Functional Analysis – Range of Normal Operator and Its Adjoint Are Equal

Question

On Wikipedia it is written that bounded normal operator in Hilbert space has the same range and kernel as its adjoint. I've been able to show equality of kernels and closures of ranges: $\|Af\|=\|A^*f\|$ so kernels must be equal. Furthermore kernel of an operator is equal to orthogonal complement of its adjoint so those must also be equal. By taking orthogonal complements we get equality of closures of ranges. However as far as I understand ranges need not be closed (but maybe in this case they are? I haven't been able to show it though) so equality of ranges does not follow. I would appreciate hints. I am also interested about generalisations to unbounded case.

Answer 1

Because $A$ is normal, then $\mathcal{N}(A)=\mathcal{N}(A^{\star})$ is invariant under $A$ and $A^{\star}$, as is its orthogonal complement $$ \overline{\mathcal{R}(A)}=\mathcal{N}(A)^{\perp}=\mathcal{N}(A^{\star})^{\perp}=\overline{\mathcal{R}(A^{\star})}. $$ That allows you to reduce to the case where $\mathcal{N}(A)=\{0\}$, and where the ranges of $A$ and $A^{\star}$ are dense in $H$.

So, assuming $A$ is normal with $\mathcal{N}(A)=\mathcal{N}(A^{\star})=\{0\}$, the map $U=A^{\star}A^{-1} : \mathcal{R}(A)\rightarrow\mathcal{R}(A^{\star})$ is linear and isometric, meaning that $\|Ux\|=\|x\|$ for all $x\in\mathcal{R}(A)$. So $U$ extends uniquely by continuity to an isometric map $U : H\rightarrow H$. In fact, this extension is unitary because $\mathcal{R}(U) =\mathcal{R}(A^{\star})$ is dense in $H$. Hence, $UA=A^{\star}$, which gives $A^{\star}U^{\star}=A$ by applying the adjoint. Because $U$ is unitary, $A^{\star}=AU$. The identities $A^{\star}U^{\star}=A$ and $A^{\star}=AU$ together imply $\mathcal{R}(A)=\mathcal{R}(A^{\star})$.