Showing derivative operator is self-adjoint Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain
$$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$
I’m showing that it’s self-adjoint. It’s not hard to show that it’s symmetric, and suffice to show that $D(A^*) \subseteq D(A)$, which I don’t know how to deal with.
Any help will be appreciated.
 A: Let $u\in D(A^\ast)$. For all $\phi\in C_c^\infty(\mathbb{R})$ we have
$$
\langle A^\ast u,\phi\rangle=\langle u,A\phi\rangle=-i\int u\phi'=i\langle u',\phi\rangle,
$$
where $u'$ is the distributional derivative of $u$.
Thus
$$
|\langle u',\phi\rangle|=|\langle A^\ast u,\phi\rangle|\leq \|A^\ast u\|_2\|\phi\|_2.
$$
Hence $u'\in L^2(\mathbb{R})$ by the Riesz representation theorem.
Now let
$$
\tilde u(x)=\int_0^x u'(y)\,dy.
$$
This function is absolutely continuous and has weak derivative $u'$. It follows that there exists a constant $C\in\mathbb{R}$ such that $u=\tilde u+C$ a.e. as explained here.
A: On the Hermite Function basis, $A$ is a tri-diagonal matrix
$$const \times
\begin{bmatrix}
0 & -i\sqrt{1} & 0 & 0 & 0 & \cdots \\
i\sqrt{1} & 0 & -i\sqrt{2} & 0 & 0 & \cdots \\
0 & i\sqrt{2} & 0 & -i\sqrt{3} & 0 & \cdots \\
0 & 0 & i\sqrt{3} & 0 & -i\sqrt{4} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}$$
(Wikipedia).
Since the off-diagonal elements are  $\mathscr O(\sqrt n)$,
by Carleman's test $A$ is essentially self-adjoint on the span of the Hermite Functions. Since the Hermite Functions are complete, $A$ is essentially self-adjoint on the larger domain.
A: Suppose $v\in\mathcal{D}(A^*)$. Then there exists $w\in L^2(\mathbb{R})$ such that, for all $u\in\mathcal{D}(A)$, one has
$$
           \int_{-\infty}^{\infty} (Au)vdx=\int_{-\infty}^{\infty}uwdx.
$$
Information about $w$ is obtained by choosing specific $u \in \mathcal{D}(A)$. For example, let
$$
         u_{a,b,\epsilon,\delta}=\int_{-\infty}^{x}\frac{1}{\epsilon}\chi_{[a-\epsilon,a]}(t)-\frac{1}{\epsilon}\chi_{[b,b+\delta]}(t)dt.
$$
This function is $0$ for $x < a-\epsilon$ and $0$ for $x > b+\delta$. It is $1$ in $[a,b]$, and it is linear on $[a-\epsilon,a]$ and on $[b,b+\delta]$. $u_{a,b,\epsilon,\delta}\in\mathcal{D}(A)$. So, the first equation applied with $u=u_{a,b,\epsilon}$ gives
$$
       \frac{1}{\epsilon}\int_{a-\epsilon}^{a}vdx-\frac{1}{\epsilon}\int_{b}^{b+\delta}vdx=\int_{-\infty}^{\infty}u_{a,b,\epsilon,\delta}wdx
$$
The limit as $\epsilon\downarrow 0$ exists, because it exists on the right. Similarly the limit as $\delta\downarrow 0$ exists. It follows that the following holds for almost all $a,b\in\mathbb{R}$:
$$
              v(a)-v(b)=\int_{a}^{b}w(x)dx
$$
Therefore, $v$ is equal a.e. to an absolutely continuous function that is in $L^2$ with a derivative $v'=w\in L^2$. So $v\in\mathcal{D}(A)$. In other words $\mathcal{D}(A^*)\subseteq\mathcal{D}(A)$. The opposite inclusion follows directly using the symmetry of $A$.
