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~
Best 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 subspace 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.