This is sort of a follow-up to Borel(X) = \sigma(X') for X non-separable

PROBLEM: Given a Banach space $E$ over $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$ that has the Grothendieck property. Does $\hat{C}(E) = \mathcal{B}(E)$ imply $E$ is reflexive? (This would in turn imply that $E$ is separable).

Some definitions:

A Banach space is a Grothendieck space if a sequence in $E'$ which is $\sigma(E', E)$-convergent is automatically $\sigma(E', E'')$-convergent. Or equivalently: every $\sigma(E', E)$ zero sequence has subsequence which is $\sigma(E', E'')$-convergent, or equivalently: every linear, bounded operator from $E$ to $c_0$ (or any separable Banach space) is automatically weakly compact.

The $\sigma$-algebra $\hat{C}(E)$ is the $\sigma$-algebra generated by sets of the form $\mathcal{C}_{u_1, \cdots, u_n; C} := \{x \in E : (u_1(x), \cdots, u_n(x)) \in C\}$ where $u_1, \cdots, u_n \in E'$, $C \in \mathcal{B}(\mathbb{K}^n)$ and $n \in \mathbb{N}$.

Notes:

The $\sigma$-algebra $\hat{C}(E)$ equals the $\sigma$-algebra of weak Baire sets $\mathcal{B}_0(E, \sigma(E, E'))$ for every locally convex space $E$ (see [2], Theorem 2.3).

The inclusion $\hat{C}(E) \subset \mathcal{B}(E)$ is trivially true. If $E$ is separable then $\hat{C}(E) = \mathcal{B}(E)$. [To see this use the Hahn-Banach theorem to show that $\mathcal{B}_E \in \hat{C}(E)$. As translations and scalar multplications are measurable with regard to the cylindrical $\sigma$-algebra the other inclusion follows.]

A reflexive space is automatically Grothendieck.

For a separable Grothendieck space $E$ we have that the identity is weakly compact so $E$ becomes reflexive

A reflexive space $E$ with $\hat{C}(E) = \mathcal{B}(E)$ is automatically separable ([1], Prop. 2.6, p.19). Without reflexivity the equality $\hat{C}(E) = \mathcal{B}(E)$ does not imply $E$ is separable in general.

The example $E = \ell^2(\mathbb{R})$ shows that there is a reflexive and non-separable space with $\mathcal{B}(E) \not= \hat{C}(E)$

Edgar's example below or $E = \ell^{\infty} = C(\beta \mathbb{N})$ gives a non-reflexive Grothendieck space with $\mathcal{B}(E) \not= \hat{C}(E)$. The question therefore: does every non-reflexive Grothendieck space have that property?

There are non-reflexive Grothendieck spaces which do not contain $\ell^{\infty}$ (cf. [3]). So we can't simply reduce to this case. I don't know much more about Grothendieck spaces though or characterizations of them that might be helpful.

REFERENCES:

[1] N. N. VAKHANIA, V. I. TARIELADZE, S. A. CHOBANYAN, Probability Distributions on Banach Spaces, Mathematics and its applications (D. Reidel Publishing Company), 1987

[2] http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1977/26/26053

[3] R. HAYDON, A non-reflexive Grothendieck Space that does not contain $\ell^{\infty}$, Israel Journal of Mathematics, Vol 40, No. 1, 1981

EDIT: I rephrased the question and added some information.