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Math Help - Regulated Function

  1. #1
    Super Member Showcase_22's Avatar
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    Regulated Function

    Let f:[-1,1] \rightarrow \mathbb{R} be an odd regulated function. Prove that:

    \int_{-1}^1f=0.
    Here's my answer.

    By definition of a regulated function \int_{-1}^1f= \lim_{n \rightarrow \infty} \phi_n where \phi_n is a sequence of step functions converging uniformly to f.

    Claim: \phi_n is odd.

    We already know that \lim_{n \rightarrow \infty} \phi_n(x)=f(x). (*)

    We can multiply both sides by -1 to get - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x) for some x \in [-1,1].

    Since (*) is valid \forall \ x \in [-1,1], we also know that:

    \lim_{n \rightarrow \infty}\phi_n(-x)=f(-x).

    But we can rewrite the first one as - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)=f(-x) since we know that f is odd.

    We can now equate the two to get:

    - \lim_{n \rightarrow \infty} \phi_n(x)=\lim_{n \rightarrow \infty} \phi_n(-x)

    So we know that \phi_n is odd.

    My next claim is:

    \phi_k:[-1,1] \rightarrow \mathbb{R} is an odd step function so \int_{-1}^1 \phi_k=0 where k=0,1, \ldots.

    \int_{-1}^1 \phi_k==\int_{-1}^0 \phi_k+\int_0^1 \phi_k=\int_{-1}^0 \phi_k(x)~dx+\int_0^1 \phi_k(x)~dx.

    We can use the fact \phi_k is odd:

    =\int_{-1}^0 \phi_k(x)~dx-\int_0^1 - \phi_k(x)~dx=\int_{-1}^0 \phi_k(x)~dx-\int_0^1 -\phi_k(-x)~dx

    We can then rewrite this as:

    =\int_{-1}^0 \phi_k(x)~dx-\int_{-1}^0 \phi_k(x)~dx=0

    Is this the right?
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  2. #2
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    Isn't \int_{-a}^a f(x)dx = 0 for all odd functions? Why do you need it to be regulated?
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  3. #3
    Super Member Showcase_22's Avatar
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    I don't really think it's that necessary for it to be regulated, it's just what the question (and my course) is about.

    I imagine it's a case of proving existing theories from a regulated/step function perspective.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    H

    Claim: \phi_n is odd.

    We already know that \lim_{n \rightarrow \infty} \phi_n(x)=f(x). (*)

    We can multiply both sides by -1 to get - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x) for some x \in [-1,1].

    Since (*) is valid \forall \ x \in [-1,1], we also know that:

    \lim_{n \rightarrow \infty}\phi_n(-x)=f(-x).

    But we can rewrite the first one as - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)=f(-x) since we know that f is odd.

    We can now equate the two to get:

    - \lim_{n \rightarrow \infty} \phi_n(x)=\lim_{n \rightarrow \infty} \phi_n(-x)

    So we know that \phi_n is odd.
    I agree that the regulated part (I haven't even heard this term used except rarely in books on Banach spaces!) is overkill. But, for the above, why can't you say that 0=f(x)+f(-x)=\lim_{n\to\infty}\left(\phi_n(x)+\phi_n(-x)\right) and thus eventually \phi_n(x)=-\phi_n(-x). I don't see why that implies that \phi_n(x) is odd for all n.

    For example \phi_n(x)=\begin{cases} \frac{x}{n} & \mbox{if} n\geqslant 10^{10^{10}} \\ \pi & \mbox{if} \quad n<10^{10^{10}}\end{cases}
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  5. #5
    Super Member Showcase_22's Avatar
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    That's actually very true! I've come up with something else:

    Claim: \phi_p, \ 1 \leq p \leq n is odd.

    By definition of a regulated function, we can choose \phi_p, as defined above, s.t |\phi_p(x)-f(x)| < \frac{\epsilon}{2} where x \in [-1,1].

    Therefore:

    -\frac{\epsilon}{2}<\phi_p(x)-f(x)<\frac{\epsilon}{2} (*)

    Since x \in [-1,1], and \phi_p(x) and f(x) are defined at all values of x:

    -\frac{\epsilon}{2}<\phi_p(-x)-f(-x)<\frac{\epsilon}{2}

    But since f is odd:

    -\frac{\epsilon}{2}<\phi_p(-x)+f(x)<\frac{\epsilon}{2} (**)

    So we can add (*) and (**) together to get:


    -\epsilon<\phi_p(x)+\phi_p(-x)<\epsilon

    and this implies that \phi_p(x)=-\phi_p(-x) or, in a more conventional way, -\phi_p(x)=\phi_p(-x)
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