# Large radius limit for a differential operator on a circle

Let $$(A_r)_{r\geq0}$$ be a one-parameter family of linear operators, with $$A_r$$ being the (weak) first derivative operator on $$L^2(S_r)$$, $$S_r$$ being the one-dimensional circle, with a multiplicative factor $$-i$$ to ensure symmetry. Each $$A_r$$ is known to be self-adjoint on its Hilbert space.

Intuitively, one may expect $$A_r$$ to "converge" to the first-derivative operator on the real line (again with a factor $$-i$$), say $$A_\infty$$, which is self-adjoint as well. However, all operators $$A_r$$ act on distinct Hilbert spaces, so all usual notions of operator convergence (e.g. resolvent ones), as to my knowledge, are of no use.

Is there a suitable definition of convergence of operators acting on distinct Hilbert spaces, such that $$A_r\to A_\infty?$$

The solution is to write down explicit expressions in coordinates. The circle of radius $$r$$ is parametrized as $$r\mathbb S^1=\{(r\cos \theta, r \sin \theta)\, :\, \theta\in (-\pi, \pi)\},$$ so the operator you refer to is $$i\partial_\theta f := i\partial_\theta [f(r\cos \theta, r \sin \theta)],$$ where $$f\in C^\infty(\mathbb R^2)$$. The chain rule shows that $$i\partial_\theta= ir (-\sin \theta, \cos \theta)\cdot \nabla,$$ so, in some sense, it diverges as $$r\to \infty$$.
Thus, I expect that the right operator to consider is the normalized one, $$\frac{i\partial_\theta}{r}.$$