It is well-known that any bounded normal operator $A$ can be written as $A= A_1+ i A_2$ where $A_1$ and $A_2$ are commuting bounded self-adjoint operators. This leads to a proof of spectral theorem of bounded normal operators using spectral theorem of bounded self-adjoint operators.

My question is, for an unbounded normal operator $A$, is there also such a decompostion $A=A_1+i A_2$ with $A_1$, $A_2$ (unbounded) self-adjoint? If so, is there an elementary proof of this without using spectral theorem of unbounded normal operators?

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    $\begingroup$ Just to remind for unbounded operator, normal means closed and $AA^*=A^*A$ and self-adjoint means $A=A^*$(in the sense that their domain coincides), not merely symmetric $\endgroup$ – Yuchen Liao Oct 7 '18 at 15:37

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