About unital star-homomorphisms

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Apr 6th 2011, 07:27 PM
bkarpuz

About unital *-homomorphisms
Dear MHF members,

I have a functional analysis problem as follows.

Problem. Let denote the map on the unit circle of the complex plane .
For , let be the mapping for .
1. Show that the map $M$ is a unital $\ast$ -homomorphism of $\mathrm{C}(\mathbb{T})$ to $\mathcal{B}(L^{2}(\mathbb{T}))$ .
  Conclude that $M_{z}$ is a unitary.
2. Show that $\sigma(M_{z})=\mathbb{T}$ , while $\sigma_{\mathrm{p}}(M_{z})=\emptyset$ .
3. Show that $\mathrm{Im}(M_{z}-\mathrm{I})$ is a proper dense subspace of $L^{2}(\mathbb{T})$ .
Thanks.

Notations.
is the set of continuous functions.
is the set of bounded operators.
is the set of square integrable functions.
and stand for the spectrum and the point spectrum, respectively.
-homomorphism - Star-Homomorphism -- from Wolfram MathWorld
-algebra - C*-Algebra -- from Wolfram MathWorld
Apr 6th 2011, 08:49 PM
Drexel28
Quote:
Originally Posted by bkarpuz

Dear MHF members,

I have a functional analysis problem as follows.

Problem. Let $z$ denote the map $z\mapsto z$ on the unit circle $\mathbb{T}$ of the complex plane $\mathbb{C}$ .
For $f\in\mathrm{C}(\mathbb{T})$ , let $M_{f}\in\mathcal{B}(L^{2}(\mathbb{T}))$ be the mapping $g\mapsto fg$ for $g\in L^{2}(\mathbb{T})$ .

Show that the map $M$ is a unital $\ast$ -homomorphism of $\mathrm{C}(\mathbb{T})$ to $\mathcal{B}(L^{2}(\mathbb{T}))$ .
Conclude that $M_{z}$ is a unitary.
Show that $\sigma(M_{z})=\mathbb{T}$ , while $\sigma_{\mathrm{p}}(M_{z})=\emptyset$ .
Show that $\mathrm{Im}(M_{z}-\mathrm{I})$ is a proper dense subspace of $L^{2}(\mathbb{T})$ .

Thanks.
bkarpuz
I think you should define some of your notation.
Apr 6th 2011, 10:10 PM
bkarpuz

Quote:

Originally Posted by Drexel28

I think you should define some of your notation.

Okay, updated the post.
Thanks.
Apr 7th 2011, 12:08 AM
Opalg
Quote:
Originally Posted by bkarpuz

Dear MHF members,

I have a functional analysis problem as follows.

Problem. Let $z$ denote the map $z\mapsto z$ on the unit circle $\mathbb{T}$ of the complex plane $\mathbb{C}$ .
For $f\in\mathrm{C}(\mathbb{T})$ , let $M_{f}\in\mathcal{B}(L^{2}(\mathbb{T}))$ be the mapping $g\mapsto fg$ for $g\in L^{2}(\mathbb{T})$ .

Show that the map $M$ is a unital $\ast$ -homomorphism of $\mathrm{C}(\mathbb{T})$ to $\mathcal{B}(L^{2}(\mathbb{T}))$ .
Conclude that $M_{z}$ is a unitary.
Show that $\sigma(M_{z})=\mathbb{T}$ , while $\sigma_{\mathrm{p}}(M_{z})=\emptyset$ .
Show that $\mathrm{Im}(M_{z}-\mathrm{I})$ is a proper dense subspace of $L^{2}(\mathbb{T})$ .

Thanks.

Notations.
$\mathrm{C}$ is the set of continuous functions.
$\mathcal{B}$ is the set of bounded operators.
$L^{2}$ is the set of square integrable functions.
$\sigma$ and $\sigma_{\mathrm{p}}$ stand for the spectrum and the point spectrum, respectively.
$\ast$ -homomorphism - Star-Homomorphism -- from Wolfram MathWorld
$\ast$ -algebra - C*-Algebra -- from Wolfram MathWorld
The proof of 1. is a simple verification. First, show that is a bounded operator on . Then check that and .

For 2., you should show that the spectrum of is the range of f, This follows fairly easily from 1., because the function is invertible provided that it never vanishes. The point spectrum of , on the other hand, is the set of points such that the function f takes the constant value on some interval of positive length.

For 3., I think I would use the fact that the closure of the image of a bounded operator is the orthogonal complement of the kernel of the adjoint. So try to show that
Apr 7th 2011, 01:13 PM
bkarpuz

Quote:

Originally Posted by Opalg

The proof of 1. is a simple verification. First, show that $M_f$ is a bounded operator on $L^2(\mathbb T)$ . Then check that $M_{fg} = M_fM_g$ and $M_f^* = M_{\overline f}$ .

For 2., you should show that the spectrum of $M_f$ is the range of f, $\sigma(M_f) = f(\mathbb T).$ This follows fairly easily from 1., because the function $f-\lambda1$ is invertible provided that it never vanishes. The point spectrum of $M_f$ , on the other hand, is the set of points $\lambda$ such that the function f takes the constant value $\lambda$ on some interval of positive length.

For 3., I think I would use the fact that the closure of the image of a bounded operator is the orthogonal complement of the kernel of the adjoint. So try to show that $\ker(M_{\overline z-1}) = \{0\}.$

Thanks a lot Opalg, let me try...
Apr 15th 2011, 11:17 PM
bkarpuz

Below, I give an answer for the question.
I would be very glad if one can confirm these.

Quote:

Originally Posted by opalg

the proof of 1. Is a simple verification. First, show that $m_f$ is a bounded operator on $l^2(\mathbb t)$ . Then check that $m_{fg} = m_fm_g$ and $m_f^* = m_{\overline f}$ .

1. http://sciencesoft.at/image/latex?index=134&id=21998888 and http://sciencesoft.at/image/latex?in...&id=1812191282 are both unital Banach algebras. By the definition of the operator http://sciencesoft.at/image/latex?in...6&id=488712098, we have http://sciencesoft.at/image/latex?in...9&id=704777319, http://sciencesoft.at/image/latex?in...0&id=587518786 and http://sciencesoft.at/image/latex?in...&id=1250409092.
Next, to show that http://sciencesoft.at/image/latex?in...6&id=488712098 is a unital *-homomorphism, we will show that http://sciencesoft.at/image/latex?in...&id=1301122670. Let http://sciencesoft.at/image/latex?in...&id=1296439660 and http://sciencesoft.at/image/latex?in...&id=1332170848, then http://sciencesoft.at/image/latex?in...4&id=923155260, where http://sciencesoft.at/image/latex?in...5&id=247848569. This leads to http://sciencesoft.at/image/latex?in...2&id=889287185, which implies http://sciencesoft.at/image/latex?in...0&id=213171330 or equivalently http://sciencesoft.at/image/latex?in...&id=1633646273. Thus, learn that http://sciencesoft.at/image/latex?in...&id=1695816014, i.e., http://sciencesoft.at/image/latex?in...8&id=356336255. Finally, http://sciencesoft.at/image/latex?in...6&id=488712098 is a unital *-homomorphism of http://sciencesoft.at/image/latex?index=134&id=21998888 to http://sciencesoft.at/image/latex?in...&id=1812191282.

Quote:

Originally Posted by opalg

for 2., you should show that the spectrum of $m_f$ is the range of f, $\sigma(m_f) = f(\mathbb t).$ this follows fairly easily from 1., because the function $f-\lambda1$ is invertible provided that it never vanishes. The point spectrum of $m_f$ , on the other hand, is the set of points $\lambda$ such that the function f takes the constant value $\lambda$ on some interval of positive length.

2.a. Since http://sciencesoft.at/image/latex?in...4&id=934555237 is unitary, we have http://sciencesoft.at/image/latex?in...&id=2085739236. Next, we show that http://sciencesoft.at/image/latex?in...&id=2135930075. Let http://sciencesoft.at/image/latex?in...&id=1998697653, to prove that http://sciencesoft.at/image/latex?in...&id=1068562478, assume the contrary that http://sciencesoft.at/image/latex?in...&id=1280169494, i.e., http://sciencesoft.at/image/latex?in...1&id=805247431 is invertible. Then, there exists http://sciencesoft.at/image/latex?in...&id=1200383310 such that http://sciencesoft.at/image/latex?index=85&id=82242425, i.e., http://sciencesoft.at/image/latex?in...&id=1651357113 a.e. http://sciencesoft.at/image/latex?in...7&id=549588809. This yields that http://sciencesoft.at/image/latex?in...&id=1781845750 a.e. http://sciencesoft.at/image/latex?in...7&id=549588809, which contradicts the fact that http://sciencesoft.at/image/latex?in...&id=1441506450. This contradiction yields http://sciencesoft.at/image/latex?in...7&id=566034439, i.e., http://sciencesoft.at/image/latex?in...&id=2135930075.
Therefore, we have http://sciencesoft.at/image/latex?in...7&id=519940329.

2.b. For the next part, we use http://sciencesoft.at/image/latex?in...8&id=114631060. In order to show that http://sciencesoft.at/image/latex?in...9&id=659686021, we will show that for any http://sciencesoft.at/image/latex?in...0&id=975206151, we have http://sciencesoft.at/image/latex?index=151&id=54354236, i.e., http://sciencesoft.at/image/latex?in...&id=1881582488. Let http://sciencesoft.at/image/latex?in...&id=1415156539 and assume that http://sciencesoft.at/image/latex?in...&id=1493138290, i.e., http://sciencesoft.at/image/latex?in...6&id=619727299 a.e. http://sciencesoft.at/image/latex?in...7&id=549588809. Since http://sciencesoft.at/image/latex?in...8&id=760268813 a.e. http://sciencesoft.at/image/latex?in...7&id=549588809, we see that http://sciencesoft.at/image/latex?in...3&id=348704221 a.e. http://sciencesoft.at/image/latex?in...7&id=549588809. Thus, http://sciencesoft.at/image/latex?in...&id=1278709888, and thushttp://sciencesoft.at/image/latex?in...1&id=393540992, which proves http://sciencesoft.at/image/latex?in...&id=1569049078.

Quote:

Originally Posted by opalg

for 3., i think i would use the fact that the closure of the image of a bounded operator is the orthogonal complement of the kernel of the adjoint. So try to show that $\ker(m_{\overline z-1}) = \{0\}.$

3. Following similar arguments to that in 2.b., we have http://sciencesoft.at/image/latex?in...&id=2044220797. Then http://sciencesoft.at/image/latex?index=87&id=1168496 http://sciencesoft.at/image/latex?in...&id=1756278081 http://sciencesoft.at/image/latex?index=94&id=555170779.
Apr 17th 2011, 10:26 AM
bkarpuz

I see that the images in my previous post expired very soon.
Now, I resend the solution.

1. $C(T)$ and $B(L^{2}(T))$ are both unital Banach algebras. By the definition of the operator $M$, we have $M_{f+h}g=(f+h)g=fg+hg=(M_{f}+M_{h})g$, $M_{cf}g=(cf)g=c(fg)=cM_{f}g$ and $M_{fh}g=(fh)g=(M_{f}M_{h})g$ for $f,h\in C(T)$, $g\in L^{2}(T)$ and $c\in C$.

Next, to show that $M$ is a *-homomorphism, we will show that $M_{f}^{*}=M_{\overline{f}}$. Let $f\in C(T)$ and $g,h\in L^{2}(T)$, then $\langle M_{f}g,h\rangle=\langle g,s\rangle$, where $s=M_{f}^{\ast}h$. This leads to $\int_{T}fg\overline{h}=\int_{T}g\overline{s}$, which is true if $\overline{s}=f\overline{h}$ or equivalently $s=\overline{f}h$. Thus, we learn that $M_{f}^{*}h=\overline{f}h=M_{\overline{f}}h$, i.e., $M_{f}^{*}=M_{\overline{f}}$.

Finally, $M$ is a unital *-homomorphism of $C(T)$ to $B(L^{2}(T))$.

2.a. Since $M_{z}$ is unitary, we have $\sigma(M_{z})\subset T$. Next, we show that $T\subset\sigma(M_{z})$. Let $\lambda\inT$, to prove $\lambda\in\sigma(M_{z})$, assume the contrary that $\lambda\not\in\sigma(M_{z})$, i.e., $M_{z}-\lambda I$ is invertible. Then there exists $g\in L^{2}(T)$ such that $(M_{z}-\lambda I)g=1$, i.e., $(z-\lambda)g(z)=1$ a.e. on $T$. This yields that $g(z)=(z-\lambda)^{-1}$ a.e. on $T$, which contradicts the fact that $g\in L^{2}(T)$. This contradiction yields $\lambda\in\sigma(M_{z})$, i.e., $T\subset\sigma(M_{z})$.

Therefore, we have $\sigma(M_{z})=T$.

2.b. For the next part, we use $\sigma_{p}(M_{z})\subset\sigma(M_{z})\subset T$. In order to prove that $\sigma_{p}(M_{z})=\emptyset$, we will show that for any $\lambda\inT$, we have $\ker(M_{z}-\lambda I)=\{0\}$, i.e., $\lambda\notin\sigma_{p}(M_{z})$. Let $g\in L^{2}(T)$ and assume that $(M_{z}-\lambda I)g=0$, i.e., $(z-\lambda)g(z)=0$ a.e. on $T$. Since $z-\lambda\neq0$ a.e. on $T$, we see that $g(z)=0$ a.e. on $T$. Thus, $\ker(M_{z}-\lambda I)=0$, and thus $\lambda\notin\sigma_{p}(M_{z})$, which proves $\sigma_{p}(M_{z})=\emptyset$.

3. Following similar arguments to that in 2.b., we have $\ker(M_{\overline{z}}-I)=\{0\}$. Then $\overline{Im(M_{z}-I)}=\ker(M_{z}-I)^{\perp}=\{0\}^{\perp}=L^{2}(T)$.