• Apr 6th 2011, 07:27 PM
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 $g\in L^{2}(\mathbb{T})$.

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.
$\mathrm{C}$ 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
• 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 $g\in L^{2}(\mathbb{T})$.

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.
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 $g\in L^{2}(\mathbb{T})$.

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.
$\mathrm{C}$ 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 $M_f$ 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 $\ker(M_{\overline z-1}) = \{0\}.$
• 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=1168496http://sciencesoft.at/image/latex?in...&id=1756278081http://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 $\overline{Im(M_{z}-I)}=\ker(M_{z}-I)^{\perp}=\{0\}^{\perp}=L^{2}(T)$.