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The notion of a strongly regular Banach space was introduced and studied in [Some topological and geometrical structures in Banach spaces, Ghoussoub et al., Memoirs of the American Mathematical Society, (1987), No. 378], see here.

A Banach space $X$ is called strongly regular if for every $\varepsilon>0$ and every nonempty convex bounded subset $C\subset X$ there exist positive reals $t_1,\dots, t_k$ with $\sum_{i=1}^n t_i=1$ and nonempty relatively weak-open subsets $U_1,\dots ,U_n \subset C$ such that the norm diameter of $\sum_{i=1}^n t_i U_i$ is less than $\varepsilon$.

Let $C^k(M)$ be the Banach space of the $k$-times continuously differentable real-valued functions on a smooth compact manifold $M$ with the usual norm. I wish to show that $C^k(M)$ is not strongly regular (because it is an assumption in a theorem I would like to quote). Is this known?

I suspect that no Banach space that contains an isomorphic copy of $c_0$ is strongly regular. Is this true?

Is there a slick way to see that $C^k(M)$ contains a copy of $c_0$? I think I can prove by hand by embedding $c_0$ to $C([0,1])$, and then embedding the latter into $C^k(M)$ by integrating $k$ times and using spherical coordinates, but I would rather quote a reference.

Disclaimer: Banach spaces in not my area of expertise.

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    $\begingroup$ For an embedded $c_0$, how about this? Choose countably many pairwise disjoint open sets $U_n \subset M$. For each one, let $f_n \in C^\infty_c(U_n)$ with $C^k$-norm one. Consider the map which sends $(a_n) \in c_0$ to $\sum a_n f_n$. $\endgroup$
    – Nate Eldredge
    Commented Mar 5, 2017 at 21:12
  • $\begingroup$ Oh wait, that still needs a little work. Take $k=1$ for instance. If $|f_1|$ is small and $|\nabla f_1|$ is large, but for $f_2$ it is the other way around, then $f_1 + f_2$ might have $C^1$ norm larger than 1. $\endgroup$
    – Nate Eldredge
    Commented Mar 5, 2017 at 21:41
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    $\begingroup$ @NateEldredge: Thank you. Your argument does work for the maximum norm $\|f\|=\max_{j,\alpha} |f^\alpha |_{C(D_j)}$ where $\alpha$ is a multi-index of order $\le k$, and $\{D_j\}$ is a finite cover of $M$ by compact disks each lying in a coordinate chart. We can assume that your sets $U_n$ lie in exactly one disk $D_j$. $\endgroup$
    – Igor Belegradek
    Commented Mar 5, 2017 at 23:45

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$C^k(M)$ is isomorphic to $C(M)$ and hence, by Milutin's Theorem, to $C[0,1]$. The first statement is more or less obvious since the norm on $C^k(M)$ is equivalent on a $k$-codimensional subspace to the sup norm of the $k$-th derivative and $C(M)$ contains a (necessarily complemented) subspace isomorphic to $c_0$.

I don't remember the definition of strongly regular and don't have the Ghoussoub et al Memoir here to look it up.

EDIT 3/6/17: As Mikhail Ostrovskii mentioned in a comment, $C^k(M)$ is NOT isomorphic to $C(M)$ when the dimension of $M$ is two or more, at least for some $M$.

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  • $\begingroup$ Strongly regular is an isomorphic property that passes to subspaces. $C[0,1]$ is isomorphically universal for separable spaces and there exist non strongly regular separable spaces ($L_1(0,1)$ is the most obvious, I guess), so $C[0,1]$ is not strongly regular. $\endgroup$
    – Bill Johnson
    Commented Mar 5, 2017 at 22:01
  • $\begingroup$ Thank you, this explains a number of things for me. I am still struggling making rigorous the isomorphism of $C^k(M)$ and $C(M)$. The $k$th derivative does not quite make sense when $M$ has dimension $>1$. But ultimately all one needs is a copy of $c_0$ inside $C^k(M)$ which is easily produced. $\endgroup$
    – Igor Belegradek
    Commented Mar 5, 2017 at 23:47
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    $\begingroup$ I would like to mention in this connection that Kislyakov [Funkcional. Anal. i Priložen. 9 (1975), no. 4, 22–27] proved that the space $C^{\ell}(I^n)$ of all $\ell$-times continuously differentiable functions on the $n$-dimensional cube $(\ell≥1,n≥2)$ is not isomorphic to any quotient space of any space $C(S)$. $\endgroup$
    – Mikhail Ostrovskii
    Commented Mar 6, 2017 at 5:49
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    $\begingroup$ @Igor: Sorry about that. Of course Mikhail is correct. I made a correction to my answer. $\endgroup$
    – Bill Johnson
    Commented Mar 6, 2017 at 15:42

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