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Let $X=\{z \in \mathbb C:1\leq|z|\leq2\}\text{ and } f(z)=z,z\in X.$ Let $A$ be the smallest closed subalgebra of $C(X)$ generated by $1$ and $f,$ and $B$ be the smallest closed subalgebra of $C(X)$ generated by $f$ and $\frac 1f.$ Determine the spectra $\sigma_A(f)$ and $\sigma_B(f).$ What happens if $X$ is a circle?

My attempt: Clearly, $\sigma_{C(X)}(f)=X$ and $A\subseteq B \subseteq C(X).$ So $X \subseteq \sigma_B(f)\subseteq \sigma_A(f).$

Now, $A=\{p(f):p\text{ is a polynomial}\}.$ Therefore, $\dfrac{1}{\lambda-f}\not \in A$ for all $\lambda \in \mathbb C.$ So, $\lambda-f$ is not invertible in $A$ for all $\lambda \in \mathbb C.$ This implies $\sigma_A(f)=\mathbb C.$

Similarly $B$ is the set of all polynomials in $f$ and $\frac 1f.$ Therefore, $\dfrac{1}{\lambda-f} \in B$ for all $\lambda \in \mathbb C\setminus X.$ So, $\lambda-f$ is not invertible in $B$ if and only if $\lambda \in X.$ This implies $\sigma_B(f)=X.$

My questions: $1.$ Is the above solution correct? $2.$ What happens if $X$ is a circle. I don't see how anything would change.

Note: This question has previously been asked here and here. However, my question is not a duplicate since it shows my attempt and I want to know if my solution is correct.

Sahiba Arora
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    Well, $\sigma_A(f)=\mathbb C$ can't be right - we know that $\sigma(x)$ is compact for any element of any Banach algebra. It's not true that $A={p(f)}$, because that set is not closed. – David C. Ullrich Dec 10 '17 at 14:12
  • Hint: If $p_n\to g$ uniformly on $X$ then $p_n-p_m\to0$ uniformly on $X$, and so the Maximum Modulus Theorem shows that in fact $p_n-p_m\to0$ uniformly on $Y$, where $Y=\dots$. – David C. Ullrich Dec 10 '17 at 14:24
  • @DavidC.Ullrich So, $A=\overline{{p(f)}}?$ So for $\frac{1}{\lambda-f} \not \in A$ if $\lambda \in X?$ To answer your question $Y=\overline{X}?$ – Sahiba Arora Dec 10 '17 at 14:29
  • No, $Y$ is much larger than $\overline X=X$. Ok, I'll tell you: $Y$ is the closed unit disk. – David C. Ullrich Dec 10 '17 at 14:54
  • @DavidC.Ullrich Is it correct that $\sigma_A(f)=X?$ – Sahiba Arora Dec 10 '17 at 14:59
  • No, $\sigma_A(f)\ne X$. It's true that $\sigma_B(f)=X$, but your explanation is wrong. First, $B$ is not what you say it is. Second, $1/(\lambda - f)$ is not a polynomial in $f$ and $1/f$. Finally, it doesn't really make any sense: You say (incorrectly) that $1/(\lambda-f)\in B$ for all $\lambda\in\mathbb C$. If that were true it would say $\sigma_B(f)=\emptyset$. Your "proof" that $\sigma_B(f)=X$ has nothing to do with $X$, so it can't be right. – David C. Ullrich Dec 10 '17 at 15:09
  • @DavidC.Ullrich Actually, I say that $1/(\lambda-f) \in B$ for all $\lambda \in \mathbb{C} \setminus X.$ – Sahiba Arora Dec 10 '17 at 15:14
  • Fine. But read the rest of what I wrote: What does "$1/(\lambda-f)$ is a polynomial in $f$ and $1/f$" have to do with $X$? The statement doesn't mention $X$ - if it were true for all $\lambda\in\mathbb C\setminus X$ it would have to be also true for all $\lambda\in\mathbb C$. – David C. Ullrich Dec 10 '17 at 15:30
  • If $\lambda \in X,$ then $\lambda -f$ vanishes at $\lambda.$ So $1/(\lambda-f) \not \in C(X).$ I understood your point that $1/(\lambda-f)$ is not a polynomial in $f$ and $1/f.$ – Sahiba Arora Dec 10 '17 at 15:36
  • That last comment is correct. Yippee. What does it imply about $\sigma_A(f)$ and $\sigma_B(f)$? – David C. Ullrich Dec 10 '17 at 15:45
  • @DavidC.Ullrich Won't $1/(\lambda-f)$ be limit of polynomial in $f$ and $1/f$ and hence in $B?$ – Sahiba Arora Dec 10 '17 at 15:50
  • If $\lambda\notin X$ and "limit" means "uniform limit on $X$" then yes. But you have to prove that... – David C. Ullrich Dec 10 '17 at 15:53
  • @DavidC.Ullrich Yes. But I'm still confused about $\sigma_A(f).$ – Sahiba Arora Dec 10 '17 at 15:54
  • $\sigma_A(f)$ is the closed unit disk. This has to do with the Maximum Modulus Theorem. – David C. Ullrich Dec 10 '17 at 16:03
  • There'd be no change to $\sigma_B(f)$ if $X$ is a circle. – Mark Dec 10 '17 at 23:03
  • @Mark What about $\sigma_A(f)?$ – Sahiba Arora Dec 10 '17 at 23:08
  • @Mark So if $X$ is a circle $\sigma_B(f)$ is still an annulus? Heh: The meaning of "$X$" changes, but actually it's $\sigma_A(f)$ that doesn't change; $\sigma_B(f)=X$ in both cases, so it does change. – David C. Ullrich Dec 11 '17 at 05:09
  • @DavidC.Ullrich I think Mark meant that $\sigma_B(f)$ would still be $X$(circle not annulus). – Sahiba Arora Dec 11 '17 at 07:16
  • @DavidC.Ullrich Since $X=\sigma_B(f) \subseteq \sigma_A(f),$ so how can $\sigma_A(f)$ be the closed unit disk? – Sahiba Arora Dec 11 '17 at 16:46
  • @SahibaArora Huh? $X$ is a subset of the disk - what's the problem? – David C. Ullrich Dec 11 '17 at 17:14
  • @DavidC.Ullrich In $X, |z| \geq 1$ while in closed unit disk $|z| \leq 1$. – Sahiba Arora Dec 11 '17 at 18:14
  • @SahibaArora Sorry, I missed the $2$. Every time here I've said anything about the closed unit disk I should have said the closed disk $|z|\le2$. – David C. Ullrich Dec 11 '17 at 18:21
  • @DavidC.Ullrich Oh, now things make more sense. – Sahiba Arora Dec 11 '17 at 18:22
  • do anyone please suggest me a way on how to get that uniform convergence in $B$ i.e. how to show there exist sequence of polynomials in $f$ and $\frac{1}{f}$ which converges uniformly to $\frac{1}{\lambda-f}$ for $\lambda \in {K}^c$ – bunny Feb 05 '19 at 12:39
  • also how can one show that ${\lambda \in \mathbb C : |\lambda| \le 1}$ is in $\sigma_{A}(f)$ – bunny Feb 05 '19 at 12:41

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