# Confusion about the definition of self adjoint and formally self-adjoint

I have some confusion about the definition of self-adjoint operators and formally self-adjoint operators. Let me write down the background information.

Let $H$ be a infinite dimensional complex Hilbert space and $T:D(T)\to H$ a (not necessarily bounded) linear operator , where $D(T)$ is dense in $H$. The operator $T$ is said to be formally self-adjoint if for all $x, y\in D(T)$, we have

$\langle Tx, y\rangle=\langle x, Ty\rangle.$

The operator $T$ is said to be self-adjoint if $T^*=T$, where $T^*$ is the adjoint of $T$. Of course if $D(T)=H$ and $T$ is formally self-adjoint in the above sense, then $T$ must be bounded and therefore $T$ is also self adjoint.

Let $M$ be a closed Riemannian manifold and $P:C^\infty(M)\to C^\infty(M)$ an elliptic pseudodifferential operator. We know $C^\infty(M)$ is dense in $L^2(M)$ (and even in $H_s(M)$).

My confusion is the following. Suppose $P$ is formally self-adjoint, i.e., for all $f, g\in C^\infty(M)$, we have

$\langle Pf, g\rangle_{L^2(M)}=\langle f, Pg\rangle_{L^2(M)}.$

We also know that the extension of $P$ to $P:H_s(M)\to H_{s-d}(M)$ is a bounded linear operator (indeed it's Fredholm). Well, then I don't know how to continue my question, perhaps I am really confused by my confusion. Anyway, I would appreciate if someone can clear my confusion for which I couldn't even explain.

Thanks~

• If $P$ is a differential operator, it usually doesn't have a continuous (bounded) extension $L^2(M)\to L^2(M)$. – Daniel Fischer Nov 8 '14 at 15:57
• oh, ok. let me edit and ask for the case of elliptic operator – Ho Man Ho Nov 8 '14 at 16:12
• Elliptic operators still usually aren't bounded. The standard example of an elliptic operator, the Laplace operator, is not. – Neal Nov 8 '14 at 16:18
• (By the way, @indextheory, I changed < and > to \langle and \rangle. These are the standard latex symbols for inner product brackets.) – Neal Nov 8 '14 at 16:24

## 1 Answer

I think you are confusing the framework for studying a symmetric operator. The setting is $$T : \mathcal{D}(T)\subseteq X\rightarrow X.$$ If you consider $T : \mathcal{D}(T) \rightarrow X$ where you put the graph norm on $\mathcal{D}(T)$, then $T$ always becomes bounded, but that's not the framework for a symmetric operator; you can't even take the inner product of $Tx$ with $y$ in this setting because $x$, $Tx$ no longer lie in the same space. And that's what you're doing when you consider $P : H_{s}(M)\rightarrow H_{s-d}(M)$. The proper setting is that you consider $\mathcal{D}(P)=H_{s}(M)$ not in its native norm, but as a subsapce of $H_{s-d}(M)$. Then you can form the inner product of $x$ and $Px$ for $x\in\mathcal{D}(P)$.

Any time you have a closed operator $T : \mathcal{D}(T)\subseteq X\rightarrow X$, you can always define $T$ to become bounded by defining a new space $Y=\mathcal{D}(T)$ endowed with the graph norm $\|y\|_{Y}=\|Tx\|_{X}+\|x\|_{X}$. The new map $T : Y\rightarrow X$ now becomes continuous because $$\|Ty\|_{X} \le \|Ty\|_{X}+\|y\|_{X}= \|y\|_{Y},$$ but the map is now from $Y$ to $X$ instead of from a subspace of $X$ to $X$.

This becomes particularly confusing in $L^{2}$ Sobolev spaces, where the norms are essentially defined in terms of the domains of fractional powers of the Laplacian. You're tempted to want to do exactly what you did.

• thanks for your explanation. I guess I am confused with the term "let $P:C^\infty(M)\to C^\infty(M)$ be a formally self-adjoint elliptic pseudodifferential operator". This sentence makes me think the author is considering $C^\infty(M)$ as a dense subspace of $L^2(M)$ or $H_s(M)$. – Ho Man Ho Nov 9 '14 at 6:56
• @indextheory : In all likelihood that is the meaning. However, some people deal directly for operators when they say formally selfadjoint. In that setting, the formal adjoint of the operator is obtained by manipulation of the operator symbols. For example, $(\partial_{x} a \partial_{y})^{\dagger}=(-\partial_{y})\overline{a}(-\partial_{x})$, etc. I doubt that's what they mean because of specifying a domain, but you never now for sure if they're not trying to be clear. – DisintegratingByParts Nov 9 '14 at 13:06
• Thanks. I was reading one of the famous about pseudodifferential operator, and that is precisely what the author says. I guess he is known to be.....not care about things like this. – Ho Man Ho Nov 10 '14 at 1:04