# Inner Product Space

1. Oct 8, 2011

### Thunder_Jet

Hi everyone!

I would like to ask how would you verify if functions form an inner product space? For example, if one has functions of the form q(x)e^-(x^2/2) where q(x) is a polynomial of degree < N in x, on the interval -∞ < x < ∞. Also, how would you specify the dimension of the space, if it exists?

Thank you!

2. Oct 8, 2011

### mathman

The dimension is N. It forms an inner product space, since the integral of the product of any two such functions is finite (the exponential term insures this).

Last edited: Oct 8, 2011
3. Oct 9, 2011

### Noblee

Check using the definition of an inner product would be my initial suggestion.

1) $(u|u)\geq 0$ and $0$ iff $u = 0$
2) $(\alpha u+ \beta v|w) = \alpha (u|w) + \beta (v|w)$ Aka that it is linear

If this holds true $\forall u,v,w$ then the inner product is defined for the said space and is thus a inner product space (given of course that it is in a vector space to begin with)

4. Oct 9, 2011

### Fredrik

Staff Emeritus
Note that vector spaces whose elements are functions usually fail the condition $\langle f,f\rangle=0\Rightarrow 0$. (Define f by f(x)=1 when x=0 and f(x)=0 otherwise; then $\langle f,f\rangle=0$ but f≠0). Such a space is sometimes called a semi-inner product space.

5. Oct 9, 2011

### mathman

Since f = 0 almost everywhere, this will be an inner product space. Moreover it is complete, so it is L2.

6. Oct 9, 2011

### Fredrik

Staff Emeritus
By definition of inner product, it's not. But you can define an equivalence relation by saying that f~g if f=g almost everywhere, and then define an inner product on the set of equivalence classes by $\langle [f],[g]\rangle=\langle f,g\rangle$. That's an inner product on the left and a semi-inner product on the right.

7. Oct 14, 2011

### Thunder_Jet

Thank you so much for the insights! All of your comments gave me an idea on how to attack the problem! Thanks once again!