I just wanted to check if the conclusion below is true, and whether the following reasoning works:
Let $H^i$ be the Sobolev spaces on a compact manifold $M$ and $D$ a self-adjoint (in the $H^0$-inner product) first-order differential operator on $M$. $D$ is initially defined only on $C_c^\infty(M)$ but can be extended to a bounded operator $$H^{i+1}\rightarrow H^i$$ for all $i\geq 0$. When viewed this way, $D$ is self-adjoint, in the sense of unbounded operators, in every $H^i$-inner product. That is,
$$\langle Dx,y\rangle_i = \langle x,Dy\rangle_i,$$
for all $x,y\in H^{i+1}\subseteq H^i$. This follows from self-adjointness of $D:H^1\rightarrow H^0$. By the criterion for self-adjointness, we know that $D^2+1:H^2\rightarrow H^0$ is a bounded invertible operator.
Conclusion: The bounded operator
$$D^2+1:H^{i+2}\rightarrow H^{i}$$
is invertible for every $i\geq 0$.
Thanks!