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Is it possible to define a basis for the space of continuous functions?

  1. Oct 5, 2012 #1
    In analogy to vector spaces, can we define a set of "basis functions" from which any continuous function can be written as a (possibly infinite) linear combination of the basis functions?

    I know the trigonometric functions 1, sin(nx), cos(nx) can be used for monotonic continuous functions, but not every Fourier series leads to a convergent solution (sinx+sin2x+sin3x+... for example diverges).

    Is there a set of functions (not necessarily orthogonal) that spans all continuous functions and does not contain divergent series?
  2. jcsd
  3. Oct 5, 2012 #2


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    Hey Boorglar.

    You want to consider a Hilbert-Space and the properties of those along with Banach Spaces in Functional Analysis.

    Basically Hilbert-Spaces are continuous in the inner product and Banach Spaces are continuous in the norm: so you can look at these in the context of general functionals and function spaces.
  4. Oct 5, 2012 #3
    Yes, I've heard about those spaces. But do they actually provide us with such a basis? I mean, is there an infinite set of known functions that span all the Hilbert space?
  5. Oct 5, 2012 #4


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    Why would this be true? You are asking for ##\sum_{i=0}^\infty a_n \, f_n(x)## where ##f_n## are fixed but the coefficients are arbitrary. We can choose ##a_n \rightarrow \infty## at any rate we want, so there must exist a divergent series (unless all f are zero).
  6. Oct 5, 2012 #5
    hmm yes I guess you're right.

    The reason I ask for that is that I was looking for a way to define integration over a space of functions i.e.: let F be a functional, taking a function and returning a number. Then I want to "integrate" this functional with respect to the function argument.

    For example: let F[f] = ∫10f(x)dx. Then somehow divide the space of functions into small function intervals (which might perhaps be done if you had a basis) and do a Riemann sum of the values of the functional taken at some arbitrary function in each interval.

    If the function space was simply c*e^x where c is a real number, then I could do the integral of the value of F[ce^x] with respect to c from -infinity to +infinity. (this would be a one-dimensional function space)
    Last edited: Oct 5, 2012
  7. Oct 5, 2012 #6

    Stephen Tashi

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    What is a "function interval"? Are we talking about functions defined on an interval of the real number line? Are "function intervals" going to be subintervals of that interval?

    If [itex] f(x) = \sum_{i=0}^\infty c_i g_i(x) [/itex] then the natural way to integrate [itex] f(x) [/itex] over an interval such as [0,1] would be to use an expession [itex] \sum_{i=0}^\infty c_i \int_0^1 g_i(x) dx [/itex] rather than to integrate over subintervals. Are you thinking about breaking [0,1] up into subintervals and using a different basis for each subinterval?
  8. Oct 5, 2012 #7


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    How are you defining function? Continuous functions won't be defined analytically in general (although some can be).

    The first thing I think you should do is consider a general continuous function in a fixed interval and consider that if it is square integrable (i.e. in L^2) if it has any basis in a Hilbert-Space.

    If it has a basis, then consider the properties that this basis must have if you want to go deeper.

    I'd look at the first one to "check" that a square integrable function (in L^2(R^n)) over some interval has a basis (any basis) and then you can go from there.
  9. Oct 5, 2012 #8


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    Ah you want Riemann integration on a function space. Start with Muldowney's paper on the topic (although this is specific to Black-Scholes). There should be a couple of other papers in the literature. I remember Henstock had a paper for an arbitrary metric-measure space.
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