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Jun 8, 2021 at 14:53 comment added Nik Weaver You would need $D$ to be continuous in order to evaluate it on the sum of an infinite series. You are specifically asking for a discontinuous derivation.
Jun 8, 2021 at 12:42 history edited Ken.Wong CC BY-SA 4.0
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Jun 8, 2021 at 10:55 history edited Ken.Wong
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Jun 7, 2021 at 7:56 comment added Narutaka OZAWA @Ken.Wong: That's the point; $t$ and $e^t$ algebraically generate an isomorph of ${\mathbb C}[{\mathbb Z}^2]$. Hence any assignment of $D(t)$ and $D(e^t)$ defines a derivation $D$ on that algebra.
Jun 7, 2021 at 7:51 comment added Ken.Wong @NarutakaOZAWA Why we can set $D(e^{t})=0$? If we expand $e^{t}$, then by linearity of $D$, $D(e^{t})=e^{t}$.
Jun 6, 2021 at 23:24 comment added Narutaka OZAWA Apparently, something is missing in that paper. Anyway, here's a simple example of a non closable derivation $D$ on $C[0,1]$. Put $D(t)=1$ and $D(e^t)=0$ and extend $D$ on the algebra generated by $t$ and $e^t$.
Jun 5, 2021 at 8:54 comment added Matthew Daws I think it is just a typo. In the linked PDF, the numbering jumps from Corollary 13 to Theorem 17. Notice that "Proposition 15" is referenced both in the proof of Corollary 13, and in the introduction to Section C (which is presumably where you are reading). My guess it that Proposition 12 used to be Proposition 15 (and so Corollary 16, then Theorem 17) and the numbering was changed at the last moment and not fully corrected. But I cannot see (quickly) how to make sense of the $\delta_0$ constructed in Prop 12 to obtain what is claimed.
Jun 4, 2021 at 16:45 history asked Ken.Wong CC BY-SA 4.0