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Are the $C^k[0,1]$ (k times differentiable real valued functions on the interval), with the $C^k$ norm, distinct as Banach spaces?

I know $C^k[0,1] \cong \mathbb{R}^k \oplus C^0[0,1]$, but I think this isomorphism is at best a homeomorphism, and not a isometry. It's not clear to me how to compare the different $\mathbb{R}^k \oplus C^0[0,1]$ for varying $k$. Is it possible for $C^0[0,1]$ to "absorb" copies of $\mathbb{R}$, maybe via some good description of $C^0$?

Just curious.

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    $\begingroup$ As I recall it is closer to being a product of $k+1$ copies of $C^0$. It's not exactly that, either, but it's more like that. $\endgroup$ – Ian Jan 17 '16 at 16:35
  • $\begingroup$ Whether it's an isometry depends on the norms you choose. One of the standard norms on $C^k([0,1])$ is $$\lVert f\rVert = \sum_{n = 0}^{k-1} \lvert f^{(n)}(0)\rvert + \lVert f^{(k)}\rVert_{\infty},$$ then taking the sum-norm on $\mathbb{R}^k \oplus C^0([0,1])$ gets you an isometry. $\endgroup$ – Daniel Fischer Jan 17 '16 at 16:36
  • $\begingroup$ @Ian the map is via Taylors theorem. $\endgroup$ – Lorenzo Jan 17 '16 at 16:56
  • $\begingroup$ @DanielFischer Thanks. I agree it is an isometry in that case. Do you happen to know if $\mathbb{R} \oplus C^0$ and $C^0$, with the norm $|x| + sup f$ and $sup f$ respectively, are distinct Banach spaces? $\endgroup$ – Lorenzo Jan 17 '16 at 16:58
  • $\begingroup$ If I knew that, I would have posted an answer. I strongly suspect they aren't isomorphic, but I don't see a distinguishing property yet. $\endgroup$ – Daniel Fischer Jan 17 '16 at 16:59
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The unit ball of $C^0([0,1])$ has only two extreme points (the constant functions with value $\pm 1$) but the one of $\mathbb R \oplus C^0([0,1])$ with the norm $|t|+ \|f\|_\infty$ has (at least) four. Therefore, the spaces are not isometric.

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