## Basis for the set of all cts fns?

What is the basis for the vector space of all continuous functions?
 PhysOrg.com science news on PhysOrg.com >> A robot that runs like a cat (w/ Video)>> Global cooling as significant as global warming, research shows>> 'Chase and run' cell movement mechanism explains process of metastasis
 Blog Entries: 1 Recognitions: Gold Member Homework Help Vector spaces don't have unique bases.
 Recognitions: Gold Member Science Advisor 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.

## Basis for the set of all cts fns?

 Quote by dx Vector spaces don't have unique bases.
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.

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

 Quote by 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.
Isn't the Hamel basis just the same thing as the 'usual basis'? Anyway, from Wikipedia
 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.
 Quote by tgt 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.

 Quote by g_edgar 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.

Recognitions:
Gold Member
Staff Emeritus
 Quote by gel Isn't the Hamel basis just the same thing as the 'usual basis'? Anyway, from Wikipedia
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.

 Quote by gel 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).

 Quote by g_edgar 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.

Blog Entries: 3
 Quote by gel "' 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.

Blog Entries: 3
 Quote by 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.
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.

Blog Entries: 3
 Quote by tgt What is the basis for the vector space of all continuous functions?
How about the set of all delta functions?

 Quote by John Creighto How about the set of all delta functions?
Delta functions aren't continuous, nor are they functions.

Blog Entries: 3
 Quote by gel 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.

 Similar Threads for: Basis for the set of all cts fns? Thread Forum Replies Quantum Physics 0 Advanced Physics Homework 0 Calculus & Beyond Homework 7 Calculus & Beyond Homework 4 Introductory Physics Homework 5