Self-adjoint operator $H$ is not in the $C^*$-algebra generated by unitary operator $U$? 
Let $U, H$ be operators on Hilbert space $L^2\left(\mathbb{T}, \frac{d \theta}{2\pi}\right)$($\mathbb{T}$ is unit circle $S^1$), for any $f\in L^2\left(\mathbb{T}, \frac{d \theta}{2\pi}\right)$,
$$
(Uf)(e^{i\theta}) = e^{i\theta}f(e^{i\theta}), (Hf)(e^{i\theta}) = \theta f(e^{i\theta}), \theta\in [0,2\pi).
$$
Why $H$ is not in the $C^*$-algebra generated by $U$?

Clearly $U$ is unitary and $H$ is self-adjoint. It seems that $U=\sum_{n=0}^{\infty} \frac{(iH)^n}{n!}$. And we have a theorem(?): $C^*(U)=C(\sigma(U))$, where $C^*(U)$ is the $C^*$-algebra generated by $U$ and $I$, $C(\sigma(U))$ is the set of all continuous functions from $\sigma(U)\to \mathbb{C}$.(I am not quite sure...)
Could you please help me? Thank you in advance!
 A: Let's observe something first: $U$ is the multiplication operator by $u(e^{i\theta})=e^{i\theta}$ while $H$ is the multiplication operator by $h(e^{i\theta})=\theta$. We have a $*$-homomorphism (i.e. a map that preserves addition, multiplication and involutions)
$$L^\infty(\mathbb{T})\to\mathcal{B}\big(L^2(\mathbb{T})\big)$$
given by $\varphi\mapsto M_\varphi$, where $M_\varphi:L^2(\mathbb{T})\to L^2(\mathbb{T})$ acts as $M_\varphi(f)=\varphi\cdot f$ (pointwise multiplication). So $U=M_u, H=M_h$. This $*$-homomorphism is actually isometric (I can add the details for this, but you can find it in other posts here as well). Thus the image of this $*$-homomorphism is actually a $C^*$-algebra, say $\mathcal{M}\subset \mathcal{B}\big(L^2(\mathbb{T})\big)$. Now our $*$-homomorphism is an isomorphism of $C^*$-algebras between $L^\infty(\mathbb{T})$ and $\mathcal{M}$. Your question, "is $H$ contained in $C^*(U)$?", is equivalent to "is $h$ contained in $C^*(u)$?" via this isomorphism.
So we have to think about the $C^*$-algebra $L^\infty(\mathbb{T})$. Involution here is simply taking the conjugate function. The $C^*$-algebra generated by a normal element $u$ is the norm-closure of all the polynomials $p(u,u^*)$, where $p(z,w)\in\mathbb{C}[z,w]$. Now since $u$ is a unitary, we can immediately see that
$$C^*(u)=\overline{\{p_1(e^{i\theta})+p_2(e^{-i\theta}): p_1(z),p_2(z)\in\mathbb{C}[z]\}}=\text{ the closure of trigonometric polynomials }$$
where the closure is taken in $\|\cdot\|_\infty$. This is equal to $C(\mathbb{T})$, the space of continuous function over the unit circle. Is $e^{i\theta}\mapsto \theta$ continuous over $\mathbb{T}$? No, because as $e^{i\theta}\to 1$, $\theta$ bumbles between $0$ and $2\pi$. So $h\not\in  C^*(u)$.
