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Suppose $T:H\rightarrow H$ is a symmetric operator on a Hilbert space $H$. I want to show that $\text{Ran}(T\pm i)$ are dense in $H$ implies that $T$ is essentially self-adjoint, that is the closure of $T$ is self-adjoint.

I already know that if $\text{Ran}(T\pm i)=H$ then $T$ is self-adjoint, and I feel that the above should follow from this in a simple way, but I don’t quite see it.

Thanks.

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Since $T$ is symmetric, it is closable. We just have to prove that if $T$ is closed, symmetric and the ranges $\text{Ran}(T\pm i)$ are dense then they are actually equal to $H$. Let's do it for $\text{Ran}(T+i)$. Let $v \in H$. By density of the range, there is a sequence $(v_n)$ such that $\lim (T+i)v_n = v$. As $T$ is symmetric, you can easily compute that $\|(T+i)x\|^2 = \|Tx\|^2 + \|x\|^2$ for any $x\in H$. Since $((T+i)v_n)$ is a Cauchy sequence, it follows that also the sequences $(Tv_n)$ and $(v_n)$ are Cauchy, hence they converge; denote their limits by $u$ and $w$. As $T$ is closed, we have $u=Tw$. As $v=u+iw$, we get $v=(T+i)w$, therefore $v \in \text{Ran}(T+i)$.

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  • $\begingroup$ Cheers! The part I was missing was that for symmetric operators you can conclude that $(v_n)$ is Cauchy given that $(T+i)v_n$ is Cauchy. :) $\endgroup$ – ougoah Nov 2 '17 at 14:37

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