$\sigma(\oplus_{n=0}^\infty A_n) =\overline{\bigcup_n \sigma(A_n)}$? Suppose $\{A_n\}$ are uniformly bounded operator families on Hilbert spaces $H$,$A=\oplus_{n=0}^\infty A_n$.Let $\sigma(A)$ denote the spectrum of $A$. Does it hold that $\sigma(A) =\overline{\bigcup_n \sigma(A_n)}$?
 A: The answer is no. The main point is that the spectral radius of an operator is not equivalent with its operator norm, unless the operator is normal.
Let $S_n$ denote the shift operator on $\mathbb{C}^n$ defined on the standard basis by
$$S_ne_1=0,\qquad S_ne_k=e_{k-1},\quad 1\le k\le n$$
Then $\sigma(S_n)=\{0\}$ and $\|S_n\|=1.$ We have $$\displaylines{(I_n-S_n)^{-1}=I_n+S_n+S_n^2+\ldots +S_n^{n-1}\\
(I_n-S_n)^{-1}e_n=e_1+e_2+\ldots e_n}$$
Therefore $$\|(I_n-S_n)^{-1}\|\ge \sqrt{n}$$
We have
$$1\notin \{0\}=\overline{\bigcup_{n=1}^\infty \sigma({S_n})} $$
Nonetheless the operator $$I-\bigoplus_{n=1}^\infty S_n=\bigoplus_{n=1}^\infty(I_n-S_n),\qquad \mathcal{H}=\bigoplus_{n=1}^\infty \mathbb{C}^n$$ is not invertible as the norms of $\|(I_n-S_n)^{-1}\|$ are not uniformly bounded.
Remark The operators $S_n$ can be extended to $\tilde{S}_n$ acting on the common Hilbert space $\ell^2$ by setting $\tilde{S}_ne_k=S_ne_k$ for $1\le k\le n$ and $\tilde{S}_ne_k=0$ for $k>n,$ where $\{e_k\}_{k=1}^\infty$ denotes the standard basis in $\ell^2.$
A: Let $\lambda \in \bigcup_n \sigma(A_n)$. Then there exists $n'$ such that $\lambda \in \sigma(A_{n'})$. $A-\lambda I$ cannot be invertible because if it had an inverse $T$ we note first that $A_{n'}$ would be $T$-invariant and then that $T_{|A_{n'}}$ would be inverse to $A_{n'}-\lambda I_{A_{n'}}$. Thus $\lambda \in \sigma(A)$. Thus $\bigcup_n \sigma(A_n)\subset \sigma(A)$ and using that the spectrum of an operator is closed yields $\overline{\bigcup_n \sigma(A_n)}\subset \sigma(A)$.
The other direction holds if we assume that the operators $A_n$ are normal. Let $\lambda \notin \overline{\bigcup_n \sigma(A_n)}$. Then each operator $A_n-\lambda I_{A_n}$ has an inverse $T_n$ and there is a $\delta>0$ such that the ball $B(\lambda; \delta)$ avoids all spectra $\sigma(A_n)$. Letting $\Phi: C^*(A_n) \rightarrow C(\sigma(A_n))$ be the $*$-isomorphism in functional calculus we have $|\Phi(A_n-\lambda I_{A_n})|\geq \delta$. Thus $\Phi(T_n)\leq 1/\delta$ so that $\|T_n\|\leq 1/\delta$ for each $n$. Thus the operators $T_n$ are uniformly bounded and $T=\oplus_{n=0}^\infty T_n$ is inverse to $T-\lambda I$.
