Basis for the set of all cts fns?

[tgt]

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

 

[dx]

Vector spaces don't have unique bases.

 

[HallsofIvy]

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.

 

[tgt]

Vector spaces don't have unique bases.

ok, what is a such basis?

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

 

[tgt]

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.

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

 

[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.

 

[gel]

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.

Isn't the Hamel basis just the same thing as the 'usual basis'? Anyway, from http://en.wikipedia.org/wiki/Hamel_basis#Related_notions"

The preference of other types of bases for infinite dimensional spaces is justified by the fact that the Hamel basis becomes "too big" in Banach spaces: If X is an infinite dimensional normed vector space which is complete (i.e. X is a Banach space), then any Hamel basis of X is necessarily uncountable. This is an easy consequence of Baire category theorem.

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

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.

 

[gel]

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.

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.

 

[HallsofIvy]

Isn't the Hamel basis just the same thing as the 'usual basis'? Anyway, from http://en.wikipedia.org/wiki/Hamel_basis#Related_notions"

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

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.

 

[g_edgar]

Right. It will converge uniformly to the continuous function f if and only if f(0) = f(1).

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

 

[gel]

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

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.

 

[John Creighto]

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

Right. It will converge uniformly to the continuous function f if and only if f(0) = f(1)."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.

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

 

[John Creighto]

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.

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.

 

[John Creighto]

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

How about the set of all delta functions?

 

[gel]

How about the set of all delta functions?

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

 

[John Creighto]

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

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.