# 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?

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.

• 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 '20 at 7:47
• @Hanned Thiel: Um, that is tautological. I had presumed the statement in square bracket was to help understand property (a), but my understanding of it, as written above, then coincides precisely with the definition of Grothendieck space (as given by property (b)). I am now thinking the information in the square brackets is for property (b). Granted that the problem seems to be asking “Is a separable Banach space with property (a) a Grothendieck space”, I am tempted to understand property (a) as “spaces for which weak-star topology on dual coincides with weak topology”...(cont’d) – Jack L. Oct 8 '20 at 9:58
• .... however, that would be seemingly a backward question, as it should more likely be “Does a separable Grothendieck space have the weak-star topology coincident with the weak topology?”— i.e., property (b) implies (a). Thus, property (a) was reflexivity (if it wasn’t defined as such at the time), where the weak-star topology coincides with the weak topology on the dual. This then would make sense, as the answer may not have been known then until the work of Eberlein and Smulian on weak compactness. In any case, that’s my best guess, until someone with more familiarity can say otherwise. – Jack L. Oct 8 '20 at 10:05
• 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 '20 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 '20 at 13:03