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I denote the invertible operators of $\mathcal{L}(E)$ as $\mathcal{I}$. $E$ is Banach.

From looking at other posts, seeing that $\mathcal{I}$ is open means that I have to see that if there exists $r>0$ such that $\forall T \in \mathcal{I},\left\|S-T\right\|< r$ implies that $S\in \mathcal{I}$.

How do I go about proving this seemingly elementary fact? Where is the hypothesis that $E$ is Banach used if at all? Any hints would be appreciated

Taken Spark
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  • Hints: First prove the special case where $T$ is the identity operator $I$. The idea is to expand $(I-A)^{-1}$ as a geometric series, which converges because $E$ is Banach. Then reduce the general case to the special case by multiplying by $T^{-1}$. – Andreas Blass Jan 07 '22 at 16:10
  • More generally, this result holds for any unital Banach algebra A. Let $a \in A^{-1}$, where $A^{-1}$ is the set of invertible elements in A, and set $||a-b|| < 1/||a^{-1}||$ for $||1-a^{-1}b||<1$. You get $a^{-1}b$ invertible, so $b$ is invertible. What you have to show is that $||a||<1$ implies $1-a$ is invertible. Its inverse is given by the absolutely convergent sum $(1-a)^{-1}=\sum_{n=0}^{\infty}a^{n}$. – undefined Jan 07 '22 at 16:20

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Let $A$ be a unital Banach algebra (for example $\mathcal{L}(E)$ with operator multiplication) and let $a \in A$ satisfy $||a|| < 1$. Set $S_n=\sum_{k=0}^{n}a^k$ and note that

$||S_n-S_m||=||\sum_{k=m+1}^{n}a^k|| \leq \sum_{k=m+1}^{n}||a||^k \to 0, \quad n,m \to \infty.$

Thus, $\lbrace S_n \rbrace_{n=0}^{\infty}$ is Cauchy, and so converges in $A$, since $A$ is Banach. Denote the limit of $\lbrace S_n \rbrace_{n=0}^{\infty}$ by $s$. Then

$s(1-a)=\text{lim}_{n \to \infty} S_n(1-a)=\text{lim}_{n \to \infty} 1-a^{n+1}=1$,

and similarly you get $(1-a)s=1$. So $1-a$ is invertible with inverse

$(1-a)^{-1}=s=\sum_{k=0}^{\infty}a^k$.

Now let $a \in A^{-1}$, where $A^{-1}$ denotes the set of invertible elements in $A$. Let $b \in A$ satisfy $||a-b|| < 1/||a^{-1}||$ so that $||1-a^{-1}b||<1$. By the preceeding argument, you have that $c=a^{-1}b$ is invertible with inverse $c^{-1}$. Let $h=c^{-1}a^{-1}$. Then

$hb=(c^{-1}a^{-1})b=c^{-1}(a^{-1}b)=c^{-1}c=1$,

and also

$bh=bc^{-1}a^{-1}=acc^{-1}a^{-1}=1$.

Thus, $b \in A$ is invertible with inverse $b^{-1}=h\in A$.

undefined
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