Are there any bounded linear operators on a Hilbert space such that its spectrum is exactly the unit circle?

It is known that the spectrum of unitary operators is contained within the unit circle, but what are some operators that have the unit circle (not the unit disk) as their spectra? I'm particularly interested in examples in infinite dimensional Hilbert spaces.

• @KaviRamaMurthy how do you understand isometry in this context? An invertible isometric linear map is automatically unitary. But if the map is not invertible, then $0$ is in the spectrum. – s.harp Aug 9 at 10:36

The standard trick to make an operator $$T$$ on $$\ell^2$$ (or some other separable Hilbert space) with a given compact subset $$S$$ of $$\mathbb C$$ its spectrum is to make an operator with eigenvalues a countable dense subset of $$S$$.
For unitary operators, the spectrum is, in any case, both discrete and continuous, a subset of the unit circle. One example of a countable dense subset $$X$$ of the unit circle is to enumerate rational numbers $$r_n$$ in $$[0,2\pi)$$, and take $$X=\{e^{2\pi ir_n}:n=1,2,\ldots\}$$, and $$Te_n=e^{2\pi ir_n}$$ with standard basis elements $$e_n$$ of $$\ell^2$$.
Since the spectrum of $$T$$ is closed, it contains at least the closure of the set $$X$$, namely, the unit circle. Since the spectrum of a unitary operator is contained in the unit circle, the spectrum of $$T$$ is exactly the unit circle.
Let $$H=L^2([0,2\pi])$$, and define $$T:H\to H$$ by $$Tf(t)=e^{it}f(t).$$