# Are all smooth functions square-integrable?

1. Aug 10, 2015

### pellman

Came across this in a discussion of essential self-adjointedness:

Let P be the densely defined operator with Dom(P) = $C^{\infty}_c (\mathbb{R}) \subset L^2 ( \mathbb{R} )$ and given by $Pf = -i df/dx$. Then P is essentially self-adjoint.

It is the $C^{\infty}_c (\mathbb{R}) \subset L^2 ( \mathbb{R} )$ 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 $C^{\infty}_c (\mathbb{R})$ ? 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 $C^{\infty}_c (\mathbb{R})$ or $L^2 ( \mathbb{R} )$

2. Aug 10, 2015

### Hawkeye18

$C^\infty_c$ means smooth compactly supported functions (subsript c is for compactly supported). Compactly supported means it is 0 outside of some finite interval, so yes $C^\infty_c(\mathbb R)\subset L^2(\mathbb R)$.

3. Aug 10, 2015

### pellman

Ah! thank you. I thought the c subscript meant complex. Makes perfect sense now.