Hilbert algebra definition clarification

Question

Consider the following fragment from Takesaki's second volume:

In (c), we read that the involution $\sharp: \mathfrak{A}\to \mathfrak{A}$ is preclosed. What does this mean?

Answer 1

Preclosed is a (maybe somewhat dated) synonym of closable in this context. Here is a short crash course:

A (linear) operator $T\colon D(T)\to X$, where $D(T)$ is a subspace of the Banach space $X$, is called closed if the graph $$ G(T)=\{(x,y)\in D(T)\times X\mid y=Tx\} $$ is a closed subspace of $X\times X$. An operator $T\colon D(T)\to X$ is called closable (or preclosed) if there exists a closed operator $S\colon D(S)\to X$ such that $D(T)\subset D(S)$ and $Sx=Tx$ for $x\in D(T)$.

There are a couple of equivalent descriptions: The operator $T$ is closed if and only if whenever $(x_n)$ is a sequence in $D(T)$ such that $x_n\to x$ and $Tx_n\to y$, one has $x\in D(T)$ and $Tx=y$. It is closable if and only if whenever $(x_n)$ is a sequence in $D(T)$ suc that $x_n\to 0$ and $Tx_n\to y$, one has $y=0$.

Equivalently, $T$ is closable if and only if the closure of its graph $G(T)$ in $X\times X$ is again the graph of a linear operator. Every closable operator $T$ has a smallest closed extension, called its closure and denoted by $\overline T$, which is the operator whose graph is the closure of $G(T)$.

If $X$ is a Hilbert space, then a densely defined operator $T$ is closable if and only if $T^\ast$ is densely defined, in which case $\overline T=T^{\ast\ast}$.

In the context of Hilbert algebras, the relevant Hilbert space is the completion of $\mathfrak A$ with respect to the given inner product.

As pointed out by QuantumSpace in the comments, the involution $\sharp$ is not linear but rather anti-linear. The definition of closable and closed anti-linear operator is essentially the same.