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 .
- Show that the map is a unital -homomorphism of to .
Conclude that is a unitary.- Show that , while .
- Show that is a proper dense subspace of .
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
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
Below, I give an answer for the question.
I would be very glad if one can confirm these.
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 .
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 .
3. Following similar arguments to that in 2.b., we have . Then .
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 $\overline{Im(M_{z}-I)}=\ker(M_{z}-I)^{\perp}=\{0\}^{\perp}=L^{2}(T)$.