# The restriction of a self-adjoint operator $A$ to ${\mathcal N(\lambda-A)}^\perp$ is symmetric, but is it even self-adjoint?

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?

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.

• 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? – 0xbadf00d Oct 1 '19 at 13:21
• 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 $\lambda=0$). – s.harp Oct 1 '19 at 13:26
• 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). – 0xbadf00d Oct 1 '19 at 13:31
• 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! – s.harp Oct 1 '19 at 13:34
• 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). – 0xbadf00d Oct 1 '19 at 14:00