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Math Help - About unital star-homomorphisms

  1. #1
    Senior Member bkarpuz's Avatar
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    Angry About unital *-homomorphisms

    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
    Last edited by bkarpuz; April 6th 2011 at 10:12 PM. Reason: Notations mentioned.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bkarpuz View Post
    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.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I think you should define some of your notation.
    Okay, updated the post.
    Thanks.
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by bkarpuz View Post
    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\}.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Opalg View Post
    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...
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  6. #6
    Senior Member bkarpuz's Avatar
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    Arrow

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

    Quote Originally Posted by opalg View Post
    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 .

    Quote Originally Posted by opalg View Post
    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 .

    Quote Originally Posted by opalg View Post
    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 .
    Last edited by bkarpuz; April 16th 2011 at 11:29 AM.
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  7. #7
    Senior Member bkarpuz's Avatar
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    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 $\overline{Im(M_{z}-I)}=\ker(M_{z}-I)^{\perp}=\{0\}^{\perp}=L^{2}(T)$.
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