{\ensuremath{\otimes}}y^\dag) {\circ}\mu^\dag {\circ}z,$$ where we identify points $x \in H$ with morphisms $x \colon
{\ensuremath{\mathbb{C}}} \to H$, and similarly $${\ensuremath{\langle y\,|\,x^*z \rangle}} 
  = y^\dag {\circ}\mu {\circ}(x^* {\ensuremath{\otimes}}z).$$ Using the monoidal well-pointedness of ${\ensuremath{\mathbf{Hilb}}}$, it is easy to see that this is equivalent to the (H) condition![^6] The following two lemmas show that the assumptions (A), (C), (M) and (H) indeed result in an H\*-algebra.

A monoid in ${\ensuremath{\mathbf{Hilb}}}$ satisfying (M) is a Banach algebra.

The condition (M) implies that $P = \mu^\dag {\circ}\mu$ is a projector: $$P^2
    = \mu^\dag {\circ}\mu {\circ}\mu^\dag {\circ}\mu 
    = \mu^\dag {\circ}\mu 
    = P$$ and clearly $P = P^\dag$. Hence a monoid in ${\ensuremath{\mathbf{Hilb}}}$ satisfying (M) is a Banach algebra: $$\begin{aligned}
        \| xy \|^2 
    & = {\ensuremath{\langle xy\,|\,xy \rangle}} \\
    & = (x^\dag {\ensuremath{\otimes}}y^\dag) {\circ}\mu^\dag {\circ}\mu {\circ}(x {\ensuremath{\otimes}}y) \\
    & = {\ensuremath{\langle x {\ensuremath{\otimes}}y\,|\,P(x {\ensuremath{\otimes}}y) \rangle}} \\
    & \leq {\ensuremath{\langle x {\ensuremath{\otimes}}y\,|\,x {\ensuremath{\otimes}}y \rangle}} \\
    & = {\ensuremath{\langle x\,|\,x \rangle}}{\ensuremath{\langle y\,|\,y \rangle}} \\
    & = \| x \|^2 \| y \|^2 .
  {\tag*{$\Box$}}\end{aligned}$$

### Remark {#remark .unnumbered}

In fact, it can be shown that the multiplication of a semigroup in ${\ensuremath{\mathbf{Hilb}}}$ satisfying (H) is automatically continuous, so that after adjusting by a constant, the semigroup is a 