Two definitions of the operator $\exp(x)$ in $L^2(\mathbb R)$ The operator $x$ acts on a dense subspace of $L^2(\mathbb R)$ and is not bounded. So if we define $\exp(x)$ via the power series $\sum_{n=0}^\infty \frac {x^n}{n!}$, convergence will not follow in the operator norm topology.
However, for a few functions $f$ in $L^2$ the series $\sum_{n=0}^\infty \frac 1{n!} x^n(f)$ will converge. By taking $\mathcal H_1$ to be the subspace generated by such functions we have that the power series converges in the strong operator topology in $\mathcal H_1$. Lets call this definition of the exponential $X_1$.
Another way to define it is to simply consider those elements $f$ of $L^2$ for which $\exp(x) f(x)$ are again $L^2$ functions. Denote this space as $\mathcal H_2$ and define the operator $X_2$ as the pointwise multiplication of $\exp(x)$ with $f$ on $\mathcal H_2$.
Note that $\mathcal H_1 \neq \mathcal H_2$, as for example with
$$g(x):=\begin{cases}\frac1x &x<-1\\
0 &\text{else}\end{cases}$$
One has $g \in L^2(\mathbb R)$ and that $\exp(x) g(x) \in L^2(\mathbb R)$, so $g \in \mathcal H_2$. But $g$ does not lie in the domain of $x$ and as such also not in $\mathcal H_1$.

If one sees the definitions of the two operators, one would expect $X_1$ to be the more general and natural construction, as the definition of $X_2$ relies on specifically choosing to view the elements of the Hilbert space as functions. Yet somehow this way yields a more broadly defined and probably also more useful operator.
My question: Is there any natural/canonical way to extend the domain of $X_1$ after we have done this construction to recover $X_2$?
 A: Almost a year later, the answer comes from spectral theory.
The operator $x$ is a maximal self-adjoint operator, and the spectral theorem for unbounded operators tells us that such an operator carries with it a resolution of unity, meaning a system of projections $E_\lambda, \lambda\in\Bbb R$ so that:

  
*
  
*$\lambda'≤\lambda\implies E_\lambda'≤E_\lambda$
  
*$E_\lambda=\inf_{\lambda'>\lambda}E_\lambda'$ and $0=\lim\limits_{\lambda\to-\infty}E_\lambda$, $\Bbb1=\lim\limits_{\lambda\to+\infty} E_\lambda$ (limits in the strong operator topology) 
  
*$f\in D(x)\iff\int_{-\infty}^{\infty}\lambda^2 d\|E_\lambda\, f\|^2 <\infty$
  

One then has for $f\in D(x)$:
$$x(f) = \int_{-\infty}^\infty \lambda\ d(E_\lambda\, f)$$
The most natural way to define the domain of $\exp(x)$ would be:
$$f\in D(\exp(x))\iff\int_{-\infty}^\infty \exp(\lambda)^2 d\|E_\lambda\, f\|^2<\infty$$
And the operator works like:
$$\exp(x)(f) = \int_{-\infty}^\infty \exp(\lambda) d(E_\lambda\,f)$$
The result is a self-adjoint operator that should agree with $\sum_n \frac{x^n}{n!}$ where the latter is defined and be maximal in the sense that it admits no self-adjoint extension.
