Can We Expand a Non-L^2 Function into an Orthonormal Basis?

[eljose]

Let,s suppose we have a function f(x) which is not on $$L^{2}$$ space but that we choose a basis of orthononormal functions so the coefficients:

$$c_{n}=\int_{0}^{\infty}dxf(x)\phi_{n}(x)$$ are finite.

would be valid to expand the series into this basis in the form:

$$f(x)=\sum_{n=0}^{\infty}\phi_{n}(x)$$ of course the sum:

$$\sum_{n=0}^{\infty}|c_{n}|^{2}$$ would diverge

 

[mathman]

I presume you meant to include the coefficients (c_n) in the infinite series for f(x). Depending on the properties of f(x), the series may or may not converge for any specific x.

 

[matt grime]

1. Pick a basis of what?

2. No that function is not equal to that sum for any reason at all, now wouldit be if you even put the c_n in as you meant to

3. It is not even true that an L^2 function is equal to its Fourier series

 

[eljose]

-I said a basis of orthonormal function (they are on L^{2} but f(x) isn,t)

-If the integral $$c_{n}=\int_{0}^{\infty}dxf(x)\phi_{n}(x)$$ is finite then every c coefficient exist.

-then when it would the equality hold?...$$f(x)\sim\sum_{n=0}^{\infty}c_{n}\phi_{n}(x)$$

perhapsh we could consider it to be an "asymptotic" representation of the function by means of eigenfunctions in the sense that you take only a few finite coefficients to approximate the function.

 

[matt grime]

1. You didn't, and still haven't said what they are a basis of. L^2 of what?

2. Why say equal and then write ~? Define your equivalence relation.