\omega}$ is equicontinuous.
Therefore, by Lemma 14.13, the linear map $S:E\to\ell_{\infty}$ defined by
$S(x):=\big{(}\langle\chi_{n},x\rangle\big{)}_{n\in\infty}$ is a
$p$-convergent operator.

We show that the operator $ST:Y\to\ell_{\infty}$ is compact. Indeed, let
$\\{x_{n}\\}_{n\in\omega}\subseteq B_{Y}$. Since $T$ is weakly sequentially
$p$-precompact, the sequence $\big{\\{}T(x_{n})\big{\\}}$ contains a weakly
$p$-Cauchy subsequence $\big{\\{}T(x_{n_{k}})\big{\\}}_{k}$. As $S$ is
$p$-convergent, Lemma 13.4 implies that $ST(x_{n_{k}})$ converges to some
element of $\ell_{\infty}$. Thus $ST(B_{Y})$ has compact closure in
$\ell_{\infty}$, and hence $ST$ is compact.

By the Schauder theorem (for its formulation, see Theorem 15.4 below), the
adjoint map $T^{\ast}S^{\ast}$ is also compact. Now, if $\\{e^{\ast}_{n}\\}$
is the standard unit basis of $\ell_{1}\subseteq(\ell_{\infty})^{\prime}$,
then

$\langle S^{\ast}(e_{n}^{\ast}),x\rangle=\langle
e_{n}^{\ast},S(x)\rangle=\langle\chi_{n},x\rangle\quad\mbox{ for every }\;x\in
E,$

and hence $S^{\ast}(e_{n}^{\ast})=\chi_{n}$. Therefore the sequence
$\big{\\{}T^{\ast}(\chi_{n})\big{\\}}_{n\in\omega}=\big{\\{}T^{\ast}S^{\ast}(e^{\ast}_{n})\big{\\}}_{n\in\omega}$
has a convergent subsequence, as desired.

###### Corollary 14.15

Let $p\in[1,\infty]$, $E$ be a locally convex space such that
$E^{\prime}_{\beta}$ is an $\ell_{\infty}$-$V_{p}$-barrelled space, and let
$T:E\to Y$ be an operator into a Banach space $Y$ whose adjoint operator
$T^{\ast}:Y^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is weakly sequentially
$p$-precompact. Then for every $p$-$(V^{\ast})$ set $A\subseteq E$, its image
$T(A)$ is relatively compact in $Y$.

###### Proof 105

Since $A$ is a $p$-$(V^{\ast})$ set, the set $J_{E}(A)$ is a $p$-$(V)$ set in
$E^{\prime\prime}$ (recall that $J_{E}:E\to E^{\prime\prime}$ denotes the
canonical inclusion). Therefore, by Proposition 14.14,
$T^{\ast\ast}\big{(}J_{E}(A)\big{)}$ is relatively compact in the Banach space
$Y^{\prime\prime}$. Since $J_{Y}$ is an embedding of $Y$ onto the closed
subspace $J_{Y}(Y)$ of $Y^{\prime\prime}$ and
$T^{\ast\ast}\big{(}J_{E}(A)\big{)}=J_{Y}\big{(}T(A)\big{)}\subseteq
J_{Y}(Y)$, we obtain that $J_{Y}\big{(}T(A)\big{)}$ is relatively compact in
$J_{Y}(Y)$. Thus $T(A)$ is relatively compact in $Y$.

Below we apply the obtained results to give necessary and sufficient
conditions for subsets of $E$ and $E^{\prime}$ to be a $p$-$(V^{\ast})$ set or
a $p$-$(V)$ set, respectively. The next result generalizes Corollary 20 of
[42].

###### Theorem 14.16

Let $1<p<\infty$, and let $E$ be a quasibarrelled space such that
$E^{\prime}_{\beta}$ is an $\ell_{\infty}$-$V_{p}$-barrelled space. Then for a
bounded subset $A$ of $E$ the following assertions are equivalent:

  1. (i)

$A$ is a $p$-$(V^{\ast})$ set;

  2. (ii)

$T(A)$ is relatively compact whenever $Y$ is a Banach space and $T:E\to Y$ is
an operator whose adjoint operator $T^{\ast}:Y^{\prime}_{\beta}\to
E^{\prime}_{\beta}$ is weakly sequentially $p$-precompact;

  3. (iii)

$T(A)$ is relatively compact in $\ell_{p}$ for every operator
$T:E\to\ell_{p}$.

###### Proof 106

(i)$\Rightarrow$(ii) follows from Corollary 14.15.

(ii)$\Rightarrow$(iii) Let $T\in\mathcal{L}(E,\ell_{p})$. Then
$T^{\ast}\in\mathcal{L}(\ell_{p^{\ast}},E^{\prime}_{\beta})$. By (ii) of
Corollary 13.11, the identity operator $\mathsf{id}_{\ell_{p^{\ast}}}$ of
$\ell_{p^{\ast}}$ is weakly sequentially $p$-compact. Hence the operator
$T^{\ast}=T^{\ast}\circ\mathsf{id}_{\ell_{p^{\ast}}}$ is weakly sequentially
$p$-precompact. Therefore, by (ii), $T(A)$ is relatively compact in
$\ell_{p}$.

(iii)$\Rightarrow$(i) Let $\\{\chi_{n}\\}_{n\in\omega}$ be a weakly
$p$-summable sequence in $E^{\prime}_{\beta}$. Since $E^{\prime}_{\beta}$ is
quasi-complete by Proposition 11.2.3 of [52], we apply Proposition 4.14 to
find an operator $S\in\mathcal{L}(\ell_{p^{\ast}},E^{\prime}_{\beta})$ such
that $S(e_{n}^{\ast})=\chi_{n}$ for every $n\in\omega$. Set $T:=S^{\ast}\circ
J_{E}$, where $J_{E}:E\to E^{\prime\prime}$ is the canonical map. Since $E$ is
quasibarrelled, $J_{E}$ is continuous, and hence $T$ is an operator from $E$
into $\ell_{p}$. By assumption of (iii), we obtain that $T(A)$ is relatively
compact in $\ell_{p}$. In particular, by Proposition 2.10, we have

$\sup_{a\in A}|\langle\chi_{n},a\rangle|=\sup_{a\in A}|\langle
J_{E}(a),S(e_{n}^{\ast})\rangle|=\sup_{a\in A}|\langle
T(a),e_{n}^{\ast}\rangle|\to 0\;\mbox{ as }n\to\infty.$

Thus $A$ is a $p$-$(V^{\ast})$ set.

The case $p=1$ was considered by Emmanuele in [25]. For Banach spaces the next
theorem was proved in Theorem 1.1 of [25].

###### Theorem 14.17

Let $E$ be a barrelled space. Then a bounded subset $A$ of $E$ is a
$(V^{\ast})$ set if and only if $T(A)$ is relatively compact in $\ell_{1}$ for
every operator $T:E\to\ell_{1}$.

###### Proof 107

Assume that $A$ is a $(V^{\ast})$ set. Then, by (iv) of Lemma 7.2, $T(A)$ is a
$(V^{\ast})$ subset of $\ell_{1}$. Since, by Proposition 9 of [71], $\ell_{1}$
has the property $V^{\ast}$ it follows that $T(A)$ is relatively weakly
compact. Therefore, by the Schur property, $T(A)$ is a relatively compact
subset of $\ell_{1}$. Conversely, assume that $T(A)$ is relatively compact in
$\ell_{1}$ for every operator $T:E\to\ell_{1}$. Let
$\\{\chi_{n}\\}_{n\in\omega}$ be an arbitrary weakly $1$-summable sequence in
$E^{\prime}_{\beta}$. Then, by Proposition 4.19, the linear map
$T:E\to\ell_{p}$ defined by $T(x):=(\langle\chi_{n},x\rangle)$ is continuous.
By assumption, $T(A)$ is relatively compact. Therefore, by Proposition 2.10,
we obtain $\lim_{n\to\infty}\sup_{a\in A}|\langle\chi_{n},x\rangle|=0.$ Thus
$A$ is a $(V^{\ast})$ set.

In Corollary 1.3 of [8] Bombal proved that every relatively weakly
sequentially compact subset of a Banach space is a $(V^{\ast})$ set. Below we
generalize this result.

###### Corollary 14.18

Let $E$ be a barrelled space. Then every weakly sequentially precompact subset
$A$ of $E$ is a $(V^{\ast})$ set.

###### Proof 108

Let $T:E\to\ell_{1}$ be an operator. Then $T(A)$ is weakly sequentially
precompact in $\ell_{1}$. Since $\ell_{1}$ is weakly sequentially complete and
has the Schur property, $T(A)$ is relatively compact in $\ell_{1}$. Thus, by
Theorem 14.17, $A$ is a $(V^{\ast})$ set.

Following [35], we shall say that a sequence $A=\\{a_{n}\\}_{n\in\omega}$ of
an lcs $E$ is equivalent to the unit basis $\\{e_{n}:n\in\omega\\}$ of
$\ell_{1}$ or is an $\ell_{1}$-sequence if there exists a linear topological
isomorphism $R$ from the closure $\overline{\mathsf{span}}(A)$ of
$\mathsf{span}(A)$ onto a subspace of $\ell_{1}$ such that $R(a_{n})=e_{n}$
for every $n\in\omega$ (we do not assume that $\overline{\mathsf{span}}(A)$ is
complete). The following corollary generalizes Theorem 1.2 of [25].

###### Corollary 14.19

Let $E$ be a barrelled space satisfying the following condition:

  1. $(\dagger)$

if a bounded subset $A$ of $E$ is not relatively weakly sequentially compact,
then there is an $\ell_{1}$-sequence $\\{a_{n}\\}_{n\in\omega}\subseteq A$
such that $\overline{\mathsf{span}}\big{(}\\{a_{n}\\}_{n\in\omega}\big{)}$ is
complemented in $E$.

Then $E$ has the property $sV^{\ast}$.

###### Proof 109

Suppose for a contradiction that there is a $(V^{\ast})$ subset $A$ of $E$
which is not relatively weakly sequentially compact. Then, by $(\dagger)$,
there are a sequence $\\{a_{n}\\}_{n\in\omega}\subseteq A$ and a linear
topological isomorphism $R$ from $\overline{\mathsf{span}}(A)$ onto a subspace
of $\ell_{1}$ such that $R(a_{n})=e_{n}$ for every $n\in\omega$ and the
subspace $\overline{\mathsf{span}}\big{(}\\{a_{n}\\}_{n\in\omega}\big{)}$ is
complemented in $E$. Let $P$ be a projection from $E$ onto
$\overline{\mathsf{span}}\big{(}\\{a_{n}\\}_{n\in\omega}\big{)}$. Then $R\circ
P:E\to\ell_{1}$ is an operator such that $R\circ P(a_{n})=e_{n}$ for every
$n\in\omega$. By Theorem 14.17, the canonical unit basis
$\\{e_{n}\\}_{n\in\omega}$ of $\ell_{1}$ is relatively compact, a
contradiction.

###### Corollary 14.20

Let $p\geq 1$ and a locally convex space $E$ satisfy one of the following
conditions

  1. (i)

$1<p<2$ and $E$ is a quasibarrelled space such that $E^{\prime}_{\beta}$ is an
$\ell_{\infty}$-$V_{p}$-barrelled space,

  2. (ii)

$p=1$ and $E$ is barrelled.

Then every relatively weakly sequentially $p$-compact subset of $E$ is a
$p$-$(V^{\ast})$ set.

###### Proof 110

Let $A$ be a relatively weakly sequentially $p$-compact subset of $E$. Then,
by Lemma 13.2, $T(A)$ is a relatively weakly sequentially $p$-compact subset
of $\ell_{p}$ for every operator $T:E\to\ell_{p}$. If $1<p<2$, we obtain
$p<p^{\ast}$ and hence, by Proposition 6.5, $\ell_{p}$ has the $p$-Schur
property. Recall that $\ell_{1}$ has the Schur property. Therefore in both
cases (i) and (ii), the set $T(A)$ is relatively sequentially compact in
$\ell_{p}$. Thus, by Theorems 14.16 and 14.17, $T(A)$ is a $p$-$(V^{\ast})$
subset of $E$.

###### Remark 14.21

The condition $1\leq p<2$ in Corollary 14.20 is essential even for Banach
spaces. Indeed, let $p=2$ and $E=\ell_{2}$. Then the standard unit basis
$S=\\{e_{n}\\}_{n\in\omega}$ of $E$ is weakly $2$-summable (indeed, if
$\chi=(a_{n})\in E^{\prime}=E$, then
$\sum_{n\in\omega}|\langle\chi,e_{n}\rangle|^{2}=\sum_{n\in\omega}a_{n}^{2}<\infty$).
Therefore $S$ is a relatively weakly sequentially $2$-compact subset of $E$.
However, for $T=\mathsf{id}_{E}$, the set $T(S)=S$ is not relatively compact
in $E$. Thus, by Theorem 14.16, $S$ is not a $2$-$(V^{\ast})$ set.

For the case when $Y=\ell_{p^{\ast}}^{0}$ and $p\not=\infty$ we can reverse
Proposition 14.14 as follows.

###### Proposition 14.22

Let $p\in[1,\infty)$, $E$ be a locally convex space, and let $B$ be a subset
of $E^{\prime}$. Assume that one of the following conditions holds:

  1. (i)

$T^{\ast}(B)$ is relatively compact in $\ell_{p}$ for every
$T\in\mathcal{L}(\ell_{p^{\ast}}^{0},E)$ $($or $T\in\mathcal{L}(c_{0}^{0},E)$
if $p=1$$)$;

  2. (ii)

$E$ is sequentially complete and $T^{\ast}(B)$ is relatively compact in
$\ell_{p}$ for every $T\in\mathcal{L}(\ell_{p^{\ast}},E)$ $($or
$T\in\mathcal{L}(c_{0},E)$ if $p=1$$)$.

Then $B$ is a $p$-$(V)$ set.

###### Proof 111

Let $\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in $E$. Then,
by Proposition 4.14, there is $T\in\mathcal{L}(\ell_{p^{\ast}}^{0},E)$ or
$T\in\mathcal{L}(c_{0}^{0},E)$ (resp., $T\in\mathcal{L}(\ell_{p^{\ast}},E)$ or
$T\in\mathcal{L}(c_{0},E)$ if $E$ is sequentially complete) such that
$T(e_{n}^{\ast})=x_{n}$ for every $n\in\omega$. Since $T^{\ast}(B)$ is
relatively compact in $\ell_{p}$, Proposition 2.10 implies

$\sup_{\chi\in B}|\langle\chi,x_{n}\rangle|=\sup_{\chi\in B}|\langle
T^{\ast}(\chi),e_{n}^{\ast}\rangle|\to 0,$

which means that $B$ is a $p$-$(V)$ set in $E^{\prime}$, as desired.

The next theorem generalizes Theorem 19 of [42], for the case $1<p<\infty$,
Theorem 3.10 of [16] and, for the case $p=1$, Proposition 5 of [19].

###### Theorem 14.23

Let $1\leq p<\infty$, and let $E$ be an $\ell_{\infty}$-$V_{p}$-barrelled
space. Then for a weak∗ bounded subset $B$ of $E^{\prime}$ the following
assertions are equivalent:

  1. (i)

$B$ is a $p$-$(V)$ set;

  2. (ii)

for every normed space $Y$ and each weakly sequentially $p$-precompact
operator $T:Y\to E$, the set $T^{\ast}(B)$ is relatively compact in the Banach
space $Y^{\prime}_{\beta}$;

  3. (iii)

$T^{\ast}(B)$ is relatively compact in $\ell_{p}$ for every
$T\in\mathcal{L}(\ell_{p^{\ast}}^{0},E)$ $($or $T\in\mathcal{L}(c_{0}^{0},E)$
if $p=1$$)$.

If additionally $E$ is sequentially complete, then (i)-(iii) are equivalent to

  1. (iv)

$T^{\ast}(B)$ is relatively compact in $\ell_{p}$ for every
$T\in\mathcal{L}(\ell_{p^{\ast}},E)$ $($or $T\in\mathcal{L}(c_{0},E)$ if
$p=1$$)$.

###### Proof 112

(i)$\Rightarrow$(ii) follows from Proposition 14.14.

(ii)$\Rightarrow$(iii) Let $T\in\mathcal{L}(\ell_{p^{\ast}}^{0},E)$ (or
$T\in\mathcal{L}(c_{0}^{0},E)$ if $p=1$). By (i) of Corollary 13.11, the
identity operator $\mathsf{id}_{\ell_{p^{\ast}}^{0}}$ (or
$\mathsf{id}_{c_{0}^{0}}$ if $p=1$) is weakly sequentially $p$-precompact, and
hence, by Lemma 13.2, also $T=T\circ\mathsf{id}_{\ell_{p^{\ast}}^{0}}$ (or
$T=T\circ\mathsf{id}_{c^{0}_{0}}$ if $p=1$) is weakly sequentially
$p$-precompact. By (ii), $T^{\ast}(B)$ is relatively compact in $\ell_{p}$.

(iii)$\Rightarrow$(i) follows from (i) of Proposition 14.22.

Assume now that $E$ is additionally sequentially complete.

(iii)$\Rightarrow$(iv) Let $T\in\mathcal{L}(\ell_{p^{\ast}},E)$ (or
$T\in\mathcal{L}(c_{0},E)$ if $p=1$). Denote by $S$ the restriction of $T$
onto $\ell_{p^{\ast}}^{0}$ (or onto $c_{0}^{0}$ if $p=1$). Since
$\ell_{p^{\ast}}^{0}$ (resp., $c_{0}^{0}$) is dense in $\ell_{p^{\ast}}$
(resp., $c_{0}$) it follows that $T^{\ast}=S^{\ast}$. Therefore, by (iii),
$T^{\ast}(B)=S^{\ast}(B)$ is relatively compact in $\ell_{p}$.

(iv)$\Rightarrow$(i) follows from (ii) of Proposition 14.22.

##  15 Gantmacher’s property for locally convex spaces and an operator
characterization of Pełczyński’s property $sV_{p}$

The aim of this section is to give an operator characterization of the
property $sV_{p}$ and an independent and short proof of Pełczyński’s Theorem
1.2. First we recall some well known constructions, we use standard notations
from [52].

Let $E$ be a locally convex space. Then every $U\in\mathcal{N}_{0}^{c}(E)$
defines a seminorm $q_{U}$ by

$q_{U}(x):=\inf\\{\lambda>0:x\in\lambda U\\}\quad(x\in E).$

Set $N(U):=q_{U}^{-1}(0)=\bigcap_{n\in\mathbb{N}}\tfrac{1}{n}U$. Observe that
$N(U)$ is a closed subspace of $E$ such that $U=U+N(U)$. Therefore $q_{U}(x)$
defines a norm on the quotient vector space

$E_{(U)}:=E/N(U)\quad\mbox{ by }\;\;\|x+N(U)\|_{U}:=q_{U}(x)\;\;(x\in E).$

We denote by $\Phi_{U}$ the quotient linear map $E\to E_{(U)}=E/N(U)$. Observe
that $U=\\{x\in E:q_{U}(x)\leq 1\\}$. It follows that
$\Phi_{U}(U)=B_{E_{(U)}}$ and hence $\Phi_{U}$ is continuous. Moreover, the
co-restriction of the adjoint map $\Phi_{U}^{\ast}$ onto
$E^{\prime}_{U^{\circ}}:=(E^{\prime})_{U^{\circ}}$ which is also denoted by
$\Phi_{U}^{\ast}$ is an isometric isomorphism of
$\big{(}E_{(U)}\big{)}^{\prime}_{\beta}$ onto $E^{\prime}_{U^{\circ}}$. In
particular, $E^{\prime}_{U^{\circ}}$ is a Banach space.

The next assertion generalizes Proposition 17.1.2 of [52] and gives a
characterization of compact-type operators defined in Definition 3.2. Recall
(see Section 3) that the topology $\mathcal{T}_{\mathcal{A}}$ of
$E^{\prime}_{\mathcal{A}}:=(E^{\prime},\mathcal{T}_{\mathcal{A}})$ is the
polar topology on $E^{\prime}$ of uniform convergence on the members of
$\mathcal{A}$.

###### Proposition 15.1

Let $E$ and $L$ be locally convex spaces,
$\mathcal{A}(L)\subseteq\mathsf{Bo}(L)$ be a bornology on $L$, and let
$T\in\mathcal{L}(E,L)$. Consider the following assertions:

  1. (i)

$T\in\mathcal{LA}(E,L)$;

  2. (ii)

there are $U\in\mathcal{N}_{0}^{c}(E)$ and
$T_{U}\in\mathcal{LA}(E_{(U)},L)\cap\mathcal{L}(E_{(U)},L_{T(U)})$ such that
$T=T_{U}\circ\Phi_{U}$;

  3. (iii)

there is $U\in\mathcal{N}_{0}^{c}(E)$ such that the adjoint operator
$T^{\ast}$ maps $L^{\prime}_{\mathcal{A}}$ continuously into the Banach space
$E^{\prime}_{U^{\circ}}$; in particular, the set
$T^{\ast}\big{(}T(U)^{\circ}\big{)}\subseteq E^{\prime}$ is equicontinuous.

Then (i)$\Leftrightarrow$(ii)$\Rightarrow$(iii). The implication
(iii)$\Rightarrow$(i) is satisfied if $\mathcal{A}(L)$ is saturated.

###### Proof 113

(i)$\Rightarrow$(ii) Let $T\in\mathcal{LA}(E,L)$. Choose
$U\in\mathcal{N}_{0}^{c}(E)$ such that $T(U)\in\mathcal{A}(L)$. Then $T(U)$ is
a bounded absolutely convex subset of $L$, and hence for every $x\in N(U)$, we
obtain $T(x)\in\bigcap_{n\in\mathbb{N}}\tfrac{1}{n}T(U)=\\{0\\}$. Therefore
$T$ induces a linear map $T_{U}:E_{(U)}\to L$. Observe that $T_{U}$ is
continuous as a linear map from $E_{(U)}$ into $L_{T(U)}$ because
$T_{U}\big{(}B_{E_{(U)}}\big{)}\subseteq T(U)$ and $T(U)$ is contained in the
closed unit ball of $L_{T(U)}$. Since, by Proposition 3.6, the norm topology
of $L_{T(U)}$ is finer than the induced original topology on $L$ it follows
that $T_{U}$ is continuous. The inclusion
$T_{U}\big{(}B_{E_{(U)}}\big{)}\subseteq T(U)$ and the property (c) of the
bornology $\mathcal{A}(L)$ imply also that $T_{U}\in\mathcal{LA}(E_{(U)},L)$.
It remains to note that $T=T_{U}\circ\Phi_{U}$.

The implication (ii)$\Rightarrow$(i) immediately follows from the equality
$T_{U}\big{(}B_{E_{(U)}}\big{)}=T(U)$ (which holds because
$\Phi_{U}(U)=B_{E_{(U)}}$).

(ii)$\Rightarrow$(iii) Assume that there are $U\in\mathcal{N}_{0}^{c}(E)$ and
an operator $T_{U}\in\mathcal{LA}(E_{(U)},L)$ such that
$T=T_{U}\circ\Phi_{U}$. Then $T^{\ast}=\Phi_{U}^{\ast}\circ T_{U}^{\ast}$. As
we noticed above, the co-restriction of $\Phi_{U}^{\ast}$ is an isometric
isomorphism of $\big{(}E_{(U)}\big{)}^{\prime}_{\beta}$ onto
$E^{\prime}_{U^{\circ}}$. Observe that $B_{E_{(U)}}=U+N(U)$, where the right
hand side contains conjugate classes of $x\in U$. Therefore, by the properties
(b)-(c) of the bornology $\mathcal{A}(L)$ and since
$T_{U}\in\mathcal{LA}(E_{(U)},L)$, we obtain
$T_{U}\big{(}U+N(U)\big{)}\in\mathcal{A}(L)$. Hence, by the definition of the
topology $\mathcal{T}_{\mathcal{A}}$ on $L^{\prime}$, the polar
$W:=T_{U}\big{(}U+N(U)\big{)}^{\circ}$ is a neighborhood of zero in
$L^{\prime}_{\mathcal{A}}$. Now, for every $x\in U$ and each $\chi\in W$, we
obtain

$|\langle
T^{\ast}(\chi),x\rangle|=|\langle\chi,T_{U}\circ\Phi_{U}(x)\rangle|=|\langle\chi,T_{U}\big{(}x+N(U)\big{)}\rangle|\leq
1$

which means that $T^{\ast}(W)\subseteq U^{\circ}=B_{E^{\prime}_{U^{\circ}}}$.
Thus $T^{\ast}:L^{\prime}_{\mathcal{A}}\to E^{\prime}_{U^{\circ}}$ is
continuous. Finally, since $T(U)^{\circ}=T_{U}\big{(}U+N(U)\big{)}^{\circ}=W$
and $T^{\ast}\big{(}T(U)^{\circ}\big{)}=T^{\ast}(W)\subseteq U^{\circ}$ the
set $T^{\ast}\big{(}T(U)^{\circ}\big{)}$ is equicontinuous.

(iii)$\Rightarrow$(i) Assume additionally that $\mathcal{A}(L)$ is saturated,
and let $T^{\ast}:L^{\prime}_{\mathcal{A}}\to E^{\prime}_{U^{\circ}}$ be
continuous. Then there is $A\in\mathcal{A}$ such that
$T^{\ast}(A^{\circ})\subseteq U^{\circ}$. This means that for every $x\in U$
and each $\chi\in A^{\circ}$, we have $|\langle\chi,T(x)\rangle|=|\langle
T^{\ast}(\chi),x\rangle|\leq 1$ and hence $T(U)\subseteq A^{\circ\circ}$. As
$\mathcal{A}(L)$ is saturated, we have $A^{\circ\circ}\in\mathcal{A}(L)$.
Therefore, by (c) of the bornology $\mathcal{A}(L)$, it follows
$T(U)\in\mathcal{A}(L)$. Thus $T\in\mathcal{LA}(E,L)$.

The next notion is motivated by (iii) of Proposition 15.1.

###### Definition 15.2

Let $E$ and $L$ be locally convex spaces. A linear mapping $T:L\to E^{\prime}$
is called equicontinuous if for some $U\in\mathcal{N}_{0}(L)$, the set
$T(U)\subseteq E^{\prime}$ is equicontinuous.

Let $E$ be a locally convex space. If $U\in\mathcal{N}_{0}(E)$, then, by
Theorem 11.3.5 of [67], $U^{\circ}$ is strongly bounded and hence, by
Proposition 3.6, the canonical inclusion $E^{\prime}_{U^{\circ}}\to
E^{\prime}_{\beta}$ is continuous. Therefore there exists a finest locally
convex topology denoted by $\mathcal{T}_{n}$ on the dual space $E^{\prime}$
such that for every $U\in\mathcal{N}_{0}(E)$, the canonical inclusion
$E^{\prime}_{U^{\circ}}\to(E^{\prime},\mathcal{T}_{n})$ is continuous.
Clearly, $\beta(E^{\prime},E)\subseteq\mathcal{T}_{n}$. In the case when $E$
is a normed space, the continuity of
$E^{\prime}_{B_{E}^{\circ}}=E^{\prime}_{\beta}\to(E^{\prime},\mathcal{T}_{n})$
implies that $\mathcal{T}_{n}\subseteq\beta(E^{\prime},E)$, and hence
$\mathcal{T}_{n}=\beta(E^{\prime},E)$ is a Banach norm topology (which
motivates the subscript $n$ in the notation $\mathcal{T}_{n}$).

###### Theorem 15.3

Let $E$ and $L$ be locally convex spaces, and assume that
$\mathcal{A}(L)\subseteq\mathsf{Bo}(L)$ is a saturated bornology on $L$. Then
$\mathcal{L}(E,L)=\mathcal{LA}(E,L)$ if and only if for every
$T\in\mathcal{L}(E,L)$, the adjoint mapping
$T^{\ast}:L^{\prime}_{\mathcal{A}}\to E^{\prime}_{\beta}$ is equicontinuous
and hence continuous.

###### Proof 114

Taking into account that $\beta(E^{\prime},E)\subseteq\mathcal{T}_{n}$, the
necessity is proved in Proposition 15.1. To prove the sufficiency, let
$T\in\mathcal{L}(E,L)$. Since $T^{\ast}$ is equicontinuous, there are
$A\in\mathcal{A}$ and $U\in\mathcal{N}_{0}^{c}(E)$ such that
$T^{\ast}(A^{\circ})\subseteq U^{\circ}$. This means that for every $x\in U$
and each $\chi\in A^{\circ}$, we have $|\langle\chi,T(x)\rangle|=|\langle
T^{\ast}(\chi),x\rangle|\leq 1$ and hence $T(U)\subseteq A^{\circ\circ}$. As
$\mathcal{A}(L)$ is saturated, we have $A^{\circ\circ}\in\mathcal{A}(L)$.
Therefore $T(U)\in\mathcal{A}(L)$ and hence $T\in\mathcal{LA}(E,L)$.

To show that $T^{\ast}$ is also continuous, fix a bounded subset $B$ of $E$.
As $T^{\ast}$ is equicontinuous, there are $A\in\mathcal{A}$ and
$U\in\mathcal{N}_{0}^{c}(E)$ such that $T^{\ast}(A^{\circ})\subseteq
U^{\circ}$. Since, by Theorem 11.3.5 of [67], $U^{\circ}$ is strongly bounded,
there is $\lambda>0$ such that $U^{\circ}\subseteq\lambda B^{\circ}$, and
hence
$T^{\ast}(\tfrac{1}{\lambda}A^{\circ})\subseteq\tfrac{1}{\lambda}U^{\circ}\subseteq
B^{\circ}$. Thus $T^{\ast}:L^{\prime}_{\mathcal{A}}\to E^{\prime}_{\beta}$ is
continuous.

Below we consider two cases when the adjoint operator has some of compact-type
properties. First we recall the next two well known and important results in
Banach space theory.

######  Theorem 15.4 (Schauder)

Let $T:E\to L$ be an operator between Banach spaces $E$ and $L$. Then $T$ is
compact if and only if its adjoint $T^{\ast}$ is compact.

######  Theorem 15.5 (Gantmacher)

Let $T:E\to L$ be an operator between Banach spaces $E$ and $L$. Then $T$ is
weakly compact if and only if its adjoint $T^{\ast}$ is weakly compact.

The next proposition is a version of the “direct part” of Schauder’s Theorem
15.4.

###### Proposition 15.6

Let $T$ be a precompact operator from a locally convex space $E$ to a normed
space $L$. Then $T^{\ast}(B_{L^{\prime}_{\beta}})$ is a compact subset of
$E^{\prime}_{\beta}$, and hence the adjoint operator
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is compact.

###### Proof 115

Consider the bornology $\mathcal{PC}(L)$ of precompact subsets of $L$. Since
$T$ is precompact, Proposition 15.1 implies that for some
$U\in\mathcal{N}_{0}^{c}(E)$, the map
$T^{\ast}:L^{\prime}_{\mathcal{PC}(L)}\to E^{\prime}_{U^{\circ}}$ is
continuous. As $U^{\circ}$ is strongly bounded by Theorem 11.3.5 of [67],
Proposition 3.6 implies that the norm topology of $E^{\prime}_{U^{\circ}}$ is
stronger than the topology induced from $E^{\prime}_{\beta}$. Therefore the
map $T^{\ast}:L^{\prime}_{\mathcal{PC}(L)}\to E^{\prime}_{\beta}$ is
continuous. By the Alaoglu–Bourbaki Theorem 8.5.2 of [52], the dual unit
closed ball $B_{L^{\prime}_{\beta}}=(B_{L})^{\circ}$ is compact in
$L^{\prime}_{\mathcal{PC}(L)}$. Thus $T^{\ast}(B_{L^{\prime}_{\beta}})$ is a
compact subset of $E^{\prime}_{\beta}$, as desired.

In what follows we shall use the following remark without mentioning.

###### Remark 15.7

Let $T$ be an operator from an lcs $E$ to a Banach space $L$. Then, by the
(weak) angelicity of $L$, the operator $T$ is (weakly) sequentially compact if
and only if it is (weakly) compact.

The next lemma immediately follows from the corresponding definitions.

###### Lemma 15.8

Let $T$ be an operator from a normed space $E$ to a Banach space $L$. Then $T$
is compact (resp., sequentially compact, weakly compact or weakly sequentially
compact) if and only if so is its unique extension $\bar{T}:\overline{E}\to L$
to the completion $\overline{E}$ of $E$.

The following assertion is a version of the necessity in Gantmacher’s Theorem
15.5.

###### Proposition 15.9

Let $T$ be an operator from a locally convex space $E$ to a Banach space $L$.
If $T$ is compact (resp., weakly compact), then its adjoint operator
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is compact and
sequentially compact (resp., weakly compact and weakly sequentially compact).

###### Proof 116

Applying Proposition 15.1 to the family $\mathcal{A}(L)=\mathcal{RC}(L)$
(resp., $\mathcal{RWC}(L)$) we obtain that there are
$U\in\mathcal{N}_{0}^{c}(E)$ and an operator $T_{U}\in\mathcal{LA}(E_{(U)},L)$
such that $T=T_{U}\circ\Phi_{U}$. Since $E_{(U)}$ is a normed space, Lemma
15.8 implies that the extension $\overline{T_{U}}$ of $T_{U}$ onto the
completion $\overline{E_{(U)}}$ of $E_{(U)}$ is a compact (resp., weakly
compact) operator. Note also that $L^{\prime}_{\beta}$ is a Banach space and
hence it is weakly angelic. Therefore, by the Schauder Theorem 15.4 and
Gantmacher’s Theorem 15.5, the adjoint operator
$\overline{T_{U}}^{\,\ast}=T_{U}^{\ast}$ is also compact and sequentially
compact (resp., weakly compact and weakly sequentially compact), and hence so
is $T^{\ast}=\Phi_{U}^{\ast}\circ\overline{T_{U}}^{\,\ast}$.

Using Proposition 15.9 we give below a sufficient condition on an lcs $E$ to
have the property $sV_{p}$, cf. Theorem 14.11.

###### Proposition 15.10

Let $p\in[1,\infty]$, and let $E$ be an $\ell_{\infty}$-$V_{p}$-barrelled
space. If each $p$-convergent operator $T:E\to\ell_{\infty}$ is weakly
compact, then $E$ has the property $sV_{p}$.

###### Proof 117

Let $B\subseteq E^{\prime}$ be a $p$-$(V)$ set. We prove that $B$ is
relatively weakly sequentially compact. To this end, fix an arbitrary sequence
$S=\\{\chi_{n}\\}_{n\in\omega}$ in $B$. By (iii) of Lemma 8.2, $S$ is also a
$p$-$(V)$ subset of $E^{\prime}$. Consider the linear map
$T:E\to\ell_{\infty}$ defined by

$T(x):=\big{(}\langle\chi_{n},x\rangle\big{)}_{n\in\omega}\quad(x\in E).$

Since $E$ is $\ell_{\infty}$-$V_{p}$-barrelled, the sequence $S$ is
equicontinuous. Therefore, by (i) of Lemma 14.13, $T$ is continuous. As $S$ is
a a $p$-$(V)$ set, (ii) of Lemma 14.13 implies that $T$ is $p$-convergent.
Hence, by assumption, $T$ is weakly sequentially compact. By Proposition 15.9,
the adjoint operator $T^{\ast}:\ell_{\infty}^{\prime}\to E^{\prime}_{\beta}$
is also weakly sequentially compact.

For every $n\in\omega$, let $e_{n}^{\ast}$ be the $n$th coordinate functional
on $\ell_{\infty}$, and observe that $T^{\ast}(e_{n}^{\ast})=\chi_{n}$
(indeed, if $x\in E$, then $\langle T^{\ast}(e_{n}^{\ast}),x\rangle=\langle
e_{n}^{\ast},T(x)\rangle=\langle\chi_{n},x\rangle$). Since $T^{\ast}$ is
weakly sequentially compact and all $e_{n}^{\ast}$ belong to the closed unit
ball of $\ell_{\infty}^{\prime}$, there is a subsequence
$\\{e^{\ast}_{n_{k}}\\}_{k\in\omega}$ of $\\{e^{\ast}_{n}\\}_{n\in\omega}$
such that the sequence
$\\{T^{\ast}(e_{n_{k}}^{\ast})=\chi_{n_{k}}\\}_{k\in\omega}$ weakly converges
to some functional $\chi\in E^{\prime}_{\beta}$. Thus $B$ is relatively weakly
sequentially compact in $E^{\prime}_{\beta}$.

Being motivated by Schauder’s Theorem 15.4 and Gantmacher’s Theorem 15.5 one
can naturally introduce the following notions.

###### Definition 15.11

A locally convex space $E$ is said to have

  1. $\bullet$

the (sequential) Schauder property if for every Banach space $L$, an operator
$T:E\to L$ is compact if and only if its adjoint
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is (resp., sequentially)
compact;

  2. $\bullet$

the strong (sequential) Schauder property if for every lcs $L$, an operator
$T:E\to L$ is (resp., sequentially) compact if and only if so is its adjoint
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$;

  3. $\bullet$

the (sequential) Gantmacher property if for every Banach space $L$, an
operator $T:E\to L$ is weakly compact if and only if
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is weakly (resp.,
sequentially) compact;

  4. $\bullet$

the strong (sequential) Gantmacher property if for every lcs $L$, an operator
$T:E\to L$ is weakly (resp., sequentially) compact if and only if so is
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$.

The introduced notions are well-behaved under taking products.

###### Proposition 15.12

Let $\\{E_{i}\\}_{i\in I}$ be a non-empty family of locally convex spaces.
Then the product $E=\prod_{i\in I}E_{i}$ has the Schauder property (resp., the
sequential Schauder property, the Gantmacher property or the sequential
Gantmacher property) if and only if all $E_{i}$ have the same property.

###### Proof 118

Assume that $E$ has the Schauder property (resp., the sequential Schauder
property, the Gantmacher property or the sequential Gantmacher property). Fix
$j\in I$, and let $T_{j}:E_{j}\to L$ be an operator from $E_{j}$ to a Banach
space $L$. If $T_{j}$ is compact (resp., sequentially compact, weakly compact
or weakly sequentially compact), then, by Proposition 15.9, the adjoint
operator $T_{j}^{\ast}:L^{\prime}_{\beta}\to(E_{j})^{\prime}_{\beta}$ has the
same property. Conversely, assume that
$T_{j}^{\ast}:L^{\prime}_{\beta}\to(E_{j})^{\prime}_{\beta}$ is compact
(resp., sequentially compact, weakly compact or weakly sequentially compact).
Let $0_{j}:\prod_{i\in I{\setminus}\\{j\\}}E_{i}\to L$ be the zero operator,
and let $I_{j}:(E_{j})^{\prime}_{\beta}\to E^{\prime}_{\beta}$ be the identity
embedding onto the $j$coordinate. Define an operator $T:E\to L$ by
$T=T_{j}\times 0_{j}$. Then for every $(x_{i})\in E$ and each $\chi\in
L^{\prime}$, we have

$\langle
T^{\ast}(\chi),(x_{i})\rangle=\langle\chi,\big{(}T(x_{i})\big{)}_{i\in
I}\rangle=\langle\chi,T_{j}(x_{j})\rangle=\langle
T_{j}^{\ast}(\chi),x_{j}\rangle=\big{\langle}\big{(}I_{j}\circ
T_{j}^{\ast}\big{)}(\chi),(x_{i})\big{\rangle},$

which means that $T^{\ast}=I_{j}\circ T_{j}^{\ast}$. Since $T_{j}^{\ast}$ is
compact (resp., sequentially compact, weakly compact or weakly sequentially
compact) so is the operator $T^{\ast}$. By the Schauder property (resp., the
sequential Schauder property, the Gantmacher property or the sequential
Gantmacher property) of $E$, the operator $T=T_{j}\times 0_{j}$ is compact
(resp., sequentially compact, weakly compact or weakly sequentially compact),
and hence so is $T_{j}$. Thus $E_{j}$ has the Schauder property (resp., the
sequential Schauder property, the Gantmacher property or the sequential
Gantmacher property).

Assume now that all spaces $E_{i}$ have the Schauder property (resp., the
sequential Schauder property, the Gantmacher property or the sequential
Gantmacher property). Let $T$ be an operator from $E$ to a Banach space $L$.
If $T$ is compact (resp., sequentially compact, weakly compact or weakly
sequentially compact), then, by Proposition 15.9, so is also the adjoint
operator $T^{\ast}$. Conversely, assume that $T^{\ast}:L^{\prime}_{\beta}\to
E^{\prime}_{\beta}$ is compact (resp., sequentially compact, weakly compact or
weakly sequentially compact). By Lemma 2.6, there is a finite subset $F$ of
$I$ such that $\prod_{i\in I{\setminus}F}E_{i}$ is in the kernel of $T$.
Therefore to show that $T$ is compact (resp., sequentially compact, weakly
compact or weakly sequentially compact) it suffices to assume that $I$ has
only two elements. So, consider an operator $T:E=H\times Z\to L$ such that
$T^{\ast}:L^{\prime}_{\beta}\to H^{\prime}_{\beta}\times Z^{\prime}_{\beta}$
is compact (resp., sequentially compact, weakly compact or weakly sequentially
compact). Let $I_{H}:H\to H\times Z$ and $I_{Z}:Z\to H\times Z$ be the
coordinate embeddings. Then

$T(h,z)=T(h,0)+T(0,z)=T\circ I_{H}(h)+T\circ I_{Z}(z).$ (15.1)

Since $T^{\ast}$ is compact (resp., sequentially compact, weakly compact or
weakly sequentially compact) so are $I_{H}^{\ast}\circ T^{\ast}$ and
$I_{Z}^{\ast}\circ T^{\ast}$. Since $H$ and $Z$ have the Schauder property
(resp., the sequential Schauder property, the Gantmacher property or the
sequential Gantmacher property), it follows that the operators $T\circ
I_{H}:H\to L$ and $T\circ I_{Z}:Z\to L$ are compact (resp., sequentially
compact, weakly compact or weakly sequentially compact). Thus, by (15.1), also
the operator $T$ has the same property.

It is well known that any locally convex space embeds into the product of a
family of Banach spaces. This fact and Schauder’s Theorem 15.4, Gantmacher’s
Theorem 15.5 and Proposition 15.12 immediately imply the following corollary.

###### Corollary 15.13

Any locally convex space embeds into a locally convex space with the Schauder
property (resp., the sequential Schauder property, the Gantmacher property or
the sequential Gantmacher property).

###### Proposition 15.14

Let $p\in[1,\infty]$, $E$ be a locally convex space with the sequential
Gantmacher property, and let $\Gamma$ be an infinite set. If $E$ has the
property $sV_{p}$, then each $p$-convergent operator
$T:E\to\ell_{\infty}(\Gamma)$ is weakly (sequentially) compact.

###### Proof 119

Let $T:E\to\ell_{\infty}(\Gamma)$ be a $p$-convergent operator. To show that
$T$ is weakly sequentially compact, by the sequential Gantmacher property, it
suffices to prove that the adjoint operator $T^{\ast}$ is weakly sequentially
compact. Let $\\{x_{i}\\}_{i\in\omega}$ be an arbitrary weakly $p$-summable
sequence in $E$, and let $B$ be the closed unit ball of
$\ell^{\prime}_{\infty}(\Gamma)$. Since $T$ is $p$-convergent, $T(x_{i})\to 0$
in $\ell_{\infty}(\Gamma)$. Therefore

$\sup_{\eta\in B}\big{|}\langle
T^{\ast}(\eta),x_{i}\rangle\big{|}=\sup_{\eta\in
B}\big{|}\langle\eta,T(x_{i})\rangle\big{|}\leq\|T(x_{i})\|\to 0,$

which means that $T^{\ast}(B)$ is a $p$-$(V)$ set in $E^{\prime}$. By the
property $sV_{p}$, we obtain that $T^{\ast}(B)$ is a relatively weakly
sequentially compact subset of $E^{\prime}_{\beta}$. Thus $T^{\ast}$ is weakly
sequentially compact.

Joining Proposition 15.10 and Proposition 15.14 we obtain the following
operator characterization of the property $sV_{p}$ which generalizes
Pełczyński’s operator characterization of the property $V$ for Banach spaces,
see [71].

###### Theorem 15.15

Let $p\in[1,\infty]$, and let $E$ be an $\ell_{\infty}$-$V_{p}$-barrelled
space with the sequential Gantmacher property. Then the following assertions
are equivalent:

  1. (i)

for every Banach space $L$, each $p$-convergent operator $T:E\to L$ is weakly
compact;

  2. (ii)

each $p$-convergent operator $T:E\to\ell_{\infty}$ is weakly compact;

  3. (iii)

$E$ has the property $sV_{p}$.

###### Proof 120

The implication (i)$\Rightarrow$(ii) is trivial, and (ii)$\Rightarrow$(iii)
follows from Proposition 15.10.

(iii)$\Rightarrow$(i) It is well known that there is an infinite set $\Gamma$
such that the Banach space $L$ embeds into $\ell_{\infty}(\Gamma)$. Let
$R:L\to\ell_{\infty}(\Gamma)$ be an embedding. Then $R\circ T$ is also a
$p$-convergent operator. Therefore, by Proposition 15.14, $R\circ T$ is weakly
compact. Since $R(L)$ is a closed subspace of $\ell_{\infty}(\Gamma)$, it is
also weakly closed. It follows that if $U\in\mathcal{N}_{0}(E)$ is such that
$R\circ T(U)$ is relatively weakly compact in $\ell_{\infty}(\Gamma)$, then
$T(U)$ is relatively weakly compact in $L$. Thus $T$ is a weakly compact
operator.

Below we provide a short proof of Pełczyński’s Theorem 1.2.

######  Theorem 15.16 ([71])

For every $p\in[1,\infty]$ and each compact space $K$, the Banach space $C(K)$
has the property $V_{p}$.

###### Proof 121

By Gantmacher’s Theorem 15.5 and Theorem 15.15, it suffice to prove that each
$p$-convergent operator $T:C(K)\to\ell_{\infty}$ is weakly compact. Suppose
for a contradiction that there is a $p$-convergent operator
$T:C(K)\to\ell_{\infty}$ which is not weakly compact. Since $T$ is not weakly
compact, Pełczyński’s Theorem 5.5.3 of [1] implies that $C(K)$ has a subspace
$H$ isomorphic to $c_{0}$ such that the restriction $T{\restriction}_{H}$ is
an isomorphism. Identifying $H$ with $c_{0}$, we observe that the standard
unit basis $\\{e_{n}\\}_{n\in\omega}$ of $c_{0}$ is weakly $p$-summable
(indeed, if $x=(a_{n})\in\ell_{1}$, then $\sum_{n\in\omega}\big{|}\langle
x,e_{n}\rangle\big{|}^{p}=\sum_{n\in\omega}|a_{n}|^{p}<\infty$). Since
$e_{n}\not\to 0$ in $c_{0}$ and $T{\restriction}_{H}$ is an isomorphism, it
follows that $T(e_{n})\not\to 0$ in $L$. Hence $T$ is not $p$-convergent, a
contradiction.

##  16 $(q,p)$-convergent operators and characterizations of Pełczyński’s type
sets

In Section 14 we obtained operator characterizations of $p$-$(V^{\ast})$ sets
and $p$-$(V)$ sets using weakly sequentially $p$-precompact operators, see
Theorems 14.16 and 14.23. The purpose of this section is to find operator
characterizations of $(p,q)$-$(V^{\ast})$ sets and $(p,q)$-$(V)$ sets. For
this we recall first some definitions.

Let $L$ be a Banach space and $p\in[1,\infty]$. Recall that a sequence
$\\{x_{n}\\}_{n\in\omega}$ in $L$ is called strongly $p$-summable if
$\big{(}\|x_{n}\|)\in\ell_{p}$ (or $\big{(}\|x_{n}\|)\in c_{0}$ if
$p=\infty$). The family of all strongly $p$-summable sequences in $L$ is
denoted by $\ell_{p}^{s}(L)$ (or by $c_{0}^{s}(L)$ if $p=\infty$). Recall also
that $\ell_{p}^{s}(L)$ and $c_{0}^{s}(L)$ are Banach spaces under pointwise
operations and the norm defined by

$\big{\|}(x_{n})\big{\|}^{strong}_{p}:=\Big{(}\sum_{n\in\omega}\|x_{n}\|^{p}\Big{)}^{1/p}\;\;\mbox{
and
}\;\;\big{\|}(x_{n})\big{\|}^{strong}_{\infty}:=\sup_{n\in\omega}\big{\|}x_{n}\big{\|},$

respectively.

The notion of strongly $p$-summable sequences in Banach spaces can be
naturally extended to the general case as follows, see § 19.4 of [52]: a
sequence $\\{x_{n}\\}_{n\in\omega}$ in an lcs $L$ is called strongly
$p$-summable if $\big{(}q_{U}(x_{n})\big{)}\in\ell_{p}$ (or
$\big{(}q_{U}(x_{n})\big{)}\in c_{0}$ if $p=\infty$) for every
$U\in\mathcal{N}_{0}^{c}(E)$, where as usual $q_{U}$ denotes the gauge
functional of $U$. Since $U\subseteq V$ implies $q_{U}\geq q_{V}$, for Banach
spaces, this notion of strong $p$-summability coincides with the above
mentioned usual notion.

Below we introduce a new class of operators which allows to characterize
$(p,q)$-$(V^{\ast})$ sets and $(p,q)$-$(V)$ sets.

###### Definition 16.1

Let $1\leq p\leq q\leq\infty$, and let $E$ and $L$ be locally convex spaces. A
linear map $T:E\to L$ is called $(q,p)$-convergent if it sends weakly
$p$-summable sequences in $E$ to strongly $q$-summable sequences in $L$.

###### Remark 16.2

(i) Our terminology is based on the evident fact that an operator $T:E\to L$
is $(\infty,p)$-convergent if and only if it is $p$-convergent and the
important notion of $(q,p)$-summing operators between Banach spaces (see [24],
[75] and Section 17 below).

(ii) The condition $p\leq q$ in Definition 16.1 is important because if $q<p$,
then only the zero linear map can be $(q,p)$-convergent. Indeed, if $T\not=0$,
take $(\lambda_{n})\in\ell_{p}{\setminus}\ell_{q}$ (or $\in
c_{0}{\setminus}\ell_{q}$ if $p=\infty$) and $U\in\mathcal{N}_{0}^{c}(L)$ such
that $q_{U}\big{(}T(x)\big{)}\not=0$ for some $x\in E$. Then the sequence
$\\{x_{n}=\lambda_{n}x\\}_{n\in\omega}$ is weakly $p$-summable. However, the
series
$\sum_{i\in\omega}q_{U}\big{(}T(x_{i})\big{)}^{q}=q_{U}\big{(}T(x)\big{)}^{q}\sum_{i\in\omega}|\lambda_{n}|^{q}$
diverges.

Using $(p,q)$-$(V)$ sets we can characterize $(q,p)$-convergent operators into
normed spaces as follows.

###### Theorem 16.3

Let $1\leq p\leq q\leq\infty$, $E$ be a locally convex space, and let $T$ be
an operator from $E$ to a normed space $L$. Then $T$ is $(q,p)$-convergent if
and only if $T^{\ast}(B_{L^{\prime}})$ is a $(p,q)$-$(EV)$ subset of
$E^{\prime}$.

###### Proof 122

First we observe that for every $x\in E$, we have

$\|T(x)\|=\sup_{\chi\in B_{L^{\prime}}}|\langle\chi,T(x)\rangle|=\sup_{\chi\in
B_{L^{\prime}}}|\langle T^{\ast}(\chi),x\rangle|.$

Now, for every weakly $p$-summable sequence $\\{x_{n}\\}_{n\in\omega}$ in $E$,
we obtain that $\big{(}\|T(x_{n})\|)_{n}\in\ell_{q}$ (or $\in c_{0}$ if
$q=\infty$) if and only if

$\Big{(}\sup_{\chi\in B_{L^{\prime}}}|\langle
T^{\ast}(\chi),x_{n}\rangle|\Big{)}\in\ell_{q}\;\;\;\;\big{(}\mbox{or }\in
c_{0}\;\mbox{ if $q=\infty$}\big{)}.$

Taking into account that $B_{L^{\prime}}$ and hence also
$T^{\ast}(B_{L^{\prime}})$ are equicontinuous, it follows that
$\big{(}\|T(x_{n})\|)_{n}\in\ell_{q}$ (or $\in c_{0}$ if $q=\infty$) if and
only if $T^{\ast}(B_{L^{\prime}})$ is a $(p,q)$-$(EV)$ subset of $E^{\prime}$.

The next corollary complements Corollary 14.7.

###### Corollary 16.4

Let $1\leq p\leq q\leq\infty$, $E$ be a locally convex space with the property
$EV_{(p,q)}$ (resp., the property $sEV_{(p,q)}$), and let $T$ be an operator
from $E$ to a normed space $L$. If $T$ is $(q,p)$-convergent, then
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is weakly (resp.,
sequentially) compact. The converse is true if every weakly compact,
absolutely convex and equicontinuous subset of $E^{\prime}_{\beta}$ is a
$(p,q)$-$(EV)$ set.

###### Proof 123

By Theorem 16.3, the absolutely convex set $T^{\ast}(B_{L^{\prime}})$ is a
$(p,q)$-$(EV)$ subset of $E^{\prime}$. Therefore, by the property $EV_{(p,q)}$
(resp., the property $sEV_{(p,q)}$), $T^{\ast}(B_{L^{\prime}})$ is a
relatively weakly (resp., sequentially) compact subset of
$E^{\prime}_{\beta}$, as desired.

Conversely, assume that every weakly compact, absolutely convex and
equicontinuous subset of $E^{\prime}_{\beta}$ is a $(p,q)$-$(EV)$ set and
$T^{\ast}(B_{L^{\prime}})$ is a relatively weakly (resp., sequentially)
compact subset of $E^{\prime}_{\beta}$. Since $T^{\ast}(B_{L^{\prime}})$ is
also equicontinuous we obtain that its closure
$\mathrm{cl}_{E^{\prime}_{\beta}}\big{(}T^{\ast}(B_{L^{\prime}})\big{)}$ and
hence also $T^{\ast}(B_{L^{\prime}})$ are $(p,q)$-$(EV)$ sets. Thus, by
Theorem 16.3, $T$ is $(q,p)$-convergent.

The following assertion is dual to Theorem 16.3.

###### Theorem 16.5

Let $1\leq p\leq q\leq\infty$, $E$ be a locally convex space, and let $T$ be
an operator from a normed space $L$ to $E^{\prime}_{\beta}$. Then $T(B_{L})$
is a $(p,q)$-$(V)$ subset of $E^{\prime}$ if and only if the linear map
$T^{\ast}{\restriction}_{E}$ from $E$ to $L^{\prime}_{\beta}$ is
$(q,p)$-convergent.

###### Proof 124

First we note that for every $x\in E$, we have (recall that $J_{E}:E\to
E^{\prime\prime}$ denotes the canonical map)

$\|T^{\ast}(x)\|_{L^{\prime}_{\beta}}=\|T^{\ast}(J_{E}(x))\|_{L^{\prime}_{\beta}}=\sup_{y\in
B_{L}}|\langle T^{\ast}(J_{E}(x)),y\rangle|=\sup_{y\in B_{L}}|\langle
T(y),x\rangle|.$ (16.1)

Let now $\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in $E$.
By (16.1), we have
$\big{(}\|T^{\ast}(x_{n})\|_{L^{\prime}_{\beta}}\big{)}_{n\in\omega}=\big{(}\sup_{y\in
B_{L}}|\langle T(y),x_{n}\rangle|\big{)}_{n\in\omega}$. Then the theorem
follows from the definition of $(p,q)$-$(V)$ sets and the definition of
$(q,p)$-convergent linear maps.

The next theorem characterizes barrelled lcs for which $(p,q)$-$(V)$ subsets
of $E^{\prime}$ have additional compact-type topological property.

###### Theorem 16.6

Let $1\leq p\leq q\leq\infty$, let $E$ be a (resp., an $\aleph_{0}$-barrelled)
locally convex space. Then the following assertions are equivalent:

  1. (i)

if $L$ is a normed space and $T:E\to L$ is a $(q,p)$-convergent operator, then
$T^{\ast}:L^{\prime}_{\beta}\to E^{\prime}_{\beta}$ is weakly sequentially
compact (resp., sequentially compact, weakly sequentially $p$-compact or
weakly sequentially $p$-precompact);

  2. (ii)

the same as (i) with $L=\ell_{\infty}$;

  3. (iii)

each $(p,q)$-$(EV)$ (resp., $(p,q)$-$(V)$) subset of $E^{\prime}$ is
relatively weakly sequentially compact (resp., relatively sequentially
compact, relatively weakly sequentially $p$-compact or weakly sequentially
$p$-precompact) in $E^{\prime}_{\beta}$.

###### Proof 125

(i)$\Rightarrow$(ii) is clear.

(ii)$\Rightarrow$(iii) Let $B$ be a $(p,q)$-$(EV)$ (resp., $(p,q)$-$(V)$)
subset of $E^{\prime}$. Fix an arbitrary sequence
$S=\\{\chi_{n}\\}_{n\in\omega}$ in $B$. We claim that $S$ is equicontinuous.
This clear if $B$ be a $(p,q)$-$(EV)$ set. Assume that $E$ is
$\aleph_{0}$-barrelled. Since $B$ is weak∗ bounded (see (ii) of Lemma 8.2),
the $\aleph_{0}$-barrelleddeness of $E$ implies that $S$ is equicontinuous.
Therefore we can apply Lemma 14.13 to get that the linear map
$T:E\to\ell_{\infty}$ defined by

$T(x):=\big{(}\langle\chi_{n},x\rangle\big{)}_{n\in\omega}\quad(x\in E).$

is continuous.

Let now $\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in $E$.
Since $\\{\chi_{n}\\}_{n\in\omega}$ is also a $(p,q)$-$(V)$ set, we obtain

$\big{(}\sup_{i\in\omega}|\langle\chi_{i},x_{n}\rangle|\big{)}_{n\in\omega}\in\ell_{q}\quad(\mbox{or
$\in c_{0}$ if $q=\infty$}).$ (16.2)

Observe that
$\|T(x_{n})\|_{\ell_{\infty}}=\sup_{i\in\omega}|\langle\chi_{i},x_{n}\rangle|$
for every $n\in\omega$. Therefore (16.2) implies

$\big{(}\|T(x_{n})\|_{\ell_{\infty}}\big{)}_{n\in\omega}\in\ell_{q}\quad(\mbox{or
$\in c_{0}$ if $q=\infty$}),$

which means that $T$ is a $(q,p)$-convergent operator. Therefore, by (ii), the
adjoint operator $T^{\ast}$ is weakly sequentially compact (resp.,
sequentially compact, weakly sequentially $p$-compact or weakly sequentially
$p$-precompact).

For every $n\in\omega$, if $e_{n}^{\ast}$ is the $n$th coordinate functional
of $\ell_{\infty}$ and $x\in E$, then $\langle
T^{\ast}(e_{n}^{\ast}),x\rangle=\langle
e_{n}^{\ast},T(x)\rangle=\langle\chi_{n},x\rangle$ and hence
$T^{\ast}(e_{n}^{\ast})=\chi_{n}$. Therefore the sequence
$\\{\chi_{n}\\}_{n\in\omega}$ has a subsequence which weakly converges (resp.,
converges, weakly sequentially $p$-converges or weakly sequentially
$p$-Cauchy). Thus $B$ is relatively weakly sequentially compact (resp.,
relatively sequentially compact, relatively weakly sequentially $p$-compact or
weakly sequentially $p$-precompact).

(iii)$\Rightarrow$(i) Let $T:E\to L$ be a $(q,p)$-convergent operator to a
normed space $L$. Then, by Theorem 16.3, $T^{\ast}(B_{L^{\prime}})$ is a
$(p,q)$-$(EV)$ subset of $E^{\prime}$ (and hence also a $(p,q)$-$(V)$ set).
Now (iii) implies that $T^{\ast}$ is a weakly sequentially compact (resp.,
sequentially compact, weakly sequentially $p$-compact or weakly sequentially
$p$-precompact) operator.

###### Theorem 16.7

Let $1\leq p\leq q\leq\infty$, let $E$ be a locally convex space. Then the
following assertions are equivalent:

  1. (i)

if $L$ is a normed space and $T:L\to E^{\prime}_{\beta}$ is an operator such
that $T^{\ast}{\restriction}_{E}:E\to L^{\prime}_{\beta}$ is a
$(q,p)$-convergent linear map, then $T$ is weakly sequentially compact (resp.,
sequentially compact, weakly sequentially $p$-compact or weakly sequentially
$p$-precompact);

  2. (ii)

the same as (i) with $L=\ell_{1}^{0}$;

  3. (iii)

each bounded subset of $E^{\prime}_{\beta}$ which is a $(p,q)$-$(V)$ set is
relatively weakly sequentially compact (resp., relatively sequentially
compact, relatively weakly sequentially $p$-compact or weakly sequentially
$p$-precompact).

Moreover, if $E^{\prime}_{\beta}$ is locally complete, then (i)-(iii) are
equivalent to

  1. (iv)

the same as (i) with $L=\ell_{1}$.

###### Proof 126

(i)$\Rightarrow$(ii) and (i)$\Rightarrow$(iv) are clear.

(ii)$\Rightarrow$(iii) and (iv)$\Rightarrow$(iii): Let $B$ be a bounded
$(p,q)$-$(V)$ subset of $E^{\prime}_{\beta}$. Fix an arbitrary sequence
$S=\\{\chi_{n}\\}_{n\in\omega}$ in $B$, so $S$ is a bounded subset of
$E^{\prime}_{\beta}$. Therefore, by Proposition 14.9, the linear map
$T:\ell_{1}^{0}\to E^{\prime}_{\beta}$ (or $T:\ell_{1}\to E^{\prime}_{\beta}$
if $E^{\prime}_{\beta}$ is locally complete) defined by

$T(a_{0}e_{0}+\cdots+a_{n}e_{n}):=a_{0}\chi_{0}+\cdots+a_{n}\chi_{n}\quad(n\in\omega,\;a_{0},\dots,a_{n}\in\mathbb{F}).$

is continuous. For every $n\in\omega$ and each $x\in E$, we have $\langle
T^{\ast}{\restriction}_{E}(x),e_{n}\rangle=\langle
T(e_{n}),x\rangle=\langle\chi_{n},x\rangle$ and hence
$T^{\ast}{\restriction}_{E}(x)=\big{(}\langle\chi_{n},x\rangle\big{)}_{n}\in\ell_{\infty}$.
In particular,
$\|T^{\ast}{\restriction}_{E}(x)\|_{\ell_{\infty}}=\sup_{n\in\omega}|\langle\chi_{n},x\rangle|$.

Let now $\\{x_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in $E$.
Since $B$ and hence also $S$ are $(p,q)$-$(V)$ sets we obtain
$\big{(}\|T^{\ast}{\restriction}_{E}(x_{n})\|_{\ell_{\infty}}\big{)}=\big{(}\sup_{i\in\omega}|\langle\chi_{i},x_{n}\rangle|\big{)}\in\ell_{q}$
(or $\in c_{0}$ if $q=\infty$). Therefore $T^{\ast}{\restriction}_{E}$ is
$(q,p)$-convergent, and hence, by (ii) or (iv), $T$ is weakly sequentially
compact (resp., sequentially compact, weakly sequentially $p$-compact or
weakly sequentially $p$-precompact). Therefore $S$ has a weakly convergent
(resp., convergent, weakly $p$-convergent, or weakly $p$-Cauchy) subsequence,
as desired.

(iii)$\Rightarrow$(i) Let $T:L\to E^{\prime}_{\beta}$ be an operator from a
normed space such that $T^{\ast}{\restriction}_{E}:E\to L^{\prime}_{\beta}$ is
a $(q,p)$-convergent linear map. Then, by Theorem 16.5, $T(B_{L})$ is a
$(p,q)$-$(V)$ set and hence it is relatively weakly sequentially compact
(resp., relatively sequentially compact, relatively weakly sequentially
$p$-compact or weakly sequentially $p$-precompact). Thus $T$ is weakly
sequentially compact (resp., sequentially compact, weakly sequentially
$p$-compact or weakly sequentially $p$-precompact).

We need the following result which is similar to Theorem 16.5.

###### Theorem 16.8

Let $1\leq p\leq q\leq\infty$, $E$ be a locally convex space, and let $T$ be
an operator from a normed space $L$ to $E$. Then $T(B_{L})$ is a
$(p,q)$-$(V^{\ast})$ subset of $E$ if and only if
$T^{\ast}:E^{\prime}_{\beta}\to L^{\prime}_{\beta}$ is $(q,p)$-convergent.

###### Proof 127

First we note that for every $\chi\in E^{\prime}$, we have

$\|T^{\ast}(\chi)\|_{L^{\prime}_{\beta}}=\sup_{y\in B_{L}}|\langle
T^{\ast}(\chi),y\rangle|=\sup_{y\in B_{L}}|\langle\chi,T(y)\rangle|.$ (16.3)

Let $\\{\chi_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in
$E^{\prime}_{\beta}$. Then, by (16.3), we have
$\big{(}\|T^{\ast}(\chi_{n})\|_{L^{\prime}_{\beta}}\big{)}_{n\in\omega}=\big{(}\sup_{y\in
B_{L}}|\langle\chi_{n},T(y)\rangle|\big{)}_{n\in\omega}$. Now the theorem
follows from the definition of $(p,q)$-$(V^{\ast})$ sets and the definition of
$(q,p)$-convergent operators.

###### Theorem 16.9

Let $1\leq p\leq q\leq\infty$, and let $E$ be a locally convex space. Then the
following assertions are equivalent:

  1. (i)

if $L$ is a normed space and $T:L\to E$ is an operator such that
$T^{\ast}:E^{\prime}_{\beta}\to L^{\prime}_{\beta}$ is $(q,p)$-convergent,
then $T$ is weakly sequentially compact (resp., sequentially compact, weakly
sequentially $p$-compact or weakly sequentially $p$-precompact);

  2. (ii)

the same as (i) with $L=\ell_{1}^{0}$;

  3. (iii)

each $(p,q)$-$(V^{\ast})$ subset of $E$ is relatively weakly sequentially
compact (resp., relatively sequentially compact, relatively weakly
sequentially $p$-compact or weakly sequentially $p$-precompact).

Moreover, if $E$ is locally complete, then (i)-(iii) are equivalent to

  1. (iv)

the same as (i) with $L=\ell_{1}$.

###### Proof 128

(i)$\Rightarrow$(ii) and (i)$\Rightarrow$(iv) are clear.

(ii)$\Rightarrow$(iii) and (iv)$\Rightarrow$(iii): Let $A$ be a
$(p,q)$-$(V^{\ast})$ subset of $E$. Fix an arbitrary sequence
$S=\\{x_{n}\\}_{n\in\omega}$ in $A$, so $S$ is a bounded subset of $E$.
Therefore, by Proposition 14.9, the linear map $T:\ell_{1}^{0}\to E$ (or
$T:\ell_{1}\to E$ if $E$ is locally complete) defined by

$T(a_{0}e_{0}+\cdots+a_{n}e_{n}):=a_{0}x_{0}+\cdots+a_{n}x_{n}\quad(n\in\omega,\;a_{0},\dots,a_{n}\in\mathbb{F}).$

is continuous. For every $n\in\omega$ and each $\chi\in E^{\prime}$, we have
$\langle
T^{\ast}(\chi),e_{n}\rangle=\langle\chi,T(e_{n})\rangle=\langle\chi,x_{n}\rangle$
and hence
$T^{\ast}(\chi)=\big{(}\langle\chi,x_{n}\rangle\big{)}_{n}\in\ell_{\infty}$.
In particular,
$\|T^{\ast}(\chi)\|_{\ell_{\infty}}=\sup_{n\in\omega}|\langle\chi,x_{n}\rangle|$.

Let now $\\{\chi_{n}\\}_{n\in\omega}$ be a weakly $p$-summable sequence in
$E^{\prime}_{\beta}$. Since $A$ and hence also $S$ are $(p,q)$-$(V^{\ast})$
sets we obtain
$\big{(}\|T^{\ast}(\chi_{n})\|_{\ell_{\infty}}\big{)}=\big{(}\sup_{i\in\omega}|\langle\chi_{n},x_{i}\rangle|\big{)}\in\ell_{q}$
(or $\in c_{0}$ if $q=\infty$). Therefore $T^{\ast}$ is $(q,p)$-convergent,
and hence, by (ii) or (iv), $T$ is weakly sequentially compact (resp.,
sequentially compact, weakly sequentially $p$-compact or weakly sequentially
$p$-precompact). Therefore $S=\\{T(e_{n})\\}_{n\in\omega}$ has a weakly
convergent (resp., convergent, weakly $p$-convergent, or weakly $p$-Cauchy)
subsequence, as desired.

(iii)$\Rightarrow$(i) Let $T:L\to E$ be an operator from a normed space such
that $T^{\ast}:E^{\prime}_{\beta}\to L^{\prime}_{\beta}$ is a
$(q,p)$-convergent operator. Then, by Theorem 16.8, $T(B_{L})$ is a
$(p,q)$-$(V^{\ast})$ set and hence it is relatively weakly sequentially
compact (resp., relatively sequentially compact, relatively weakly
sequentially $p$-compact or weakly sequentially $p$-precompact). Thus $T$ is
weakly sequentially compact (resp., sequentially compact, weakly sequentially
$p$-compact or weakly sequentially $p$-precompact).

To obtain another characterization of $(p,q)$-$(V^{\ast})$ sets we need the
next assertion. For a set $A$, we denote by $\ell^{0}_{p}(A)$ the subspace of
the Banach space $\ell_{p}(A)$ consisting of all vectors with finite support.

###### Proposition 16.10

Let $A$ be a bounded subset of a locally convex space $(E,\tau)$,
$B=\overline{\mathrm{acx}}(A)$, and let $i_{B}:(E_{B},\|\cdot\|)\to E$ be the
identity operator. Let $T_{A}:\ell_{1}^{0}(A)\to(E_{B},\|\cdot\|)$ be the
identity inclusion defined by

$T_{A}(\lambda_{0}a_{0}+\cdots+\lambda_{n}a_{n}):=\lambda_{0}a_{0}+\cdots+\lambda_{n}a_{n}\quad(n\in\omega,\;\lambda_{0},\dots,\lambda_{n}\in\mathbb{F},\;a_{0},\dots,a_{n}\in
A).$

Then $T_{A}$ is bounded. Moreover, if $E$ is locally complete, then $T_{A}$
can be extended to a continuous operator ${\tilde{T}}_{A}$ from $\ell_{1}(A)$
to $(E_{B},\|\cdot\|)$. In particular, $i_{B}\circ T_{A}:\ell_{1}^{0}(A)\to E$
(or $i_{B}\circ{\tilde{T}}_{A}:\ell_{1}(A)\to E$ if $E$ is locally complete)
is bounded and hence continuous.

###### Proof 129

First we recall that, by Proposition 3.6, $i_{B}$ is indeed an operator.

By the definition of $(E_{B},\|\cdot\|)$ we observe that $\|b\|\leq 1$ for
every $b\in B$. Therefore, if $|\lambda_{0}|+\cdots+|\lambda_{n}|\leq 1$ and
$a_{0},\dots,a_{n}\in A$, we obtain

$\big{\|}T_{A}(\lambda_{0}a_{0}+\cdots+\lambda_{n}a_{n})\big{\|}=\|\lambda_{0}a_{0}+\cdots+\lambda_{n}a_{n}\|\leq
1$

and hence $T:\ell_{1}^{0}(A)\to(E_{B},\|\cdot\|)$ is bounded.

Assume that $E$ is locally complete. Then, by Proposition 5.1.6 of [72], the
space $E_{B}$ is a Banach space. Therefore the operator
$T:\ell_{1}^{0}(A)\to(E_{B},\|\cdot\|)$ can be extended to an operator
${\tilde{T}}_{A}:\ell_{1}(A)\to(E_{B},\|\cdot\|)$.

###### Theorem 16.11

Let $1\leq p\leq q\leq\infty$. Then a subset $A$ of a locally convex space $E$
is a $(p,q)$-$(V^{\ast})$ set if and only if the adjoint operator
$T_{A}^{\ast}:E^{\prime}_{\beta}\to\ell_{\infty}(A)$ is $(q,p)$-convergent.

###### Proof 130

By Proposition 16.10, every bounded subset $A$ of $E$ defines an operator
$T_{A}:\ell_{1}^{0}(A)\to E$ by

$T_{A}(\lambda_{0}a_{0}+\cdots+\lambda_{n}a_{n}):=\lambda_{0}a_{0}+\cdots+\lambda_{n}a_{n}\quad(n\in\omega,\;\lambda_{0},\dots,\lambda_{n}\in\mathbb{F},\;a_{0},\dots,a_{n}\in
A).$

Observe that for each $\chi\in E^{\prime}$, the $a$th coordinate
$T_{A}^{\ast}(\chi)(a)$ of $T_{A}^{\ast}(\chi)$ is

$T_{A}^{\ast}(\chi)(a)=\langle
T_{A}^{\ast}(\chi),a\rangle=\langle\chi,T_{A}(a)\rangle=\langle\chi,a\rangle,$

and hence

$\|T^{\ast}_{A}(\chi)\|_{\ell_{\infty}(A)}=\sup_{a\in
A}|T^{\ast}_{A}(\chi)(a)|=\sup_{a\in A}|\langle\chi,a\rangle|$ (16.4)

Now, by definition, a subset $A$ of $E$ is a $(p,q)$-$(V^{\ast})$ set if and
only if $\big{(}\sup_{a\in A}|\langle\chi_{n},a\rangle|\big{)}\in\ell_{q}$ (or
$\in c_{0}$ if $q=\infty$) for every weakly $p$-summable sequence
$\\{\chi_{n}\\}_{n\in\omega}$ in $E^{\prime}_{\beta}$, and hence, by (16.4),
if and only if
$\big{(}\|T^{\ast}_{A}(\chi_{n})\|_{\ell_{\infty}(A)}\big{)}\in\ell_{q}$ (or
$\in c_{0}$ if $q=\infty$) for every weakly $p$-summable sequence
$\\{\chi_{n}\\}_{n\in\omega}$ in $E^{\prime}_{\beta}$, i.e., $T^{\ast}_{A}$ is
a $(q,p)$-convergent operator.

##  17 $(q,p)$-summing operators and applications

Let $1\leq p\leq q\leq\infty$. Recall (see Definition 16.1) that an operator
$T$ from an lcs $E$ to an lcs $L$ is $(q,p)$-convergent if it sends weakly
$p$-summable sequences in $E$ to strongly $q$-summable sequences in $L$. Using
this class of operators we characterized in Theorem 16.11 $(p,q)$-$(V^{\ast})$
subsets of $E$. Therefore it is natural to find some classes of operators from
$E$ to $L$ which are $(q,p)$-convergent. If $E$ and $L$ are Banach spaces it
is well known (see [24]) that the class of $(q,p)$-convergent operators
coincides with the class of $(q,p)$-summing operators (a detailed proof of a
more general result is given in Corollary 17.7 below). In this section we
define $(q,p)$-summing operators between arbitrary locally convex spaces, show
that the most important results for Banach spaces (as the Pietsch Domination
Theorem and the Pietsch factorization Theorem) remain true and show that any
$(q,p)$-summing operator is $(q,p)$-convergent.

First we recall the definition of $(q,p)$-summing operators between Banach
spaces which plays an essential role in many areas of Banach Space Theory, see
the classical books [24] and [62].

###### Definition 17.1

Let $1\leq p\leq q<\infty$. An operator $T:X\to Y$ between Banach spaces is
called $(q,p)$-summing if there is a constant $C>0$ such that for every
$n\in\omega$ and each $x_{0},\dots,x_{n}\in X$, we have

$\Big{(}\sum_{i=0}^{n}\|T(x_{i})\|^{q}\Big{)}^{1/q}\leq
C\cdot\sup\Big{\\{}\Big{(}\sum_{i=0}^{n}|\langle\chi,x_{i}\rangle|^{p}\Big{)}^{1/p}:\chi\in
B_{X^{\prime}}\Big{\\}}.$ (17.1)

The least $C$ for which the inequality (17.1) always holds is denoted by
$\pi_{q,p}(T)$. We shall write $\prod_{q,p}(X,Y)$ for the family of all
$(q,p)$-summing operators from $X$ into $Y$. If $q=p$ we shall say that $T$ is
$p$-summing and set $\prod_{p}(X,Y):=\prod_{p,p}(X,Y)$ and
$\pi_{p}(T)=\pi_{p,p}(T)$.

Since the both sides in (17.1) are homogenous, one can assume that all vectors
$x_{0},\dots,x_{n}$ belongs to the closed unit ball $B_{X}$ of $X$. Observe
also that $B_{X}$ is an absolutely convex closed neighborhood of $0\in X$
(that is $B_{X}\in\mathcal{N}_{0}^{c}(X)$) and $B_{X^{\prime}}=B_{X}^{\circ}$.
Therefore the notion of $(q,p)$-summing operator between Banach spaces is