TITLE: Scottish Book Problem 172
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: The answer to both Q1 (in the context given) and Q2 is yes.
Regarding the weak closure of an arbitrary set of linear functional being linear that appears to me to be sequential weak-star convergence — that is, if they converge sequentially in the weak-star topology, then they converge in the weak topology on the dual.