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Math Help - Proving a function approaches 0 given certain conditions

  1. #1
    Super Member redsoxfan325's Avatar
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    Proving a function approaches 0 given certain conditions

    Problem:
    If f(x) is a non-negative, Riemann-integrable, uniformly continuous function on [0,\infty), prove that \lim_{x\to\infty}f(x) = 0.
    Here's what I have so far.

    I want to try to prove that \sum_{n=1}^{\infty} f(n) converges. Since f is uniformly continuous, \forall~ \epsilon>0, \exists~ \delta>0 such that |f(x_i)-f(x_{i-1})|<\epsilon when |x_i-x_{i-1}|<\delta.

    Let P be a partition of [0,\infty) such that \Delta x_i < \delta. Let m_i = \inf f(x) on [x_{i-1},x_i] and let M_i = \sup f(x) on [x_{i-1},x_i]. We know that:

    0 < \sum_{i=1}^{\infty} m_i\Delta x_i < \sum_{i=1}^{\infty} m_i\delta \leq \int_0^{\infty}f(x)\,dx \leq \sum_{i=1}^{\infty} M_i\Delta x_i < \sum_{i=1}^{\infty} M_i\delta

    Now I get stuck. I was thinking that since I know |f(x_i)-f(x_{i-1})|<\epsilon, I could put some restrictions on m_i and M_i, but I'm not sure how to do this. Am I on the right track? Regardless, can someone give me a hint as to the best way to proceed?
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    Problem:
    Here's what I have so far.

    I want to try to prove that \sum_{n=1}^{\infty} f(n) converges. Since f is uniformly continuous, \forall~ \epsilon>0, \exists~ \delta>0 such that |f(x_i)-f(x_{i-1})|<\epsilon when |x_i-x_{i-1}|<\delta.

    Let P be a partition of [0,\infty) such that \Delta x_i < \delta. Let m_i = \inf f(x) on [x_{i-1},x_i] and let M_i = \sup f(x) on [x_{i-1},x_i]. We know that:

    0 < \sum_{i=1}^{\infty} m_i\Delta x_i < \sum_{i=1}^{\infty} m_i\delta \leq \int_0^{\infty}f(x)\,dx \leq \sum_{i=1}^{\infty} M_i\Delta x_i < \sum_{i=1}^{\infty} M_i\delta

    Now I get stuck. I was thinking that since I know |f(x_i)-f(x_{i-1})|<\epsilon, I could put some restrictions on m_i and M_i, but I'm not sure how to do this. Am I on the right track? Regardless, can someone give me a hint as to the best way to proceed?
    Hi

    I don't understand why you "want to try to prove that \sum_{n=1}^{\infty} f(n) converges". This is an hypothesis since f is Riemann-integrable.

    Suppose that \lim_{x \rightarrow \infty} \neq 0

    Then f being positive \lim_{x\rightarrow\infty} = L > 0, L being eventually infinite

    I suppose that L is finite

    \forall~ \epsilon>0, \exists~ A>0 such that |f(x) -L|<\epsilon when x > A

    Let \epsilon = \frac{L}{2}
    Let A > 0 such that \forall~ x, x > A \implies |f(x) - L|<\frac{L}{2}
    Then \forall~ x, x > A \implies |f(x)|>\frac{L}{2}

    Let 's take the subdivision of [0,\infty) x_i = iA starting from i = 0
    \Delta x_i = A

    Let m_i = \inf f(x) on [x_i,x_{i+1}]
    Since \forall i \geq 1 \:x_i \geq A then m_i \geq \frac{L}{2} \:\forall i \geq 1

    \int_0^{\infty}f(x)\,dx > \sum_{i=0}^{\infty} m_i\Delta x_i

    \sum_{i=0}^{n} m_i\Delta x_i = A\: \sum_{i=0}^{n} m_i = A\: \left(m_0 + \sum_{i=1}^{n} m_i\right)

    Spoiler:

    But
    \sum_{i=1}^{n} m_i \geq n \:\frac{L}{2}

    Therefore
    \sum_{i=0}^{n} m_i\Delta x_i \geq A\: \left(m_0 +  n\:\frac{L}{2}\right)

    And
    \sum_{i=0}^{\infty} m_i\Delta x_i = +\infty
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  3. #3
    Super Member redsoxfan325's Avatar
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    Where did you use the fact that f is uniformly continuous?

    And I wanted to prove that \sum_{n=1}^{\infty} f(n) converged because I thought that only followed if f is monotonically decreasing. There are unbounded functions that are integrable.
    Last edited by redsoxfan325; April 14th 2009 at 11:32 PM.
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    Where did you use the fact that f is uniformly continuous?

    And I wanted to prove that \sum_{n=1}^{\infty} f(n) converged because I thought that only followed if f is monotonically decreasing. There are unbounded functions that are integrable.


    Unbounded functions are not Riemann integrable.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Consider this function:

    f(x) = \left\{<br />
     \begin{array}{lr}<br />
      0 &: 0\leq x<1\\<br />
      -4n^7(x-n)(x-(n+n^{-3})) &: x\in[n,n+n^{-3}]\\<br />
      0 &: otherwise<br />
     \end{array}<br />
   \right\}

    It's continuous (though not uniformly), positive, unbounded, and Riemann integrable.

    I need to prove that if I add the condition that it's uniformly continuous, \lim_{x\to\infty} f(x) = 0.
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  6. #6
    Super Member redsoxfan325's Avatar
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    Oh, and runninggag, how do you know the limit L exists? What if you're looking at a function like x\sin x, which I don't think has a limit as x\to\infty.
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  7. #7
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    If f(x) is a non-negative, Riemann-integrable, uniformly continuous function on [0,\infty), prove that \lim_{x\to\infty}f(x)=0.
    Careful! This is an improper Riemann integral. You cannot analyse a Riemann integral over an infinite interval by using partitions. You need to define it as the limit of integrals over finite intervals.

    What you are told here is that \int_0^\infty\!\!\!f(x)\,dx \mathrel{\mathop=^\text{d{}ef}}  \lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx exists. This implies that \int_X^Y\!\!\!f(x)\,dx can be made arbitrarily small by taking X and Y sufficiently large (I'll make that more precise later).

    The uniform continuity of f means that given \varepsilon>0 there exists \delta>0 such that |f(x)-f(y)|<\varepsilon whenever |x-y|<\delta.

    Now suppose that f(x)\geqslant2\varepsilon for some value of x. Then (by the triangle inequality) it will follow that f(y)>\varepsilon whenever whenever |x-y|<\delta. Therefore \int_x^{x+\delta}\!\!\!f(y)\,dy>\varepsilon\delta. But the fact that \lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx exists tells you that there exists X_0 such that \int_X^Y\!\!\!f(x)\,dx<\varepsilon\delta whenever Y>X>X_0 (as in the previous paragraph). Therefore f(x)\geqslant2\varepsilon\:\Rightarrow\:x<X_0. Equivalently, f(x)<2\varepsilon whenever x>X_0. This says that f(x)\to0 as x\to\infty.
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  8. #8
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Opalg View Post
    Now suppose that f(x)\geqslant2\varepsilon for some value of x. Then (by the triangle inequality) it will follow that f(y)>\varepsilon whenever whenever |x-y|<\delta. Therefore \int_x^{x+\delta}\!\!\!f(y)\,dy>\varepsilon\delta. But the fact that \lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx exists tells you that there exists X_0 such that \int_X^Y\!\!\!f(x)\,dx<\varepsilon\delta whenever Y>X>X_0 (as in the previous paragraph). Therefore f(x)\geqslant2\varepsilon\:\Rightarrow\:x<X_0. Equivalently, f(x)<2\varepsilon whenever x>X_0. This says that f(x)\to0 as x\to\infty.
    Why should we prove that f(x)\geqslant2\varepsilon \implies x<X_0? Can't we just go straight to proving f(x)<2\varepsilon \implies x>X_0 or do we need to prove the former to prove the latter?
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