Background: I'm examining Theorem 25.2 in the book "Introduction to Hilbert Space and the Theory of Spectral Multiplicity" by Paul Halmos, which discusses the conditions for an operator $U$ to be an automorphism of a Hilbert space $\mathcal{H}$. The theorem states:
Theorem 25.2: A necessary and sufficient condition that an operator $U$ be an automorphism of $\mathcal{H}$ is that it be unitary.
The proof in the text goes as follows:
Proof: Observe that since $\langle Ux, Uy \rangle=\langle U^{\ast} Ux, y \rangle$ the equation $U^{\ast}U=I$ implies and is implied by the identity $\langle Ux, Uy \rangle=\langle x,y \rangle$. Since a unitary operator is invertible and, consequently, is a one-to-one transformation form $\mathcal{H}$ onto $\mathcal{H}$, we infer from this observation that a unitary operator is an automorphism. Since (cf. 21.3) an automorphism is also an invertible operator, we infer from the same observation that if $U$ is an automorphism then $U^{-1}=U^{\ast}$ and hence that $U$ is unitary.
Question: The proof suggests that if $U$ is an automorphism, then $U^{−1}=U^{\ast}$, thereby implying that $U$ is unitary. It's clear to me that $U^{\ast}U=I$ follows from the isometry property. However, in operator theory, to establish that an operator $A$ is invertible with inverse $B$, we typically show that $AB=BA=I$. Consequently we would also need to show that $UU^{\ast}=I$ to conclude that $U^{−1}=U^{\ast}$. What am I missing?
Additional Context: In the text, $\mathcal{H}$ is a Hilbert space, and the term "operator" refers to bounded operators by convention. Also, the term "automorphism" is used in the sense of an isometric isomorphism.
Theorem 21.3 An operator $A$ is invertible if and only if its range is dense in $\mathcal{H}$ and there exists a positive real number $\alpha$ such that $\|Ax\| \geq \alpha \|x\|$ for every vector $x$.
If $U$ is an automorphism then $U^*U=I$ and $U$ is invertible, so that (multiplying both sides of the equation on the right by $U^{-1}$) $U^*=U^{-1}$.