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Thread: Integral Inequalities

  1. #1
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    Integral Inequalities

    Suppose $\displaystyle f $ and $\displaystyle g $ are real valued functions that are Riemann integrable on $\displaystyle [a,b] $. Suppose further that $\displaystyle f(x) \leq g(x) $ for all $\displaystyle x \in [a,b] $ and that a strict inequality holds for at least one point in $\displaystyle [a,b] $. Prove that $\displaystyle \smallint_{a}^{b} f < \smallint_{a}^{b} g $ need not hold. If we add a continuity condition to $\displaystyle f $ and $\displaystyle g $, does the above inequality hold?

    Proof. We know that $\displaystyle \smallint_{a}^{b} f \leq \smallint_{a}^{b} g $. Define $\displaystyle f $ and $\displaystyle g $ as follows:
    $\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ K \ \ \ \text{if} \ x = c \end{cases} $

    $\displaystyle g(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ L \ \ \ \ \text{if} \ x = c \end{cases} $

    Here we have $\displaystyle c \in [a,b] $ and $\displaystyle K < L $. Then $\displaystyle \smallint_{a}^{b} f = \smallint_{a}^{b} g = 0 $. $\displaystyle \diamond $


    If we had a continuity condition, then I think the above inequality will hold? Do we use the following fact: There exists an open interval $\displaystyle (c,d) $ in $\displaystyle [a,b] $ such that $\displaystyle x \in (c,d) $ and $\displaystyle f(u)>0 $ for all $\displaystyle u \in (c,d) $? Is this correct?
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    Quote Originally Posted by manjohn12 View Post
    If we had a continuity condition, then I think the above inequality will hold? Do we use the following fact: There exists an open interval $\displaystyle (c,d) $ in $\displaystyle [a,b] $ such that $\displaystyle x \in (c,d) $ and $\displaystyle f(u)>0 $ for all $\displaystyle u \in (c,d) $? Is this correct?
    You have the correct idea what contunity. This is similar to the proof of this result: $\displaystyle \int_a^b f = 0 \implies f = 0 \text{ if }f\geq 0 \text{ and continous}$.

    Basically what you do is assume that $\displaystyle f(x_0) > 0$ and then you know in an interval of this point we have $\displaystyle f>0$ which leads to a contradiction. You can apply this similar argument to $\displaystyle f-g$.
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