In the chapter 6.4 on normal and self-adjoint operators, there is an example of an infinite dimensional inner product space H that has a normal operator but that has no eigenvectors.


The space is the set of functions f\_n(t) = e^(int) , t in [0,2pi]
with inner product = 1/2pi \* integral\_0\_2pi(e^(-int) e^(imt))dt


The operator T(f\_n) = f\_(n+1)


My question is what is the basis for the validity of the following statement.


any vector f in the inner product space can be represented as f = sum\_i=n\_to\_i=m(a\_i \* f\_i),
a\_m <> 0


They use this fact to prove that the operator as no eigenvectors. But it seems to imply that any member of an infinite dimensional inner product space can be represented with a finite number of basis elements, since n and m are finite.


Are there no elements of an infinite dimensional inner product space that require an infinite number of terms to represent?



Friedberg, Insel, and Spence Linear Algebra example 
The space being defined is the **span** of the functions fn, and in the definition of the span we only allow finite sums of the basis vectors.


Edit: I should also mention that the notion of infinite sum in an inner product space doesn't make sense unless the space is also complete with respect to the induced norm, i.e. is a Hilbert space. In the nice situation a Hilbert space has an "orthonormal basis," which is not a basis in the linear algebra sense but in the sense that the span of the basis is dense.


