Theorem 8.2.1.
Let \(\L\) be a Banach space. Suppose that \(A \in \L\text{.}\) If \(\norm{A} \lt 1\) then \(I - A\) is invertible and
\begin{equation*}
(I - A)\inv = \sum_{n=0}^\infty A^n,
\end{equation*}
where \(A^0 = I_V\text{.}\)
Section 8.2 Inverses
Need to get this from notes. Big theorem here is
Proof.
Suppose that \(\norm{A} \lt 1\text{.}\) Then the sequence
\begin{equation*}
S_N = \sum_{n = 0}^N A^n
\end{equation*}
is Cauchy, and since \(\L(\HH)\) is a Banach space, it converges to
\begin{equation*}
S = \sum_{n=0}^\infty A^n.
\end{equation*}
Then by computation (and appropriate application of limiting arguments),
\begin{equation*}
A S = A \sum_{n=0}^\infty A^n = \sum_{n=1}^\infty A^n = S - I.
\end{equation*}
Thus,
\begin{equation*}
S(I - A) = S - AS - I
\end{equation*}
and so
\begin{equation*}
S = (I - A)\inv.
\end{equation*}
Theorem 8.2.2.
Let \(\L\) be a Banach space. The set of invertible operators in \(\L\) is an open subset of \(\L\text{.}\)
