# Dimension of a Hilbert space

• Oct 10th 2009, 11:34 PM
tjkubo
Dimension 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, 03:37 AM
Failure
Quote:

Originally Posted by tjkubo
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.

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, 10:52 AM
HallsofIvy
For example, if are talking about the space of square integrable functions on a finite interval, $L^2([a, b])$ then all functions of theform $x^n$ 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.