What is 'completeness' (function space)

[zetafunction]

given a set of functions that depend on a parameter lambda $$f(\lambda x)$$ , how can be proved or what does it mean that this set of functions is COMPLETE in $$L^{p} (a,b)$$ do the functions $$f(\lambda x)$$ need to form an orthogonal basis or it is enough that for diffrent values of lambda ,there is a linear independence.

 

[elibj123]

I'm not sure about sets which are not orthogonal, but I think that completness generally says that there is no other non-trivial vector which is linearly independent of the set vectors.

 

[Landau]

In this context, completeness usually means 'maximal spanning', i.e. every other element of L^p(a,b) should be expressible as (not necessarily finite) linear combination of these 'functions'. It is an extension of the concept of 'basis' from finite-dimensional linear algebra to infinite dimensions. The orthogonal (or even orthonormal) requirement usually is explicitly added, as in "complete orthonormal set".

See e.g. here.

 

[rochfor1]

Picky aside: orthogonality isn't well-defined when p != 2.

 

[wofsy]

given a set of functions that depend on a parameter lambda $$f(\lambda x)$$ , how can be proved or what does it mean that this set of functions is COMPLETE in $$L^{p} (a,b)$$ do the functions $$f(\lambda x)$$ need to form an orthogonal basis or it is enough that for diffrent values of lambda ,there is a linear independence.

The L^p(a,b) is a metric space. The space of functions is complete if each Cauchy sequence converges in the L^p metric to another function in the space.

 

[Landau]

@wofsy: that's a different type of completeness; here we're talking about completeness of a set of vectors in L^p, not of the (metric) space L^p itself!
See the link in my previous post.