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Math Help - Continuous and regulated functions

  1. #1
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    Continuous and regulated functions

    (a) Let g:[a,b] \to \mathbb{R} be a continuous function. Suppose that g(x) \geq 0 ~\forall x \in [a,b] and \int_{a}^{b} g = 0.

    Prove that g(x)=0 \forall x \in [a,b] .


    Intuitively, this statement is obviously true. However, I can't seem to find a starting point for a formal proof. Where to start? Any hints?

    (b) Give an example of a regulated function h:[a,b] \to \mathbb{R} such that h(x) \geq 0 ~ \forall x \in [a,b], \int_{a}^{b} h = 0, and there is a point c\in [a,b] such that h(c) \neq 0.

    My problem here is that I don't really understand the difference between continuous and regulated functions. I do know the formal definition of a regulated function, and the fact that a continuous function is regulated but do not entirely grasp the concept.
    Also, it seems rather counterintuitive to assert that there would be a point c such that h(c) \neq 0.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by nmatthies1 View Post
    (a) Let g:[a,b] \to \mathbb{R} be a continuous function. Suppose that g(x) \geq 0 ~\forall x \in [a,b] and \int_{a}^{b} g = 0.

    Prove that g(x)=0 \forall x \in [a,b] .


    Intuitively, this statement is obviously true. However, I can't seem to find a starting point for a formal proof. Where to start? Any hints?

    (b) Give an example of a regulated function h:[a,b] \to \mathbb{R} such that h(x) \geq 0 ~ \forall x \in [a,b], \int_{a}^{b} h = 0, and there is a point c\in [a,b] such that h(c) \neq 0.

    My problem here is that I don't really understand the difference between continuous and regulated functions. I do know the formal definition of a regulated function, and the fact that a continuous function is regulated but do not entirely grasp the concept.
    Also, it seems rather counterintuitive to assert that there would be a point c such that h(c) \neq 0.
    For (a), proceed by contradiction and assume that there exists a point c\in[a,b] s.t. g(c)>0. Then use the continuity of g to show that there must exist \delta>0 s.t. \forall x\in(c-\delta,c+\delta), g(x)>0, and derive a contradiction from that.

    For (b), what about something as simple as:

    h(x)\left\{\begin{array}{lr}1:&x=(a+b)/2\\0:&else\end{array}\right\}

    The left and right limits always exist, so it's regulated, and certainly has an integral of 0.
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