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.
Best Answer
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 $r(A) > 0$.