# Are biduals of spaces of differentiable functions on hypercubes Grothendieck?

Consider the space $$E_n = C^1([0,1]^n)$$ of continuously differentiable functions with the usual norm $$\max\{ \|f\|_\infty, \|f^\prime_{x_1}\|_\infty, \ldots, \|f^\prime_{x_n}\|_\infty\}.$$ making it a Banach space. (The norm with the subscript $$\infty$$ stands for the supremum norm.)

Inspired by the recent question:

Jean Bourgain's Relatively Lesser Known Significant Contributions

let me point out that Bourgain proved in 1983 that the dual of $$E_n$$ is weakly sequentially complete for any $$n$$.

Moreover, it is an exercise on Taylor's theorem that $$E_1$$ is isomorphic to $$C[0,1]$$ in which case the dual of $$E_1$$ is an abstract $$L_1$$-space so it is weakly sequentially complete and $$E_1^{**}$$ a Grothendieck space.

Is $$E_n^{**}$$ a Grothendieck space for every $$n$$?

Note that this, if true, would imply Bourgain's result as duals of Grothendieck spaces are weakly sequentially complete so here even the tridual of $$E_n$$ would have this property.

Tracking citations of Bourgain's paper gives evidence that this is an open problem but it is not inconceivable that it follows from a combination of results proved later.

• Link to MathReview for Bourgain's paper: mathscinet.ams.org/mathscinet-getitem?mr=712065 – Yemon Choi Jan 3 at 20:40
• I think you might want to define $E_n$ as $C^1([0,1]^n)/\mathbb{C}$ (i.e. consider everything modulo the constant functions and leave out the sup-norm). With this definition, I agree that $E_1\approx C([0,1])$. I think I'm right to say that for $n>1$, $E_n$ is the space of all closed 1-forms on $[0,1]^n$ with continuous coefficients (`closed' being interpreted in the distributional sense). Just a thought! – DCM Jan 4 at 12:22
• DCM, as it is customary in Banach space theory we are talking about non necessarily isometric isomorphisms so quotienting by the constants is not really needed. (After all, spaces considered here are isomorphic to their hyperplanes.) As for the distributional picture of $E_n$ you are completely right. – Tomek Kania Jan 4 at 12:26