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Math Help - Proof involving integrals equal to 0.

  1. #1
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    Proof involving integrals equal to 0.

    Let
    f be a bounded integrable non-negative function continous on [a, b].
    Prove:
    integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

    I believe I understand why it is true, I'm just not sure how to write the formal proof.

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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by amm345 View Post
    Let
    f be a bounded integrable non-negative function continous on [a, b].
    Prove:
    integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

    I believe I understand why it is true, I'm just not sure how to write the formal proof.

    What have you tried? Here is a weird approach.

    Suppose that f\ne 0 then there exists some x\in[a,b] such f(x)\in(0,\infty) and since this set is open there exists some open ball B_{\varepsilon}(f(x))\subseteq(0,\infty) and by continuity there exists some B_{\delta}(x) such that f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq(0,\infty). In other words, there exists some open ball around the point where f(x)\ne0 which is entirely positive. Take the concentric closed ball B_{\frac{\delta}{2}}. This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since f:B_{\frac{\delta}{2}}[x]\mapsto\mathbb{R} is continuous we know that f assumes a minimum on B_{\frac{\delta}{2}}[x]. In particular, \inf_{x\in B_{\frac{\delta}{2}}[x]}f(x)=\varepsilon>0. Thus, \int_a^b\text{ }f=\int_a^{x-\frac{\delta}{2}}\text{ }f+\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}\text{ }f+\int_{x+\frac{\delta}{2}}^b \text{ }f\geqslant 0+\frac{\varepsilon \delta}{2}+0>0. This of course is a contradiction.
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    That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by amm345 View Post
    That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.
    Which theorem is that?
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    Oh sorry, the Heine-Borel theorem.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by amm345 View Post
    Oh sorry, the Heine-Borel theorem.
    Are you serious? What's the point of having tools if you can't use them! What if I proved it first?
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    I have no idea.. I couldn't think of anything else it is kinda ridiculous!
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by amm345 View Post
    I have no idea.. I couldn't think of anything else it is kinda ridiculous!
    Ok. I can think of one or two other ways....what are you currently doing in class?
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  9. #9
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    We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by amm345 View Post
    We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.
    Hmm...Ok. I see an angle. Make a modified argument of mine and use the FTC.
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  11. #11
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    Hello,

    Why can't we just say that (without HB theorem) :
    if there exists m\in[a,b] such that f(m)>0, then by definition of the continuity, \forall \epsilon>0, \exists \delta>0,\forall x\in[a,b], (|x-m|<\delta)\Rightarrow (|f(x)-f(m)|<\epsilon)

    In particular, if we let \epsilon=\frac{f(m)}{2}, |f(x)-f(m)|<\epsilon \Leftrightarrow -\tfrac{f(m)}{2}<f(x)-f(m)<\tfrac{f(m)}{2} \Leftrightarrow \tfrac{f(m)}{2}<f(x)<f(m)

    So there exists \delta>0 such that \forall x\in]m-\delta,m+\delta[, f(x)>\tfrac{f(m)}{2}>0

    Hence \int_a^b f(x) ~dx=\int_a^{m-\delta} \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m+\delta}^b \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m-\delta}^{m+\delta} \underbrace{f(x)}_{> \frac{f(m)}{2}} ~dx

    So \int_a^b f(x) ~dx>0+0+2\delta \cdot\tfrac{f(m)}{2}=\delta f(m)>0

    which is a contradiction.

    So m can't exist, and hence f(x)=0 ~,~ \forall x\in[a,b]
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