symmetric monoidal (∞,1)-category of spectra
AQFT and operator algebra
$C^*$-coalgebras are like $C^*$-algebras, but coalgebras. Their duals are $W^*$-algebras.
Let $A$ be a Banach *-coalgebra over the ground field $K$. Let $\lambda\colon A \to K$ be a bounded linear functional on $A$, and let $\lambda^*$ be the composite
(where $\bar{}$ is complex conjugation, trivial if $K$ is real?); although some of the maps in this composite may be only antilinear, the composite $\lambda^*$ is linear (over all of $K$). Now consider the composite
since $\Delta$ and $*$ are short, the norm of this functional is at most ${\|\lambda\|^2}$.
$A$ is a $C^*$-coalgebra if the norm of the map in (1) is exactly ${\|\lambda\|^2}$, for every bounded linear functional $\lambda$.
Although there is an asymmetry in (1) (in the relative placement of $\lambda$ and $\lambda^*$), if we start with $\lambda^*$ instead of $\lambda$, we see that the universally quantified definition of $C^*$-coalgebra is symmetric.
If we take the formal dual of everything in the definition above, then $A$ becomes a Banach *-algebra and $\lambda\colon K \to A$ becomes (multiplication of scalars by) an element $x$ of $A$. The formal dual of the composite (1) is (multiplication of scalars by) the element $x^* x$. The requirement that the norm of this be exactly the square of the norm of $x$ is the $B^*$-identity that defines a $C^*$-algebra.
So the definition of $C^*$-coalgebra dualises everything in the definition of $C^*$-algebra, down to using coelement?s (in this case linear functionals) instead of elements.
In general, the dual of a coalgebra is an algebra, in any closed monoidal category. In particular, the dual of a Banach coalgebra is a Banach algebra. The involution $*$ gets along with this just fine; the dual of a Banach *-coalgebra is a Banach *-algebra. Finally, the linear functional in (1) becomes the element in the $B^*$-identity for the dual, so the dual of a $C^*$-coalgebra is a $C^*$-algebra.
But we have more! If $A$ is a $C^*$-coalgebra, then (as we've just seen) $A^*$ is a $C^*$-algebra; but since $A^*$ has a predual $A$, this means that $A^*$ is actually a $W^*$-algebra as well.
Which $W^*$-algebras arise in this way?
The sequence space $l^1$ is a $C^*$-coalgebra, whose dual $W^*$-algebra is the sequence space $l^\infty$. (For details of the comultiplication on $l^1$, see the examples in Banach coalgebra.)
Although $l^\infty$ is a Banach coalgebra (under ‘coconvolution’), it is not a $C^*$-coalgebra (at least not under coconvolution).
(YC) Moreover, the TVS-isomorphism class of the predual of a $W^*$-algebra is very restricted (let alone its isomorphism class in Bang). In particular $\ell^\infty$ doesn’t have a snowball’s chance in Texas of being a $C^*$-coalgebra under any kind of choice of comultiplication, because it’s the wrong kind of Banach space.)
Although the dual of the Lebesgue space $L^1$ (on the real line $\mathbb{R}$ with Lebesgue measure) is the $W^*$-algebra $L^\infty$, $L^1$ is not a $C^*$-coalgebra, nor even a Banach coalgebra (at least not in the obvious way). Essentially, this is because the diagonal in $\mathbb{R}^2$ has measure zero (so $\Delta$ takes an element of $L^1(\mathbb{R})$, interpreted as an absolutely continuous measure on $\mathbb{R}$, to a measure on $\mathbb{R} \times \mathbb{R}$ that is not absolutely continuous and so cannot be reinterpreted as an element of $L^1(\mathbb{R} \times \mathbb{R}) \cong L^1(\mathbb{R}) {\displaystyle\hat{\otimes}_\pi} L^1(\mathbb{R})$).