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Let $A$ be an unital Banach algebra, $x \in A$ and $(x_n)$ a sequence in $A$ converging to $x$. I want to show that $$ \lim\limits_n \rho (x_n) = \rho (x).$$ I can show that $$\limsup \rho(x_n) \leq \rho(x)$$ for every Banach algebra. In addition, if A is a commutative algebra, it's easy to prove that $$\liminf \rho(x_n) \geq \rho(x),$$ so the proposition it's true in commutative Banach algebras. Can we prove it when A is non commutative? If it's not, how can we show a counterexample? I've tried to build some counterxamples in the space of non-singular matrix, but nothing seems to work. My other idea is consider the Banach algebra of operators defined in some Hilbert/Banach space, but the spectral radius is a bit difficult to calculate.

Anyone can help me? Thank you very much.

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  • $\begingroup$ It's not true in general, and it appears the question has been studied in some detail, see here for example: google.com/… $\endgroup$ – user138530 Mar 12 at 23:50
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The answer is no.

This example is due to Kakutani, found in C. E. Rickart's General Theory of Banach Algebras, page $282$.

Consider the Banach space $\ell^2$ with the canonical basis $(e_n)_n$. Define a sequence of scalars $(\alpha_n)_n$ with the relation $\alpha_{2^k(2l+1)} = e^{-k}$ for $k,l \ge 0$.

Define $A : \ell^2 \to \ell^2$ with $Ae_n = \alpha_n e_{n+1}$. We have $\|A\| = \sup_{n\in\mathbb{N}}|\alpha_n|$. Also define a sequence of operators $A_k : \ell^2 \to \ell^2$ with $$A_k e_n = \begin{cases} 0, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \\ \alpha_ne_{n+1}, &\text{ if } n =2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$ Then $A_k^{2^{k+2}} = 0$ so $A_k$ is nilpotent. We also have $A_k \to A$ since $$(A - A_k)e_n = \begin{cases} e^{-k}, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \\ 0, &\text{ if } n =2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$ so $\|A - A_k\| = e^{-k} \to 0$.

For $j \in \mathbb{N}$ we have $$A^je_n = \alpha_n\alpha_{n+1}\cdots\alpha_{n+j-1}e_{n+j}$$ Notice that $$\alpha_{1}\alpha_2\cdots\alpha_{2^t-1} = \prod_{r=1}^{t-1} \exp(-r2^{t-r-1})$$

$$r(A)= \limsup_{j\to\infty}\|A^j\|^{\frac1j} \ge \limsup_{t\to\infty} \|A^{2^t-1}\|^{\frac1{2^t-1}} \ge \limsup_{t\to\infty} \|A^{2^t-1}e_1\|_2^{\frac1{2^{t-1}}} = \limsup_{t\to\infty} |\alpha_{1}\alpha_2\cdots\alpha_{2^t-1}|^{\frac1{2^{t-1}}} \\ \ge\limsup_{t\to\infty} \left(\prod_{r=1}^{t-1} \exp(-r2^{t-r-1})\right)^{\frac1{2^{t-1}}} = \limsup_{t\to\infty}\left(\prod_{r=1}^{t-1} \exp\left(-\frac{r}{2^{r}}\right)\right) = \limsup_{t\to\infty}\exp\left(-\sum_{r=1}^{t-1}\frac{j}{2^{r}}\right) = e^{-\sum_{r=1}^\infty \frac{j}{2^{r}}}$$

Therefore $A_k \to A$ but $r(A_k) \not\to r(A)$ since $r(A_k) = 0$ but $r(A) > 0$.

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  • $\begingroup$ I guess you want to say "Banach space" and not "Banach algebra". $\endgroup$ – Martin Argerami Mar 13 at 14:39
  • $\begingroup$ @MartinArgerami Yeah, thanks, the Banach algebra in question is $\mathbb{B}(\ell^2)$. $\endgroup$ – mechanodroid Mar 13 at 14:40
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I am not sure if it is true in general, but I believe it is true if $\sigma(x)$ is discrete. We can prove the following:

For all $x\in A$ if $U$ is an open set containing $\sigma(x)$, there exists a $\delta_U>0$ such that $\sigma(x+y)\subset U$ for all $y\in A$ with $\|y\|<\delta$.

Proof: This is Theorem 10.20 in Rudin's functional analysis. I will reproduce the proof for convenience. The function $\mathbb C\backslash \sigma(x)\ni\lambda\mapsto\|(\lambda e-x)^{-1}\|$ is continuous on the resolvent set. Furthermore we know that as $|\lambda|\to \infty$ we must have $\|(\lambda e-x)^{-1}\|\to 0$. Thus we have some finite $M$ such that $\|(\lambda e-x)^{-1}\|<M$ for all $\lambda\notin U$. Hence if $y$ satisfies $\|y\|<1/M$ and $\lambda \notin U$ we have $\lambda e-(x+y)$ is invertible. This follows because $$\lambda e-(x+y)=(\lambda e-x)(e-(\lambda e-x)^{-1}y)$$ and $\|(\lambda e-x)^{-1}y\|<1$, so invertible.

We can also prove a strengthening of this:

If $U$ is an open set containing a component of $\sigma(x)$ (for any $x\in A$) and $\lim x_n=x$, then $\sigma(x_n)\cap U\neq \emptyset$ for all $n$ sufficiently large.

Proof: If $\sigma(x)$ is connected this follows from the above, so let us assume that it isn't. Thus there exists an open $V$ such that $\sigma(x)\subset U\cup V$. Assume by way of contradiction that the statement is false. Then for all $n\in \mathbb N$ there exists a $N\geq n$ such that $\sigma(x_n)\cap U=\emptyset$. In conjunction with the above theorem we thus have arbitrarily large $n$ such that $\sigma(x_n)\subset V$. As these $n$ are arbitrarily large we must be able to find a $x_n$, such that $\sigma(x_n)\subset V$ and $\|x-x_n\|<\delta_U$. The above theorem then implies that $\sigma(x)\subset V$, a contradiction.

Using this last theorem it is a fairly simple corollary that if $x$ has a discrete spectrum and $x_n\to x$ we must have $\lim \rho(x_n)=\rho(x)$. In particular this holds true for compact operators on Banach spaces.

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