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Let $H=(H,(\cdot,\cdot)_H)$ be Hilbert space and $A:D(A) \subset H \longrightarrow H$ be a self-adjoint operator, not necessarily bounded.

Question. Is true that $\text{Ker}(A)^{\perp}=\text{Range}(A)$?

Note that, given any $v \in \text{Range}(A)$ there exist $u \in D(A)$ such that $A(u)=v$. Hence, if $w \in \text{Ker}(A)$ is arbitrary, then since $A$ is self-adjoint we obtain $$ 0=(A(w),u)_H=(w,A(u))_H=(w,v)_H $$ that is, $\text{Ker}(A)^{\perp}$. Thus, $ \text{Range}(A) \subset \text{Ker}(A)^{\perp}$. But, the reverse inclusion holds? If so, would this be in a sense Fredholm Alternative for self-adjoint operators? Or is there a Fredholm Alternative for self-adjoint operators? I only know the version for compact operators.

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  • $\begingroup$ The general relationship is that $\text{Range}(A)^{\perp} = \text{Ker}(A^*)$ and $\text{Range}(A^*)^{\perp} = \text{Ker}(A)$. Since you're working with self-adjoint, $A$ and $A^*$ coincide. $\endgroup$
    – user932138
    Commented Jun 3, 2021 at 20:03
  • $\begingroup$ @O.Peters where can I find a reference to this result? $\endgroup$
    – Guilherme
    Commented Jun 3, 2021 at 20:06
  • $\begingroup$ For the bounded case it's theorem 12.10 in Rudin's functional analysis. I am probably wrong that there is exact equality in the unbounded case. The range of an operator never has to be closed. In the bounded case the kernel is always closed. In the unbounded case the kernel doesn't have to be closed so it's probably $\text{Range}(A)^{\perp} = \overline{\text{Ker}(A^*)}$ and not what I wrote. I don't see an exact analog in chapter 13 of Rudin about unbounded operators. The closest is theorem 13.10 about the relation of graphs of $T$ and $T^*$. $\endgroup$
    – user932138
    Commented Jun 3, 2021 at 20:15
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    $\begingroup$ See below for a precise statement. The kernel of a closed operator is always closed even in the unbounded case. For $T$ self-adjoint the below holds with $T = T^*$. A self-adjoint operator is densely-defined and closed. $\endgroup$
    – user932138
    Commented Jun 3, 2021 at 21:02
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    $\begingroup$ The Range of $A$ does not have to be closed. However, the closure of the range of $A$ is equal to the orthogonal complement of the null space of $A$. The orthogonal complement of the range of $A$ is equal to the null space of $A^*$. $\endgroup$ Commented Jul 9, 2021 at 23:17

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If $T$ is densely defined with domain $\mathcal{D}(T)$ then $T^*$ is closed and $$y \perp \text{Range}(T)\iff \forall x\in \mathcal{D}(T):\langle Tx,y\rangle=0 \iff y\in\mathcal{D}(T^*) \land T^*(y) = 0$$ This proves $$\text{Range}(T)^\perp = \text{Ker}(T^*)$$ and proves, inter alia that $\text{Ker}(T^*)$ is a closed subspace because the orthogonal complement to any set is a closed subspace.
If $T$ is densely defined and closable then $T^*$ is densely defined (and as before closed). Additionally, $T^{**} = \overline{T}$ the closure of $T$ obtained by taking the closure of the graph of $T$. From this we have $$\text{Range}(T^*)^\perp = \text{Ker}(T^{**})=\text{Ker}(\overline{T})$$

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  • $\begingroup$ But since $\text{Range}(T)$ is not closed, we cannot guarantee that $\text{Ker}(T)^{\perp}=\text{Range}(T)$? Right? $\endgroup$
    – Guilherme
    Commented Jun 3, 2021 at 21:09
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    $\begingroup$ That's right. The range of an operator is never guaranteed to be closed so it's not safe to say it's the orthogonal complement of something. However, the closure of $\text{Range}(T)$ is the orthogonal complement of $\text{Ker}(T)$ (for $T$ self-adjoint), because the closure of a subspace is always the orthogonal complement of the ortogonal complement. $\endgroup$
    – user932138
    Commented Jun 3, 2021 at 21:13

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