# Scottish Book Problem 172

The problem is formulated using old terminology and I want to understand what it actually says.

The problem reads: "A space $$E$$ of type (B) has the property (a) if the weak closure of an arbitrary set of linear functionals is weakly closed. [A sequence of linear functionals $$f_n(x)$$ converges weakly to $$f(x)$$ if $$f_n(x) \to f(x)$$ for every $$x$$.] The space $$E$$ of type (B) has the property (b) if every sequence of linear functionals weakly convergent converges weakly as a sequence of elements in the conjugate space $$\bar{E}$$. Question: Does every separable space of type (B) which has property (a) also possess property (b)?"

I know that "space of type (B)" means "Banach space" in modern terminology.

Q1: Does "linear functional" include the assumption of continuity?

Q2: What is the conjugate space? Does it mean the dual space of continuous linear functionals on $$E$$?

If the answers to Q1 and Q2 are yes, then this would mean Banach spaces with property (b) are exactly what nowadays are called Grothendieck spaces. [A Grothendieck space is a Banach space $$E$$ such that every sequence in the dual space $$E^*$$ that converges weak* - that is, with respect to $$\sigma(E^*,E)$$ - also converges weakly - that is, with respect to $$\sigma(E^*,E^{**})$$.]

I am also unsure how to understand the condition "the weak closure of an arbitrary set of linear functionals is weakly closed", as this seems like a tautology. Is the point here that we consider a sequential closure?

• Thank you. So the problem is: Assume that $E$ is a Banach space such that the weak* sequential closure of every subset of $E^*$ is weak* sequentially closed. Is $E$ a Grothendieck space? – Hannes Thiel Oct 8 at 7:47
• The terminology "weak* sequential closure" is counterintuitive. Tthe weak* sequential closure of a set $X\subset E^*$, lets denote it by $c(X)$, is the set of all limits of weak* convergent sequences in $X$. It is not necessarily true that $c(c(X))=c(X)$. Thus, the weak* sequential closure need not be weak* sequentially closed. I found a paper of Godun (Weak* derivatives of sets of linear functionals), which shows that a separable Banach space $E$ is quasireflexive iff every subspace $F\subseteq E^*$ satisfies $c(c(F))=c(F)$. Thus, separable Banach spaces with (a) are quasireflexive. – Hannes Thiel Oct 8 at 12:50
• In the same paper, it is also shown that a Banach space $E$ is Grothendieck if and only if $c(F)=F$ for every norm-closed subspace $F\subset E^*$. Further, for every (not necessarily norm-closed) subspace $F\subset E^*$ with norm-closure $\overline{F}$, we have $c(F)=c(\overline{F})$. – Hannes Thiel Oct 8 at 13:03