Polar decomposition of invertible elements in a unital C$ ^{*} $-algebra. If $ A $ is a unital C$ ^{*} $-algebra and $ a $ is invertible, then


*

*$ a = u|a| $ for a unique unitary element $ u $ of $ A $.

*If $ \| a \| = \| a^{-1} \| = 1 $, what can you say about $ |a| $?


I don’t know how to start!
 A: We will make use of the following basic result from the theory of Banach algebras.

Theorem 1 Let $ A $ be a Banach algebra. If $ x \in A $ is invertible, then $ 0 \notin {\sigma_{A}}(x) $ and
  $$
{\sigma_{A}}(x^{-1}) = \left\{ \frac{1}{\lambda} ~ \Bigg| ~ \lambda \in {\sigma_{A}}(x) \right\}.
$$

We also need the following theorem from the theory of C$ ^{*} $-algebras.

Theorem 2 Let $ A $ be a C$ ^{*} $-algebra. Then for any normal element $ x $ of $ A $, we have
  $$
{r_{A}}(x) = \| x \|_{A},
$$
  where $ {r_{A}}(x) $ denotes the spectral radius of $ x $.


Let $ A $ be a C$ ^{*} $-algebra. Let $ a \in A $ be invertible, and suppose that $ \| a \|_{A} = \| a^{-1} \|_{A} = 1 $. Then
\begin{align}
{\sigma_{A}}(|a|) &= {\sigma_{A}} \left( \sqrt{a^{*} a} \right) \\
                  &= \sqrt{{\sigma_{A}}(a^{*} a)} \quad
                     (\text{By the Spectral Mapping Theorem applied to $ a^{*} a $.}) \\
                  &= \{ 1 \}. \quad (\clubsuit)
\end{align}

Proof of $ (\clubsuit) $:
  Firstly, observe that $ {\sigma_{A}}(a^{*} a) \subseteq [0,\infty) $. Next, using the C$ ^{*} $-condition and the fact that involution in a C$ ^{*} $-algebra is an isometry, we have
  $$
\| a^{*} a \|_{A} = \| (a^{*}) (a^{*})^{*} \|_{A}
                  = \| a^{*} \|_{A}^{2}
                  = \| a \|_{A}^{2}
                  = 1
$$
  and
  \begin{align}
\| (a^{*} a)^{-1} \|_{A} &= \| a^{-1} (a^{*})^{-1} \|_{A} \\
                         &= \| a^{-1} (a^{-1})^{*} \|_{A} \\
                         &= \| a^{-1} \|_{A}^{2} \\
                         &= 1.
\end{align}
  By Theorems $ 1 $ and $ 2 $, we therefore obtain $ {\sigma_{A}}(a^{*} a) = \{ 1 \} $.

Let us now work with the Continuous Functional Calculus corresponding to the self-adjoint element $ a^{*} a $. As $ {\sigma_{A}}(a^{*} a) = \{ 1 \} $, applying both the constant function $ 1 $ and the identity function to $ a^{*} a $ yields
$$
\mathbf{1}_{A} = a^{*} a.
$$
Therefore, $ |a| := \sqrt{a^{*} a} = \mathbf{1}_{A} $ and $ a = u|a| = u $.

Conclusion: If $ a \in A $ is invertible and $ \| a \|_{A} = \| a^{-1} \|_{A} = 1 $, then $ |a| = \mathbf{1}_{A} $ and $ a $ is unitary.

Note: In the argument above, we first proved that $ a^{*} a = \mathbf{1}_{A} $, then used polar decomposition to deduce that $ a $ is unitary and hence $ a a^{*} = \mathbf{1}_{A} $. However, we could have proven directly that $ a $ is unitary by performing the argument with $ a^{*} $ instead of $ a $, as the given conditions already imply that $ a^{*} $ is invertible and that $ \| a^{*} \|_{A} = \| (a^{*})^{-1} \|_{A} = 1 $. Once again, we have to use the fact that involution in a C$ ^{*} $-algebra is an isometry.
