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:

Then we have:

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.