# Is the limit of the spectral radius the spectral radius of the limit?

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.

• It's not true in general, and it appears the question has been studied in some detail, see here for example: google.com/… – user138530 Mar 12 at 23:50

## 2 Answers

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 $$r(A) > 0$$.

• I guess you want to say "Banach space" and not "Banach algebra". – Martin Argerami Mar 13 at 14:39
• @MartinArgerami Yeah, thanks, the Banach algebra in question is $\mathbb{B}(\ell^2)$. – mechanodroid Mar 13 at 14:40

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 $$\|(\lambda e-x)^{-1}\| for all $$\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.