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

2. 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.

3. Originally Posted by Drexel28
I think you should define some of your notation.
Okay, updated the post.
Thanks.

4. 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 $\ker(M_{\overline z-1}) = \{0\}.$

5. 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...

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

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. and are both unital Banach algebras. By the definition of the operator , we have , and .
Next, to show that is a unital *-homomorphism, we will show that . Let and , then , where . This leads to , which implies or equivalently . Thus, learn that , i.e., . Finally, is a unital *-homomorphism of to .

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 is unitary, we have . Next, we show that . Let , to prove that , assume the contrary that , i.e., is invertible. Then, there exists such that , i.e., a.e. . This yields that a.e. , which contradicts the fact that . This contradiction yields , i.e., .
Therefore, we have .

2.b. For the next part, we use . In order to show that , we will show that for any , we have , i.e., . Let and assume that , i.e., a.e. . Since a.e. , we see that a.e. . Thus, , and thus, which proves .

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 . Then .

7. 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 $\overline{Im(M_{z}-I)}=\ker(M_{z}-I)^{\perp}=\{0\}^{\perp}=L^{2}(T)$.