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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.

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    $\begingroup$ The range of a normal operator need not be closed. Consider a multiplication operator $\mu_c \colon x \mapsto (c_0x_0,c_1x_1,\dotsc)$ on $\ell^2(\mathbb{N})$ where $c\in \ell^\infty(\mathbb{N})$. If $c_k \neq 0$ for all $k$ and $\liminf \lvert c_n\rvert = 0$, then the range is a proper dense subspace of $\ell^2(\mathbb{N})$. $\endgroup$ – Daniel Fischer Jul 2 '15 at 19:59
  • $\begingroup$ Do you know about polar decompositions of normal operators? $\endgroup$ – Daniel Fischer Jul 2 '15 at 20:19
  • $\begingroup$ I know them from linear algebra in case of finite dimension but I have not thought about it in infinite dimension yet. I will look into it $\endgroup$ – Blazej Jul 2 '15 at 21:07
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    $\begingroup$ The range of $A$ is closed if and only if the range of $A^*$ is closed by the closed range theorem. $\endgroup$ – daw Jul 3 '15 at 15:03
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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 $\mathcal{R}(A)=\mathcal{R}(A^{\star})$.

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  • $\begingroup$ Why can we restrict ourselves to the case $\mathcal N(A) = \{0\}$? Also, if the range is only dense and not the whole space, why is $A$ invertible? $\endgroup$ – Viktor Glombik Mar 26 at 23:02
  • $\begingroup$ Also the extension of $U$ is only unitary because $U$ is already, right? $\endgroup$ – Viktor Glombik Mar 26 at 23:17
  • $\begingroup$ $X=(\mathcal{N}(A)=\mathcal{N}(A^*))\oplus (\overline{\mathcal{R}(A)}=\overline{\mathcal{R}(A^*)})$ $\endgroup$ – DisintegratingByParts Mar 27 at 5:30

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