Assume $A$ is a symmetric operator in a Hilbert space, which generates a contraction semigroup (a priori it is not known, whether this semigroup is self-adjoint). Is A then self-adjoint?
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1$\begingroup$ Yes. The spectrum of a symmetric densely defined operator is either the entire plane or the upper or the lower half plane or a subset of the real line; in the latter case the operator is self-adjoint. $\endgroup$– Jochen GlueckCommented Dec 13, 2023 at 23:18
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$\begingroup$ Thanks! (I assume in the real case one can complexify and conclude likewise, right?) $\endgroup$– Mike_BoolCommented Dec 13, 2023 at 23:21
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$\begingroup$ Indeed, complexification should work in the real case $\endgroup$– Jochen GlueckCommented Dec 13, 2023 at 23:24
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