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Let $H$ be a $\mathbb R$-Hilbert space, $A$ be a densely-defined self-adjoint linear oprator on $H$, $\lambda\in\mathbb R$ and $A_\lambda:=\left.A\right|_{\mathcal D(A)\:\cap\:{\mathcal N(\lambda-A)}^\perp}$. It's easy to see that $$A_\lambda\left(\mathcal D(A)\cap{\mathcal N(\lambda-A)}^\perp\right)\subseteq{\mathcal N(\lambda-A)}^\perp\tag1$$ and hence $A_\lambda$ is a linear operator on ${\mathcal N(\lambda-A)}^\perp$. Since $\mathcal D(A)$ is dense in $H$, $\mathcal D(A_\lambda)=\mathcal D(A)\cap{\mathcal N(\lambda-A)}^\perp$ is dense in ${\mathcal N(\lambda-A)}^\perp$, i.e. $A_\lambda$ is densely-defined.

Now $$\langle A_\lambda x,y\rangle_H=\langle Ax,y\rangle_H=\langle x,A^\ast y\rangle_H\;\;\;\text{for all }x,y\in\mathcal D(A_\lambda)\tag2$$ and hence $\mathcal D(A_\lambda)\subseteq\mathcal D(A_\lambda^\ast)$ and $\left.A_\lambda^\ast\right|_{\mathcal D(A_\lambda)}=\left.A^\ast\right|_{\mathcal D(A_\lambda)}$, i.e. $A_\lambda$ is symmetric. Are we able to show that $A_\lambda$ is even self-adjoint?

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This is true. It might be true under more general assumptions on the space $\mathcal N(\lambda-A)^\perp$, but the fact that this is the complement of an eigenspace makes the relevant calculation very simple here.

Let $z\in \mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp}^*)$, that is $z\in\mathcal N(\lambda-A)^\perp$ so that $|\langle z , Ax\rangle|≤\|x\|\,C_z$ for all $x\in\mathcal D(A)\cap \mathcal N(\lambda -A)^\perp$. We want to show that $z\in\mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp})$. This, together with symmetry, shows self-adjointness of $A\lvert_{\mathcal N(\lambda-A)^\perp}$.

Now for any $y\in\mathcal D(A)$ you have $y=x+v$ with $v\in \mathcal D(A)\cap \overline{\mathcal N(\lambda-A)}= \mathcal N(\lambda -A)$, that is $v$ an eigenvector of $A$ to the eigenvalue $\lambda$, and $x\in \mathcal D(A)\cap\mathcal N(\lambda-A)^\perp$. Note that $\|x+v\|^2=\|x^2\|+\|v\|^2$ since $x$ and $v$ are perpendicular. It follows that:

$$\langle z, Ay\rangle = \langle z,Ax\rangle +\lambda\langle z,v\rangle = \langle z,Ax\rangle$$ since $z$ is perpendicular to the eigenvectors to $\lambda$ of $A$. Hence $$|\langle z, Ay\rangle| ≤ \|x\|\,C_z≤\|y\|\,C_z$$ and $z\in\mathcal D(A^*)$. But $\mathcal D(A^*)=\mathcal D(A)$, hence $z\in \mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp})$ and $A\lvert_{\mathcal N(\lambda -A)^\perp}$ is self-adjoint.

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  • $\begingroup$ Thank you for your answer! The whole point of my considerations is that I'd like to show that $\lambda\in\rho(A_\lambda)$ whenever $\lambda\in\sigma(A)$. I've asked for that here: math.stackexchange.com/q/3376495/47771. The crucial point is the boundedness of $\lambda-A_\lambda$. I guess this won't hold in general, but am willing to assume $\lambda-A$ is nonnegative. Is that sufficient? $\endgroup$
    – 0xbadf00d
    Oct 1, 2019 at 13:21
  • $\begingroup$ With $\lambda\in\rho(A_\lambda)$ you mean that $\lambda-A_\lambda$ is invertible (ie $\rho$ is the resolvent)? In that case you should be able to take some bounded, self-adjoint positive (or negative) semi-definite operator without eigenvalues so that $0\in\sigma(A)$. Then $A_\lambda=A$ for all $\lambda$, in particular you find $0\in\sigma(A_0)$ and $0\notin\rho(A_0)$. As an example look at the operator $x$ on $L^2([0,1])$ (or $-x$ if you want $\lambda -A$ to be nonnegative for $\lambda=0$). $\endgroup$
    – s.harp
    Oct 1, 2019 at 13:26
  • $\begingroup$ I mean the resolvent set, yes. As shown in the other question, $\lambda-A_\lambda$ is injective and has dense range for all $\lambda\in\mathbb R$, whenever $A$ is bounded and self-adjoint. All what's left to show for $\lambda\in\rho(A_\lambda)$ is that $(\lambda-A_\lambda)^{-1}$ is bounded. I want to show that this holds for all $\lambda\in\sigma(A)$ which are eigenvalues of $A$ (and I'm willing to assume that $\lambda-A$ is nonnegative, if that helps). $\endgroup$
    – 0xbadf00d
    Oct 1, 2019 at 13:31
  • $\begingroup$ I think that is false, the germ of an answer is in my previous comment. I'll go to the grocery store, when I come back I'll see about making it more formal! $\endgroup$
    – s.harp
    Oct 1, 2019 at 13:34
  • $\begingroup$ I guess (given a fixed $\lambda$) we need to assume that $(\lambda-\varepsilon)-A$ is nonnegative for some $\varepsilon>0$. Then we can apply math.stackexchange.com/a/3375845/47771. I've added details to the other question (math.stackexchange.com/q/3376495/47771). $\endgroup$
    – 0xbadf00d
    Oct 1, 2019 at 14:00

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