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Can we construct an unbounded derivation on abelian C* algebra which is not closable?

One of possible construction may be found in the paper by Bratteli and Robinson(Unbounded derivations of C*-algebras). In the paper they construct $\delta_{0}$ in theorem 15. The only question is that there are no "theorem 15" in this paper. I doubt this is a typo. The only related theorem seems to be theorem 12, but $\delta_{0}$ there is differentiation, which is closable. So I don't know if that counts.

Edit In the comments @Narutaka_OZAMA provide an example of non-closed derivation by setting $D(t)=1$, $D(e^{t})=0$. However I fail to see why this example is well defined. If $D(t)$ is fixed, using Taylor expansion of $e^{t}$, we can act $D$ term by term is $D$ is linear and the norm of each term is finite, thus $D(e^{t})=e^{t}\neq0$. Can anyone further explain to me?

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    $\begingroup$ 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. $\endgroup$
    – Matthew Daws
    Commented Jun 5, 2021 at 8:54
  • $\begingroup$ 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 $e^t$. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 6, 2021 at 23:24
  • $\begingroup$ @NarutakaOZAWA Why we can set $D(e^{t})=0$? If we expand $e^{t}$, then by linearity of $D$, $D(e^{t})=e^{t}$. $\endgroup$
    – Ken.Wong
    Commented Jun 7, 2021 at 7:51
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    $\begingroup$ @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. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 7, 2021 at 7:56
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    $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Commented Jun 8, 2021 at 14:53

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