Can I conclude these weak derivatives are strong

Question

I'm trying to show the momentum operator $P : \mathcal D(P)\ \dot{=}\ C^\infty_0(\Bbb R)\to L^2(\Bbb R)$ with $P=-i\hbar\partial_x$ is essentially self-adjoint, which is equivalent to saying that it is symmetric (densely defined, which is true because $\overline{C^\infty_0}(\Bbb R)=L^2(\Bbb R)$, and Hermitian) and $\ker(P^*\pm i 1) = \{0\}$. The momentum is symmetric because
$$\require{cancel}(\varphi|P\psi) = \int_\Bbb R \overline{\varphi(x)}(-i\hbar\partial_x\psi)(x)\ dx = \cancel{[-i\hbar \overline \varphi \psi]_{-\infty}^{+\infty}} + \int_\Bbb R \overline{-i\hbar \text{w-}\partial_x\varphi(x)}\psi(x)\ dx, $$
where $\text{w-}\partial_x$ indicates the weak derivative, and the equation holds for all $\psi \in C^\infty_0(\Bbb R)$ and $\varphi \in \mathcal D(P^*)$, which can (and should) be taken to be the maximal domain of definition, i.e. the Sobolev space
$$\mathcal D(P^*) = W^{1,2}(\Bbb R)=\{\varphi \in L^2(\Bbb R)\ |\ \text{w-}\partial_x\varphi \in L^2(\Bbb R)\}. $$
Now, to show the kernels of $P^*\pm i 1$ are trivial, I can take $\psi_\pm\in\ker(P^*\pm i1)$, which means that
$$\text{w-}\partial_x \psi_\pm(x) = \pm \frac 1 \hbar \psi_\pm(x). $$
At this point, I could proceed to ignore the $\text{w-}$ symbol and just treat the derivative as a strong derivative, so I can solve the ODE via $\psi_\pm(x) = a_\pm e^{\pm x/\hbar}$, which is $\in L^2$ iff $a_\pm = 0$ as needed. But why can I do this? The Sobolev embedding theorem only seems to guarantee that $\psi_\pm \in C^0(\Bbb R)$, which wouldn't be enough to argue that $\psi_\pm$ can be strongly differentiated.

Answer 1

A bootstrapping argument shows that we can worry only about classical solutions: Consider the equation \begin{equation}\tag{1} \text{w-}\partial_x \psi= \dfrac{\psi}{h}, \end{equation} given that $\psi \in W^{1,2}$ (the domain doesn't matter here). Since the right hand side is weakly differentiable, then so is the left and so w-$\partial_x \psi\in W^{1,2}$. This means two things:

1. $\psi\in C^1$ by Sobolev embedding.
2. We can (weakly) differentiate the equation (1) to obtain that w-$\partial_x^2\psi \in W^{1,2}$.

Now you basically iterate step 2 to conclude that w-$\partial_x^k\psi\in W^{1,2}$ for all $k\geq 0$, and by Sobolev embedding (as in step 1) $\psi\in C^\infty$.