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Suppose $H$ is a complex Hilbert space and $A:D(A)\to H$ is a densely defined, closed operator. Then consider the defect number $d_{\lambda}(A):=\text{dim}(\text{ran}(T-\lambda)^{\perp})$ of $A$ at $\lambda\in \Pi(A):=\{\lambda\in\mathbb{C}: A-\lambda\text{ has a bounded inverse on its range}\}$. I have the result that in this case the resolvent set $\rho(A)=\{\lambda\in\Pi(A): d_\lambda(A)=0\}$. As far as I can see, this is the same as saying that $\text{ran}(T-\lambda)$ is dense. The spectrum is the complement of the resolvent set, yet the continuous part of the spectrum consists of those $\lambda\in\mathbb{C}$ for which $T-\lambda$ is injective, not surjective but has dense range. This seems to be a contradiction. What am I missing?

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    If $A$ is symmetric, this is related to the Cayley transform of $A$. See Rudin, *Functional Analysis*, 13.20. – 2017-02-22
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    These results, as presented to me, do not assume that $A$ is symmetric. I'm trying to find the mistake in the above argument. – 2017-02-22
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    For $A$ closed, $A-\lambda$ has closed range for $\lambda\in\Pi(A)$ and if this is dense, the operator is surjective. That means the characterization of the resolvent above is correct. Is the characterization of continuous spectrum correct? I got it from Wikipedia. I guess this means that for $\lambda$ in the spectrum and $T-\lambda$ injective the inverse of $T-\lambda$ on it's range is not bounded? – 2017-02-23

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