- #1
pellman
- 684
- 5
Came across this in a discussion of essential self-adjointedness:
Let P be the densely defined operator with Dom(P) = [itex]C^{\infty}_c (\mathbb{R}) \subset L^2 ( \mathbb{R} )[/itex] and given by [itex]Pf = -i df/dx[/itex]. Then P is essentially self-adjoint.
It is the [itex]C^{\infty}_c (\mathbb{R}) \subset L^2 ( \mathbb{R} )[/itex] part that bothered me. Doesn't this say the space of smooth complex functions on R is contained in the space of square-integrable functions on R? But isn't, say, f(x) = x an element of [itex]C^{\infty}_c (\mathbb{R}) [/itex] ? And isn't f(x) = x not square-integrable on R?
I am thinking maybe I don't know what the author means by the symbols [itex]C^{\infty}_c (\mathbb{R}) [/itex] or [itex]L^2 ( \mathbb{R} )[/itex]
Let P be the densely defined operator with Dom(P) = [itex]C^{\infty}_c (\mathbb{R}) \subset L^2 ( \mathbb{R} )[/itex] and given by [itex]Pf = -i df/dx[/itex]. Then P is essentially self-adjoint.
It is the [itex]C^{\infty}_c (\mathbb{R}) \subset L^2 ( \mathbb{R} )[/itex] part that bothered me. Doesn't this say the space of smooth complex functions on R is contained in the space of square-integrable functions on R? But isn't, say, f(x) = x an element of [itex]C^{\infty}_c (\mathbb{R}) [/itex] ? And isn't f(x) = x not square-integrable on R?
I am thinking maybe I don't know what the author means by the symbols [itex]C^{\infty}_c (\mathbb{R}) [/itex] or [itex]L^2 ( \mathbb{R} )[/itex]