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Are all smooth functions square-integrable?

  1. Aug 10, 2015 #1
    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]
  2. jcsd
  3. Aug 10, 2015 #2
    ##C^\infty_c## means smooth compactly supported functions (subsript c is for compactly supported). Compactly supported means it is 0 outside of some finite interval, so yes ##C^\infty_c(\mathbb R)\subset L^2(\mathbb R)##.
  4. Aug 10, 2015 #3
    Ah! thank you. I thought the c subscript meant complex. Makes perfect sense now.
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