Inverse Mapping on Group of Units in Unital Banach Algebra is Continuous

Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.

Let $\map G A$ be the group of units of $A$.

Define $\phi : \map G A \to \map G A$ by:

$\map \phi x = x^{-1}$

for each $x \in \map G A$.

Then $\phi$ is continuous.

Proof

Let $x \in \map G A$ and $y \in A$ be such that:

$\ds \norm {x - y} < \frac 1 {\norm {x^{-1} } }$

As shown in Group of Units in Unital Banach Algebra is Open, we have $y \in \map G A$ and:

$\norm {1 - x^{-1} y} < 1$

Then, from Element of Unital Banach Algebra Close to Identity is Invertible, we have:

$\ds \norm {\paren {x^{-1} y}^{-1} } = \norm {y^{-1} x} \le \frac 1 {1 - \norm {1 - x^{-1} y} }$

Using the algebra norm property, we have:

$\ds \frac 1 {1 - \norm {1 - x^{-1} y} } = \frac 1 {1 - \norm {x^{-1} \paren {x - y} } } \le \frac 1 {1 - \norm {x^{-1} } \norm {x - y} }$

We now have:

\(\ds \norm {y^{-1} }\)

\(=\)

\(\ds \norm {\paren {x \paren {x^{-1} y} }^{-1} }\)

\(\ds \)

\(=\)

\(\ds \norm {\paren {x^{-1} y}^{-1} x^{-1} }\)

\(\ds \)

\(\le\)

\(\ds \norm {\paren {x^{-1} y}^{-1} } \norm {x^{-1} }\)

Definition of Algebra Norm

\(\ds \)

\(\le\)

\(\ds \frac {\norm {x^{-1} } } {1 - \norm {x^{-1} } \norm {x - y} }\)

Then we have:

\(\ds \norm {y^{-1} - x^{-1} }\)

\(=\)

\(\ds \norm {y^{-1} \paren {x - y} x^{-1} }\)

\(\ds \)

\(\le\)

\(\ds \norm {y^{-1} } \norm {x - y} \norm {x^{-1} }\)

Definition of Algebra Norm

\(\ds \)

\(\le\)

\(\ds \frac {\norm {x^{-1} }^2 \norm {x - y} } {1 - \norm {x^{-1} } \norm {x - y} }\)

Taking $y \to x$, we have $\norm {x - y} \to 0$ and so:

$\ds \frac {\norm {x^{-1} }^2 \norm {x - y} } {1 - \norm {x^{-1} } \norm {x - y} } \to 0$

so that:

$\norm {y^{-1} - x^{-1} } \to 0$ as $y \to x$.

So $\phi$ is continuous.

$\blacksquare$