In Sakai note, on the fourth part differentiation. Sakai stated the following:

It is an open question whether the result can be extended to $n=2,3,...$

What $n$ he is referring to? Also Sakai stated a conjecture:

Let $\delta_{0}$ be a closed *-derivation in $U$ and let $\delta$ be a *-derivation in $U$ with $D(\delta_{0})=D(\delta)$. Then $\delta$ is closable....

If $\delta_{0}$ is closed, isn't it a bounded operator? If $D(\delta_{0})=D(\delta)$, it means $D(\delta)$ is the whole space, which implies boundedness and closable. Am I missing something here?

Edit For the first question, what I really to know is which result he is referred to? Is it the unit interval one($D(\delta)=C^{n}(I)$) or the compact Hausdorff space one($\cap^{\infty}_{n=1}D(\delta_{0}^{n})$)? Are these problem solved?

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    $\begingroup$ There are closed unbounded operators as, for example, all self-adjoint operators. However, if the domain of $\delta_0$ is $U$, it follows from the closed graph theorem that $\delta_0$ is bounded. $\endgroup$ – Romain Gicquaud May 4 at 9:55
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    $\begingroup$ ``$C^n(I)$ ($n$ times continuously differentiable)'' $\endgroup$ – Nik Weaver May 4 at 11:23
  • $\begingroup$ ``Suppose that $\delta_0, \delta$ are two $*$-derivations in $C(K)$ $\ldots$ then there is a unique continuous function $\lambda$ on $K$ such that $\delta = \lambda \delta_0$. In particular, $\delta$ is closable. It is an open question whether the result can be extended to $n = 2, 3, \ldots$.'' $\endgroup$ – Nik Weaver May 4 at 15:57
  • $\begingroup$ The result stated for $C^n(I)$ is not open for $n = 2, 3, \ldots$, as it is stated, and a reference is given, for ``$n = \infty, 1, 2, 3, \ldots$''. $\endgroup$ – Nik Weaver May 4 at 16:01

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