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Math Help - Integral proof problem

  1. #1
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    Integral proof problem

    Prove that, if f is continuous on [a,b], and

    <br />
\int_{a}^{b} |f(x)| dx = 0<br />

    then f(x)=0 for all x in [a,b].

    A hint provided was to prove this via contradiction.
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  2. #2
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    Hello, Runty!

    I have an intuitive (graphic) proof . . .


    Prove that, if f(x) is continuous on [a,b], and: . \int_{a}^{b} |f(x)| dx \:=\: 0
    . . then f(x)\,=\,0 for all x \in [a,b].

    A hint provided was to prove this via contradiction.
    Suppose f(x) is not equal to 0 for all x \in [a,b].
    . . Then f(x) is not the x-axis.


    There are three cases:

    . . (1) f(x) is above the x-axis on the interval [a,b]
    . . (2) f(x) is below the x-axis on the inverval [a,b].
    . . (3) f(x) crosses the x-axis on the interval [a,b].



    For case (1), we have this graph:
    Code:
            |
            |           *
            |         *:|
            |       *:::|
            |     *:::::|
            |   *:::::::|
            |   |:::::::|
        - - | - + - - - + - -
            |   a       b
    \int^b_a|f(x)|\,dx represents the area under the curve,
    . . . which is evidently not zero.



    For case (2), we have this graph:
    Code:
            |
            |   a       b
        - - + - + - - - + - -
            |   |:::::::|
            |   *:::::::|
            |     *:::::|
            |       *:::|
            |         *:|
            |           *
    \int^b_a|f(x)|\,dx represents the area above the curve,
    . . . which also is not zero.



    For case (3), we have this graph for f(x):
    Code:
            |
            |   *
            |     *
            |       *
            |         *
        - - + - + - - - * - - + - - -
            |   a         *   b
            |               *
            |                 *


    The the graph of |f(x)| looks like this:
    Code:
            |
            |   *
            |   |:*           *
            |   |:::*       *:|
            |   |:::::*   *:::|
        - - | - + - - - * - - - + - - -
            |   a               b
            |
    \int^b_a|f(x)|\,dx represents the area under the curve,
    . . . which again is not zero.


    [This may not be considered a a satisfactory proof.]

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Runty!

    I have an intuitive (graphic) proof . . .


    Suppose f(x) is not equal to 0 for all x \in [a,b].
    . . Then f(x) is not the x-axis.


    There are three cases:

    . . (1) f(x) is above the x-axis on the interval [a,b]
    . . (2) f(x) is below the x-axis on the inverval [a,b].
    . . (3) f(x) crosses the x-axis on the interval [a,b].



    For case (1), we have this graph:
    Code:
            |
            |           *
            |         *:|
            |       *:::|
            |     *:::::|
            |   *:::::::|
            |   |:::::::|
        - - | - + - - - + - -
            |   a       b
    \int^b_a|f(x)|\,dx represents the area under the curve,
    . . . which is evidently not zero.



    For case (2), we have this graph:
    Code:
            |
            |   a       b
        - - + - + - - - + - -
            |   |:::::::|
            |   *:::::::|
            |     *:::::|
            |       *:::|
            |         *:|
            |           *
    \int^b_a|f(x)|\,dx represents the area above the curve,
    . . . which also is not zero.



    For case (3), we have this graph for f(x):
    Code:
            |
            |   *
            |     *
            |       *
            |         *
        - - + - + - - - * - - + - - -
            |   a         *   b
            |               *
            |                 *


    The the graph of |f(x)| looks like this:
    Code:
            |
            |   *
            |   |:*           *
            |   |:::*       *:|
            |   |:::::*   *:::|
        - - | - + - - - * - - - + - - -
            |   a               b
            |
    \int^b_a|f(x)|\,dx represents the area under the curve,
    . . . which again is not zero.


    [This may not be considered a a satisfactory proof.]

    It's a good proof, but it needs to be solved via calculus equations. I'm afraid I can't really use it.
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  4. #4
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    Are you allowed to use this definition of continuity of at c ?

    \forall \epsilon > 0 \: \exists \eta > 0 \:\forall x \in [a,b] \:|x-c|  < \eta \Rightarrow |f(x) - f(c)| < \epsilon

    If so suppose that f is not equal to 0 for all x in [a,b]

    This means that exists c in [a,b] such that f(c) is not equal to 0

    Let \epsilon = \frac{|f(c)|}{2}
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Runty View Post
    Prove that, if f is continuous on [a,b], and

    <br />
\int_{a}^{b} |f(x)| dx = 0<br />

    then f(x)=0 for all x in [a,b].

    A hint provided was to prove this via contradiction.
    Quote Originally Posted by running-gag View Post
    Are you allowed to use this definition of continuity of at c ?

    \forall \epsilon > 0 \: \exists \eta > 0 \:\forall x \in [a,b] \:|x-c|  < \eta \Rightarrow |f(x) - f(c)| < \epsilon

    If so suppose that f is not equal to 0 for all x in [a,b]

    This means that exists c in [a,b] such that f(c) is not equal to 0

    Let \epsilon = \frac{|f(c)|}{2}
    Are you trying to prove this using Riemann-Stieltjes integration concepts?
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Are you trying to prove this using Riemann-Stieltjes integration concepts?
    No, that was not mentioned.
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  7. #7
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    Quote Originally Posted by Runty View Post
    Prove that, if f is continuous on [a,b], and <br />
\int_{a}^{b} |f(x)| dx = 0 then f(x)=0 for all x in [a,b].
    This is a well known theorem: If g is continuous on [a,b] and \left( {\exists p \in [a,b]} \right)\left[ {g(p) > 0} \right] then  \left( {\exists [c,d] \subseteq [a,b]} \right)\left[ {\left( {\forall x \in [c,d],\;g(x) > 0} \right)} \right].

    Apply that theorem to |f|. Note that  \int_a^b {\left| f \right|dx}  = \int_a^c {\left| f \right|dx}  + \int_c^d {\left| f \right|dx}  + \int_d^b {\left| f \right|dx}  > 0 .
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