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Orthogonal complement of even functions

  1. May 29, 2014 #1
    The problem statement, all variables and given/known data.

    Let the ##\mathbb R##- vector space ##C([-1,1])=\{f:[-1,1] \to \mathbb R\ : f \space \text{is continuous}\ \}## with the inner product ##<f,g>=\int_{-1}^1 f(t)g(t)dt##. Determine the orthogonal complement of the subspace of even functions (call that subspace ##S##).

    The attempt at a solution.

    Straight from the definition of orthogonal complement, ##g \in (C([-1,1]))^\bot## if and only if ##\int_{-1}^1 g(t)f(t)dt=0## for all ##f## even function. By hypothesis, ##f(-t)=f(t)##, so ##\int_{-1}^1 g(t)f(-t)dt=0## From this condition I don't know what else to conclude. I would appreciate any suggestions.
     
    Last edited: May 29, 2014
  2. jcsd
  3. May 29, 2014 #2

    micromass

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    Some things to try:

    1) The function ##f(x) = 1## for all ##x## is an even function. We want an element of ##S## to be orthogonal to that, what does that imply for functions in ##S##?

    2) Can you give some (nontrivial) examples of functions in ##S##?
     
  4. May 29, 2014 #3
    Hmm, for 1), the only thing I can think of is something that follows trivially, which is: ##\int_{-1}^1 g(t)dt=0##.

    2) Yes, I can think of ##\sin(x)##, which is an odd function. Suppose ##g## is odd, then ##\int_{-1}^1 g(-t)f(-t)dt=-\int_{-1}^1 g(t)f(t)dt=0## since ##g(-t)=-g(t)## and ##f(-t)=f(t)##, then the set of odd functions (subspace) is included in ##S##.

    Is there any other function which belongs to ##S##? I don't know what you are trying to insinuate me in 1).
     
  5. May 29, 2014 #4

    micromass

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    Forget about (1), it was just to give you some more intuition.

    But it seems you now have a good conjecture. Is ##S## exactly the set of odd functions or not? I think you should try to prove that it is.
     
  6. May 29, 2014 #5

    pasmith

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    Every function can be written as the sum of an even function and an odd function. Thus, given an arbitrary [itex]g \in C([-1,1])[/itex], we have [itex]g = f + h[/itex] where [itex]f \in S[/itex] and [itex]h[/itex] is odd. It would be worth looking at [tex]
    \int_{-1}^1 f(x)g(x)\,dx = \int_{-1}^1 f(x)^2\,dx + \int_{-1}^1 f(x)h(x)\,dx.
    [/tex]
     
  7. May 29, 2014 #6
    I want to check if this is correct:

    Taking your suggestion, ## \int_{-1}^1 f(x)^2\,dx + \int_{-1}^1 f(x)h(x)\,dx=\int_{-1}^1 f(x)^2\,dx =\int_{-1}^0 f(x)^2\,dx + \int_0^1 f(x)^2\,dx##. Making the substitution ##x=-t##, I get ##\int_{-1}^0 f(x)^2\,dx + \int_0^1 f(x)^2\,dx=-\int_1^0 f(-u)du+ \int_0^1 f(x)^2\,dx=\int_0^1 f(u)du+ \int_0^1 f(x)^2\,dx=2\int_0^1 f(x)^2\,dx##

    But this means ##\int_0^1 f(x)^2\,dx=0 \implies f^2(x)=0 \implies f(x)=0##. From here it follows ##g=h##, but ##h## is an odd function so ##g## is odd.
     
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