The group of units of a Banach algebra.

Proposition 1

Let \(\mathcal{A}\) be a Banach algebra with unit \(\mathbf{1}\). Let \(x \in \mathcal{A}\) be such that \(\lVert x - \mathbf{1} \rVert < 1.\) Then \(x\) is a unit.

Proof

Our candidate inverse is \(\sum_{j = 0}^{\infty} (\mathbf{1} - x)^{j}.\) Define for any integer \(n > 0\) the partial sum \(S_{n} = \sum_{j = 0}^{n} (\mathbf{1} - x)^{j}.\) First, I show the sequence of partial sums is Cauchy. Next, I verify the series actually converges to an inverse.

Fix integers \(0 < m < n\). Then

\[\lVert S_{n} - S_{m} \rVert \leq \sum_{j = m + 1}^{n} \lVert \mathbf{1} - x \rVert^{j} \leq \sum_{j = m + 1}^{\infty} \lVert \mathbf{1} - x \rVert^{j} = \frac{\lVert \mathbf{1} - x \rVert^{m+1}}{1 - \lVert \mathbf{1} - x \rVert}.\]

Since the rightmost quantity tends to \(0\) as \(m\) tends to \(\infty\), the sequence is Cauchy. Hence, the \(S_{n}\) converge to an element \(y \in \mathcal{A}\).

I verify that \(xy = yx = \mathbf{1}\). I compute:

\[\lVert \mathbf{1} - xy \rVert = \lim_{n \to \infty} \lVert \mathbf{1} - xS_{n} \rVert\] \[= \lim_{n \to \infty} \lVert \mathbf{1} - (\mathbf{1} - (\mathbf{1} - x)) S_{n} \rVert\] \[= \lim_{n \to \infty} \lVert (\mathbf{1} - x)^{n+1} \rVert\] \[\leq \lim_{n \to \infty} \lVert \mathbf{1} - x \rVert^{n+1} = 0.\]

Similarly, \(\lVert \mathbf{1} - yx \rVert = 0\). \(\tag*{$\blacksquare$}\)

In other words, the unit ball centered at the unit \(\mathbf{1}\) is contained in the group of units. We can generalize this to all units in Banach algebras.

Proposition 2

Let \(\mathcal{A}\) be a Banach algebra with unit \(\mathbf{1}\). Let \(x,y \in \mathcal{A}\) be such that \(x\) is a unit and \(\lVert x - y \rVert < \frac{1}{\lVert x^{-1} \rVert}.\) Then \(y\) is a unit.

Proof

We shall see that Proposition 1 implies both \(x^{-1}y\) and \(yx^{-1}\) are units. Consider the former.

\[\lVert \mathbf{1} - x^{-1}y \rVert = \lVert x^{-1} \rVert \lVert x - y \rVert < 1.\]

Likewise, \(yx^{-1}\) is a unit. Finally, observe

\[\left( (x^{-1}y)^{-1}x^{-1} \right) y = y \left( x^{-1}(yx^{-1})^{-1} \right) = \mathbf{1}.\]

That is, \(y\) admits both a left inverse and a right inverse. \(\tag*{$\blacksquare$}\)

In sum, the group of units of a Banach algebra is a Banach manifold. Our next stop is the compatibility of this Banach manifold-structure with the group multiplication. Specifically, the hurdle is proving the continuity of taking inverses. We use the following bound.

Lemma

Let \(\mathcal{A}\) be a Banach algebra with unit \(\mathbf{1}\). Let \(x,y \in A\) be such that \(x\) is a unit and \(\lVert x - y \rVert < \frac{1}{2 \lVert x^{-1} \rVert}.\) Then \(\lVert y^{-1} \rVert < 2 \lVert x^{-1} \rVert.\)

Proof

Recall the inverse \(y^{-1}\) has the formula \(y^{-1} = (x^{-1}y)^{-1}x^{-1}\) and that

\[\lVert (x^{-1}y)^{-1} \rVert = \lVert \sum_{j = 0}^{\infty} (\mathbf{1} - x^{-1}y)^{j} \rVert \leq \frac{1}{1 - \lVert \mathbf{1} - x^{-1}y \rVert}.\]

Using

\[\lVert \mathbf{1} - x^{-1}y \lVert \leq \lVert x^{-1} \rVert \lVert x - y \rVert < \frac{1}{2},\]

we refine the bound on \(\lVert y^{-1} \rVert\) to \(2 \lVert x^{-1} \rVert\). \(\tag*{$\blacksquare$}\)

Proposition 3

Let \(\mathcal{A}\) be a Banach algebra with unit \(\mathbf{1}\). The function \(i\colon \mathcal{A}^{\times} \to \mathcal{A}^{\times}\) from the group of invertible elements to itself is continuous.

Proof

Fix \(x,y \in \mathcal{A}^{\times}\). Our lemma readily yields

\[\lVert x^{-1} - y^{-1} \rVert \leq \lVert x^{-1} \rVert \lVert y^{-1} \rVert \lVert x - y \rVert < 2 \lVert x^{-1} \rVert^{2} \lVert x - y \rVert.\]

Therefore, at any point \(x \in \mathcal{A}^{\times}\) and for any \(\epsilon > 0\), the image under \(i\) of the ball of radius \(\frac{\epsilon}{2 \lVert x^{-1} \rVert^{2}}\) centered at \(x\) is contained in the ball of radius \(\epsilon\) centered at \(x^{-1}\). \(\tag*{$\blacksquare$}\)

In conclusion, \(\mathcal{A}^{\times}\) is a Banach Lie group.