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This is an example in Rudin's Functional Analysis, in the chapter on Unbounded Operators. Consider the right shift operator $V$ on $l^2$. It is an isometry and closed, and $I-V$ is one-one and so $V$ is the Cayley transform of a closed operator $T$ in $l^2$ which is not self-adjoint (since $V$ is not unitary). The claim is that $T$ is densely defined. I am unable to show this. If $V$ had been unitary, then $R(I-V)^{\perp}=N(I-V)$, and since $I-V$ is one-one and $D(T)=R(I-V)$, this would imply that $T$ is densely-defined. However, since $V$ is an isometry that is not onto, this would not work. Is there a different way of showing that $T$ is densely defined?

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The spectrum of the right shift $V$ is the closed unit disc with the unit circle being its continuous spectrum. In particular, $I - V$ has dense range.

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