Integral Inequalities

**manjohn12** · March 15th 2009, 08:38 PM

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

Proof. We know that $\smallint_{a}^{b} f \leq \smallint_{a}^{b} g$ . Define $f$ and $g$ as follows:

$f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ K \ \ \ \text{if} \ x = c \end{cases}$

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

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

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 $(c,d)$ in $[a,b]$ such that $x \in (c,d)$ and $f(u)>0$ for all $u \in (c,d)$ ? Is this correct?

**ThePerfectHacker** · March 16th 2009, 08:18 PM

Originally Posted by manjohn12

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 $(c,d)$ in $[a,b]$ such that $x \in (c,d)$ and $f(u)>0$ for all $u \in (c,d)$ ? Is this correct?

You have the correct idea what contunity. This is similar to the proof of this result: $\int_a^b f = 0 \implies f = 0 \text{ if }f\geq 0 \text{ and continous}$ .

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

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