Timeline for Is the Banach space of continuously differential functions strongly regular?

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Mar 6, 2017 at 15:42	history	edited	Bill Johnson	CC BY-SA 3.0	added 165 characters in body
Mar 6, 2017 at 15:42	comment	added	Bill Johnson		@Igor: Sorry about that. Of course Mikhail is correct. I made a correction to my answer.
Mar 6, 2017 at 5:49	comment	added	Mikhail Ostrovskii		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 $C(S)$.
Mar 5, 2017 at 23:47	comment	added	Igor Belegradek		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.
Mar 5, 2017 at 22:01	comment	added	Bill Johnson		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.
Mar 5, 2017 at 21:38	history	answered	Bill Johnson	CC BY-SA 3.0