# Basis for the set of all cts fns?

1. Jul 18, 2009

### tgt

What is the basis for the vector space of all continuous functions?

Last edited: Jul 18, 2009
2. Jul 18, 2009

### dx

Vector spaces don't have unique bases.

3. Jul 18, 2009

### HallsofIvy

Staff Emeritus
Further, I'm inclined to suspect, although I can't prove it, that any basis would be uncountable. That is (probably) why it is more common to use a Hamel basis rather than the usual basis of Linear Algebra.

4. Jul 18, 2009

### tgt

ok, what is a such basis?

Isn't it that any cts function can be modelled by sins and cosines? I could be completely wrong.

5. Jul 18, 2009

### tgt

So you don't know what a basis could be?

6. Jul 18, 2009

### g_edgar

Let's consider $$C[0,1]$$, the space of continuous functions on the interval $$[0,1]$$. There is a natural norm making this a Banach space. (Convergence in that norm is uniform convergence of functions.) A Hamel basis for this space will, indeed, be uncountable. But also is of no practical use. Theoretical use, perhaps, but not practical.

Another type of basis is the Schauder basis, where we allow infinite-series expansions (of course they must converge according to the norm). Schauder himself in 1926 gave a basis for $$C[0,1]$$ consisting of certain piecewise-linear functions.

The family $$\sin(nx), \cos(nx)$$ is not a Schauder basis for $$C[0,1]$$, however. The Fourier series of a continuous function need not converge uniformly.

The family $$x^n$$ of powers of $$x$$ is also not a Schauder basis for $$C[0,1]$$... If a series $$\sum_{n=0}^\infty a_n x^n$$ converges uniformly, then the sum is differentiable, so not all continuous functions can be expanded this way.

7. Jul 18, 2009

### gel

Isn't the Hamel basis just the same thing as the 'usual basis'? Anyway, from http://en.wikipedia.org/wiki/Hamel_basis#Related_notions"
Assuming that you do mean a Hamel basis, then I expect that its existence relies on the axiom of choice, and that no-one could give you a specific example.

Last edited by a moderator: Apr 24, 2017
8. Jul 18, 2009

### gel

Right. It will converge uniformly to the continuous function f if and only if f(0) = f(1). As any function can be split into a linear term and term taking the same values at 0 and 1, we can extend sin(nx), cos(nx) to a Schauder basis by adding the linear basis function u(x)=x.

Alternatively, hierarchical basis functions can be used.

9. Jul 18, 2009

### HallsofIvy

Staff Emeritus
I thought I had looked at that site! But you are right. I have the "Hamel" basis and "Schauder" basis reversed.

Last edited by a moderator: Apr 24, 2017
10. Jul 18, 2009

### g_edgar

No. Not even if f(0)=f(1).

11. Jul 18, 2009

### gel

Aargh, you're right. Converges uniformly if also of finite variation. Merely continuous funtions aren't even guaranteed to converge everywhere - just almost everwhere.

Still, hierarchical basis functions such as hat functions can be used for a basis.

12. Jul 18, 2009

### John Creighto

There are an awful lot of functions which can be fit to a Fourier series. I'd be interested in hearing some counter examples.

13. Jul 18, 2009

### John Creighto

Ah. I didn't know what uniform convergence ment. Well, what about every differential function then. Would the Fourier series be sufficient basis for the interval zero to one? As for converging almost everywhere it gives an average error of zero which sounds good to me anyway for a lot of applications.

14. Jul 18, 2009

### John Creighto

How about the set of all delta functions?

15. Jul 18, 2009

### gel

Delta functions aren't continuous, nor are they functions.

16. Jul 18, 2009

### John Creighto

Yeah, but it is used in quantum mechanics as a basis. Another thing that is used is the fourier integral. I also think the delta function can be expressed as an infinite sum of sinc functions. Seems to be some possibilities.