I'm reading Quantum mechanics: a modern development by Leslie Ballentine (1998), page 19:
A simple counter-example shows that the theorem is not generally true for an infinite-dimensional space. Consider the operator $D = −id/dx$, defined on the space of differentiable functions of $x$ for $a ≤ x ≤ b$. (The limits a and b may be finite or infinite.) Its adjoint, $A^\dagger$, is identified by using (1.21), which now takes the form $$\int_a^b\phi^*(x)D^\dagger\psi(x)dx = \left(\int_a^b\psi^*(x)D\phi(x)dx\right)^*=$$ $$\int_a^b\phi^*(x)D\psi(x)dx+i[\psi(x)\phi^*(x)]|_a^b\ \ \ \textbf{(1.31)}$$ The last line is obtained by integrating by parts. If boundary conditions are imposed so that the last term vanishes, then $D$ will apparently be a Hermitian operator.
I have two questions here.
First, it seems to me (though I may be incorrect about meaning of "identified") that to write (1.31), (1.18) (definition of adjoint) is used, not (1.21), which is definition of Hermitian (this the author is only trying to establish for different boundary conditions later). Am I correct or not? See (1.18) and (1.21) below (*).
Secondly, after I try to do integration by parts, my taking of conjugate of remaining integral, as opposed to the result from the book, I get $\overline{dx}$ (where there is $dx$ in the book for $D$ operator) and also different order of $\phi$ and $\psi$ (I've marked some conjugates by overline as in Does the complex conjugate of an integral equal the integral of the conjugate?), I hope using two different marks is not confusing:
$$\overline{\int_a^b\psi^*(x)D\phi(x)dx} = \overline{\int_a^b-i\psi^*(x)\frac{\phi(x)}{dx}dx}= \overline{-i[\psi^*(x)\phi(x)]|_a^b+\int_a^b i\frac{\psi^*(x)}{dx}\phi(x)dx}=i[\psi(x)\phi^*(x)]|_a^b-\int_a^b i\frac{\psi(x)}{\overline{dx}}\phi^*(x)\overline{dx}$$
Having order in domain of x (differentiable functions of x for a ≤ x ≤ b) to me almost certainly implies x is a real number. Is it correct? Or could it be something else (the book does not define domain for $x$)?
If so, is the claim in linked QA that $\overline{dx}$ is same as $dx$ if $f$ is a function of a real variable correct?
Also, even if $x$ is just a real number, could codomains of functions be neither $\mathbb{R}$ nor $\mathbb{C}$? What allows to change order of $\phi$ and $\psi$ if so?
TIA
(*)
An operator $A$ that is equal to its adjoint $A^\dagger$ is called self-adjoint. This means that it satisfies $$\langle\phi|A|\psi\rangle=\langle\psi|A|\phi\rangle^*\ for\ all\ > \phi\ and\ \psi \ \ \ \ (1.21)$$ and that the domain of A (i.e. the set of vectors $\psi$ on which $A\psi$ is well defined) coincides with the domain of $A^\dagger$ . An operator that only satisfies (1.21) is called Hermitian
$$(A^\dagger\phi,\psi)^*=(\psi,A\phi)\ for\ all\ \phi\ and\ \psi \ \ \ (1.17)$$ This is the usual definition of the adjoint, $A^\dagger$, of the operator $A$.
$$\langle\phi|A^\dagger|\psi\rangle^*=\langle\psi|A|\phi\rangle\ for\ > all\ \phi\ and\ \psi \ \ \ \ (1.18)$$ this relation being equivalent to (1.17). Although simpler than the previous introduction of $A^\dagger$ via the Riesz theorem, this formal method fails to prove the existence of the operator $A^\dagger$.
