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Math Help - Strict

  1. #1
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    Strict

    Let  f: [a,b] \to \mathbb{R} be Riemann integrable on  [a,b] . Suppose that  f(x) \geq 0 for all  x \in [a,b] and that  f is continuous on  [a,b] . If  f>0 somewhere in  [a,b] , prove that  \smallint_{a}^{b} f > 0 .

    Proof. If we look at the most extreme case in which  f is  0 everywhere except at  c \in [a,b] then  \mathcal{R}(f,P) \geq 0 . E.g. for the "extreme
    case" we define  f as
     f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ L \ \ \ \ \text{if} \ x = c \end{cases}

    where  c \in [a,b] and  L is a constant. So  \smallint_{a}^{b} f \neq 0 . Thus  \smallint_{a}^{b} > 0 .  \diamond

    Is this correct?

    If  f were not continuous then the above statement would be false, right?
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  2. #2
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    The way you define f doesn't work since f is assumed to be continuous. Incidentally the integral of your function is 0.

    The key idea here is that if f is continuous and f(x)>0 for some x in [a,b], then there exists an open interval (c,d) contained in [a,b] such that x is in (c,d) and f(u)>0 for all u in (c,d).
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  3. #3
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    Quote Originally Posted by Diamondlance View Post
    The way you define f doesn't work since f is assumed to be continuous. Incidentally the integral of your function is 0.

    The key idea here is that if f is continuous and f(x)>0 for some x in [a,b], then there exists an open interval (c,d) contained in [a,b] such that x is in (c,d) and f(u)>0 for all u in (c,d).
    So suppose  a<c<d<b . Let  k be any non-zero number. Define  f: \mathbb{R} \to \mathbb{R} as follows:  f(x) = \begin{cases} 0 \ \ \ \text{if} \ x < c \\ k \ \ \ \text{if} \ c \leq x \leq d \\ 0 \ \ \ \text{if} \ d < x \end{cases}
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  4. #4
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    Your function still isn't continuous. Actually, you don't need to define an explicit f.

    There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?
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  5. #5
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    Quote Originally Posted by Diamondlance View Post
    Your function still isn't continuous. Actually, you don't need to define an explicit f.

    There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?
     \int_{c}^{d} f \leq \int_{a}^{b} f and  \int_{c}^{d} f > 0 .
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  6. #6
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    I think you've got it !
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  7. #7
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    The only technical point we haven't addressed is if x=a or x=b. But this isn't too tough to deal with: we still know f is continuous, so if x=a, there exists d in (a,b] such that f(u)>0 on [a,d), and the same argument then applies. The case where x=b is similar.
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  8. #8
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    Quote Originally Posted by Diamondlance View Post
    The only technical point we haven't addressed is if x=a or x=b. But this isn't too tough to deal with: we still know f is continuous, so if x=a, there exists d in (a,b] such that f(u)>0 on [a,d), and the same argument then applies. The case where x=b is similar.

    And if  f is not continuous then the statement in post #1 does not hold.
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  9. #9
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    Right, and you can use the f you came up with in post 1 to illustrate that.
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  10. #10
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    Quote Originally Posted by Diamondlance View Post
    Your function still isn't continuous. Actually, you don't need to define an explicit f.

    There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?

    This is not enough. You have to come up with a piecewise defined function  g(x) such that  f(x) \geq g(x) aka post #3. E.g. an open interval  (c,d) contained in  [a,b] such that  f(u) > z_0/2 .
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