Why is the dimension of the Hilbert space of square-integrable functions infinite? I guess I'm having trouble visualizing the basis of this space.

Printable View

- Oct 10th 2009, 10:34 PMtjkuboDimension of a Hilbert space
Why is the dimension of the Hilbert space of square-integrable functions infinite? I guess I'm having trouble visualizing the basis of this space.

- Oct 11th 2009, 02:37 AMFailure
Why not just show a concrete infinite set of linearly-independent square-integrable functions? (Although, come to think of it, we must know something more about the

*domain*of those square-integrable functions. If that domain can be*any*measure space, then the proposition might not even be true in general.) - Oct 11th 2009, 09:52 AMHallsofIvy
For example, if are talking about the space of square integrable functions on a finite interval, $\displaystyle L^2([a, b])$ then all functions of theform $\displaystyle x^n$ for positive integer n are in the space. Can you show that they are independent?

A set of functions like sin(nx) for integer n would be a natural choice also.