Verifying Inner Product Space: q(x)e^-(x^2/2)

[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!

 

[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).

 

[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)

 

[Fredrik]

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.

 

[mathman]

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.

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

 

[Fredrik]

Since f = 0 almost everywhere, this will be an inner product space.

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.

 

[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!