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Math Help - Lower integral on two functions

  1. #1
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    Lower integral on two functions

    Suppose that two bounded functions f : [a,b] \rightarrow R and g : [a,b] \rightarrow R have the property that g(x) \leq f(x) for all x in [a,b].

    For a partition P of [a,b] show that L(g,P) \leq L(f,P)

    I tried to do this by considering a partition P = \{{x_0},...{x_n}\}, and for each index i \geq 1, the number m is a lower bound for both the functions given x in [{x_i-1}, {x_i}]. so, m \leq m_i
    is this the correct path? I am stuck on this step..
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    Suppose that two bounded functions f : [a,b] \rightarrow R and g : [a,b] \rightarrow R have the property that g(x) \leq f(x) for all x in [a,b].

    For a partition P of [a,b] show that L(g,P) \leq L(f,P)

    I tried to do this by considering a partition P = \{{x_0},...{x_n}\}, and for each index i \geq 1, the number m is a lower bound for both the functions given x in [{x_i-1}, {x_i}]. so, m \leq m_i
    is this the correct path? I am stuck on this step..
    What are you having trouble with, precisely?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    What are you having trouble with, precisely?
    What I am unsure about is that if we are considering the same partition, as defined above, for both the functions, then for both the functions we end up as:

    m(b-a) = \sum_{i=1}^n m({x_i} - {x_{i-1}}) \leq \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(g,P)

    and \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(f,P)

    Is that what I need to do?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    What I am unsure about is that if we are considering the same partition, as defined above, for both the functions, then for both the functions we end up as:

    m(b-a) = \sum_{i=1}^n m({x_i} - {x_{i-1}}) \leq \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(g,P)

    and \sum_{i=1}^n {m_i}({x_i} - {x_{i-1}}) \leq L(f,P)

    Is that what I need to do?
    It seems like you're all messed up.

    Would you agree that f\leqslant g\implies \inf_{x\in I}f(x)\leqslant\inf_{x\in I}g(x) for any I\subseteq [a,b]?
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    It seems like you're all messed up.

    Would you agree that f\leqslant g\implies \inf_{x\in I}f(x)\leqslant\inf_{x\in I}g(x) for any I\subseteq [a,b]?
    you're right Drexel28. I am totally confused about this chapter. Still trying to learn.

    I think I would agree to the inequality you have presented, but I dont think I wont be able to write a proof on how that is true!

    isnt it {m_{i_g}} \leq {m_{i_f}}?
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  6. #6
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    If g\le f, you have that for any x\in I, \inf_{y\in I} g(y)\le g(x) \le f(x), so \inf_{y\in I} g(y) is a lower bound for f(y) ( y\in I). Since \inf_{y\in I} f(y) is the greatest lower bound of the f(y), it follows that it is \ge the lower bound \inf_{y\in I} g(y).
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by serious331 View Post
    Suppose that two bounded functions f : [a,b] \rightarrow R and g : [a,b] \rightarrow R have the property that g(x) \leq f(x) for all x in [a,b].

    For a partition P of [a,b] show that L(g,P) \leq L(f,P)

    I tried to do this by considering a partition P = \{{x_0},...{x_n}\}, and for each index i \geq 1, the number m is a lower bound for both the functions given x in [{x_i-1}, {x_i}]. so, m \leq m_i
    is this the correct path? I am stuck on this step..
    Hi serious, how about this?

    For any partition interval [x_{i-1}, {x_i}]:

    m_{g_i} = \mbox{inf}  \{g(x) : x \in [{x_{i-1}},{x_i}]\} \leq m_{f_i} = inf \{f(x) : x \in [{x_{i-1}},{x_i}]\}

    So, for each x \in [{x_{i-1}},{x_i}], we have g(x) \leq <br />
f(x). Suppose this is false & {m_{g_i}} > {m_{f_i}}

    Let \epsilon = \frac{{m_{g_i}} - {m_{f_i}}}{2}. Then, there exists  x \in [{x_{i-1}},{x_i}] and f(x) such that f(x) - {m_{f_i}} < \epsilon \implies f(x) < \epsilon + {m_{f_i}} = {m_{f_i}}+ \frac{{m_{g_i}} - {m_{f_i}}}{2} = \frac{{m_{g_i}} + {m_{f_i}}}{2} < {m_{g_i}} which is a contradiction!
    Last edited by harish21; May 5th 2010 at 01:11 AM.
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