TITLE: Is the limit of the spectral radius the spectral radius of the limit?
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: 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
$$A_k e_n = \begin{cases} 0, &\text{ if } n = 2^k(2l+1) \text{ for some } l \ge 0 \\
\alpha_ne_{n+1}, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$
Then $A_k^{2^{k+1}} = 0$ so $A_k$ is nilpotent. We also have $A_k \to A$ since
$$(A - A_k)e_n = \begin{cases} e^{-k}, &\text{ if } n = 2^k(2l+1) \text{ for some } l \ge 0 \\
0, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$
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 $$\alpha_{1}\alpha_2\cdots\alpha_{2^t-1} = \prod_{r=1}^{t-1} \exp(-r2^{t-r-1})$$
$$r(A)= \limsup_{j\to\infty}\|A^j\|^{\frac1j} \ge \limsup_{t\to\infty} \|A^{2^t-1}\|^{\frac1{2^t-1}} \ge \limsup_{t\to\infty} \|A^{2^t-1}e_1\|_2^{\frac1{2^{t-1}}} = \limsup_{t\to\infty} |\alpha_{1}\alpha_2\cdots\alpha_{2^t-1}|^{\frac1{2^{t-1}}} \\
\ge\limsup_{t\to\infty} \left(\prod_{r=1}^{t-1} \exp(-r2^{t-r-1})\right)^{\frac1{2^{t-1}}} = \limsup_{t\to\infty}\left(\prod_{r=1}^{t-1} \exp\left(-\frac{r}{2^{r}}\right)\right) = \limsup_{t\to\infty}\exp\left(-\sum_{r=1}^{t-1}\frac{r}{2^{r}}\right) = e^{-\sum_{r=1}^\infty \frac{r}{2^{r}}}$$
Therefore $A_k \to A$ but $r(A_k) \not\to r(A)$ since $r(A_k) = 0$ but $r(A) > 0$.