We have
\begin{align}
\lambda\in\sigma(T)&\iff \lambda I-T\ \text{not invertible }\\ \ \\
&\iff (\lambda I-T)^*=\bar\lambda I-T^*\text{not invertible }\\ \ \\
&\iff \bar\lambda\in\sigma(T^*).
\end{align}
And
$$
\bar\lambda\in\sigma_p(L)\iff \exists \text{ nonzero }v\in\ker(\bar\lambda I-L)=\text{ran}\, (\lambda I-R)^\perp\iff\lambda\in\sigma_r(R).
$$
Note that the last "if and only if" requires the fact that $\sigma_p(R)=\emptyset$, since
$$
\sigma_r(T)=\{\lambda:\ \text{ran}\,(\lambda I-T)^\perp\ne0\}\setminus\sigma_p(T).
$$
Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.
$$(Tf)(x) = x\cdot f(x).$$
Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.
Clearly $T$ has no eigenvalues, since
$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$
and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.
We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.
The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.
Best Answer
There is no "otherside containment". Any compact subset of the complex plane is the spectrum of a normal operator. So any compact subset of the circle can be the spectrum of a unitary.
For a couple of particular examples, the identity is a unitary with spectrum ${1}$. And $$U=\begin{bmatrix}1&0\\0&i\end{bmatrix}$$ is a unitary with spectrum $\sigma(U)=\{1,i\}$.