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Let $(X,||\cdot||)$ be a Banach space on $\mathbb{C}$ and $\mathcal{L}(X)$ the set of linear, continuous functions from $X$ to itself. For $T\in\mathcal{L}(X)$, define the norm $||T||_{\mathcal{L}(X)}:=\sup_{v\neq 0}\frac{||T(v)||}{||v||}$ and the continuous spectrum $\sigma(T):=\{\lambda\in\mathbb{C}\mid A-\lambda I\,\text{ has no inverse}\}$. Define the sets: $$GL(X):=\{T\in\mathcal{L}(X)\mid T\text{ is invertible}\}$$ $$\mathcal{H}(X):=\{T\in GL(X)\mid T\text{ is hyperbolic}\}$$ (where hyperbolic means $|\lambda|\neq 1$ for all $\lambda\in\sigma(T)$)

Prove that: (1) $GL(X)\subset\mathcal{L}(X)$ is open and (2) $\mathcal{H}(X)$ is open, dense in $GL(X)$.

(1) Take $A\in GL(X)$ and the open ball $B_r(A)$, where $r:=1/||A^{-1}||$. If $B\in B_r(A)$, then $||I-BA^{-1}||\leq||A-B||\cdot||A^{-1}||<r||A^{-1}||=1$. The series $\sum_{n=0}^\infty(I-BA^{-1})^n$ is therefore convergent, so $BA^{-1}$ is invertible with $(BA^{-1})^{-1}=(I-(I-BA^{-1}))^{-1}=\sum_{n=0}^\infty(I-BA^{-1})^n$. Therefore, $B$ is invertible $\Rightarrow B_r(A)\subset GL(X)$. $_\blacksquare$

(2) We prove $GL(X)\setminus\mathcal{H}(X)$ is closed in $GL(X)$. If $\{A_n\}_{n\in\mathbb{N}}\subset GL(X)\setminus\mathcal{H}(X)$ a Cauchy sequence converging to $A\in GL(X)$, we need to prove $A\notin \mathcal{H}(X)$. By definition, for all $n$ there is $\lambda_n\in \mathbb{C}$ such that $|\lambda_n|=1$ and $A_n-\lambda_nI\notin GL(X)$. Since $\mathbb{S}^1:=\{\lambda\in\mathbb{C}\mid |\lambda|=1\}$ is compact, there is a convergent subsequence $\{\lambda_{i_n}\}_{n\in\mathbb{N}}$ converging to $\lambda\in\mathbb{S}^1$, so $\lim_{n\to\infty} A_{i_n}-\lambda_{i_n}I=$ $A-\lambda I$. Since $A_{i_n}-\lambda_{i_n} I\notin GL(X)$ (which is open), $A-\lambda I\notin GL(X)$, so $A\in GL(X)\setminus\mathcal{H}(X)$.

To prove $\mathcal{H}(X)$ is dense in $GL(X)$, my idea is this: let $A\in GL(X)\setminus\mathcal{H}(X)$ and $\epsilon>0$ arbitrary, then there exist $z\in\mathbb{C}$ with $|z|<\epsilon$ such that $A+zI\in\mathcal{H}(X)$. I feel like this could work, but I don't know how to guarantee $\sigma(A+zI)\cap\mathbb{S}^1=\emptyset$.

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    $\begingroup$ Adding a small multiple of identity is not going to work. For example, the spectrum of shift operator $S$ is the closed unit disk. Let $T=S + (3/2)I$; this shifts the spectrum by $3/2$, so $T$ is invertible. But $T$, and any $T+zI$ with $|z|<1/2$, still has the spectrum intersecting the unit circle. $\endgroup$ – user357151 May 28 '18 at 6:24
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    $\begingroup$ Out of curiosity, where did this problem come from? $\endgroup$ – Wraith1995 Jun 1 '18 at 20:22

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