Conditions for self-adjointness of unbounded operator

Question

I am reading this excerpt from this book. I do not understand how the self-adjointness of the operator $iA_0$ is proven. It is shown that it has to be symmetric and the range of $I\lambda-iA_0$ is dense in the Hilbert space $L^2(\mathbb{R}^n)$ and the Fourier transform is invoked as well as the space $C_0^{\infty}$. I don't follow that.

1- I can figure out what definition of self-adjointness is being used.

2- I do not follow the argument that constructs the solution to $(\lambda I-iA_0)u = f$ using the Fourier Transform. What exactly is the method for solving such equations (it doesn't look like a typical Fourier transform to me) and why is the solution an element of $D(A_0)=H^2$?

Answer 1

There's a criterion that a densely-defined symmetric operator T is self-adjoint if and only if

• $\operatorname{Range}(T+iI)$ and $\operatorname{Range}(T-iI)$ are both dense and closed (i.e. the whole space).

There are variations of this theme such as

• If the ranges are just dense, not closed then $T$ is essentially self-adjoint, i.e. its closure is self-adjoint. If T is already closed then the range isn't just dense, but also closed.
• Instead of $\pm i$ you can take the pair $T+ \lambda I, T+\bar\lambda I$ for any non-real $\lambda$.
• The criterion for density is equivalent to $\operatorname{Ker}(T^* - i) = \operatorname{Ker}(T^* + i) = 0$

The proof is found in almost any textook on the subject of unbounded operators and their spectral theorem. Skimming through, I think it's in here. There are probably more sources on the net if you look more thoroughly.

The dimensions of the orthogonal complements of $\operatorname{Range}(T\pm iI)$ are called the deficiency indices of $T$ and are a measure of how far it is from being maximally-symmetric or self-adjoint.