 Solution

An element $a$ is a Normal element of a C-star algebra if $a^{\ast }a=a{a}^{\ast }$. It is a Unitary element of a C-star algebra if $a^{\ast }a=a{a}^{\ast }=1$. Its Banach algebra spectrum iswhere $A^{\times }$ is the group of invertible elements of a Banach algebra.

For normal $a$, the continuous functional calculus is the unique unital isometric star-isomorphismwhich sends the coordinate function $z\mapsto z$ to $a$. It preserves sums, products, complex scalars and the involution: ${\Phi }_a(\overline{f})={\Phi }_a(f{)}^{\ast }$. In particular $\Vert f(a)\Vert ={\max }_{z\in {\sigma }_A(a)}|f(z)|$. The spectral mapping theorem in this setting statesThe common unit and spectral permanence for C-star subalgebras permit computing this spectrum in $A$ itself.

For unitary $u$, the C-star identity gives $\Vert u\Vert =\Vert u^{\ast }\Vert =1$. If $|z|>1$, $u-z1=-z(1-u/z)$ has a multiplicative inverse by the Neumann series. If $|z|<1$, write $u-z1=u(1-z{u}^{\ast })$ and apply the same series. Therefore ${\sigma }_A(u)$ lies on the unit circle.

For $|t|=1$, the denominator $1+\overline{\lambda }t$ is nonzero because $|\lambda |<1$. The function is consequently continuous on the spectrum, andThus $|f(t)|=1$ throughout ${\sigma }_A(u)$, so $\overline{f}f=f\overline{f}=1$ there. Applying the star-homomorphism ${\Phi }_u$ provesHence $f(u)$ is unitary. Explicitly, the same calculus giveswith the denominator's inverse also supplied by a norm-convergent Neumann series. This is the Möbius transform of a unitary element. The conjugation on $\lambda$ in the denominator is essential to the unit-modulus identity.