How about expressing cosx in terms of span{sinkx}?
Let be the space of continuous functions with norm || . I am trying to determine whether the linearly independent set is dense in Somebody gave me a hint that it is not dense and to look at the constant function 1. So that means I am trying to show that there is some open ball around the function 1, which contains no element of .
This means that I need to show for some 0.
Has anyone got any hints regarding how to go about this? Thanks very much
Well... If you suppose that f=cosx lies in span{sinkx}, then, for every a>0 there is a finite sum S(a), such that
||f-S(a)||<a. But since f(0)=1 and all finite sums are zero at x=0, in a sufficiently small right interval about 0, we can have f-S(a)>a/pi.
Integrating gives a>a.
The ingredients for a solution are already here in the comments of Rebesques and Drexel28. Define a continuous linear functional on by . Then for all n, but . So the function is not in the kernel of , which is a closed subspace of .
[I have moved this thread, because it's really a problem in functional analysis rather than linear algebra. The solution depends on the fact that is the dual of .]
Dammit, you're right! And what's more, I'm changing my mind about whether this result is true.
I now think that every continuous function on can be approximated in the -norm by a linear combination of the functions . In fact, let be a continuous function on , and extend to an odd function on , (so that ). Then the Fourier series of consists only of sine terms. Let denote the sum of the first n terms of that series. Then as , where is the -norm, given by . By the Cauchy–Schwarz inequality, if is the -norm of h, given by , then
.
It follows that , and if you restrict the interval of integration to , this tells you that . In other words, can be approximated in the -norm by a linear combination of functions .