Strict

• Mar 15th 2009, 03:20 PM
manjohn12
Strict
Let $f: [a,b] \to \mathbb{R}$ be Riemann integrable on $[a,b]$. Suppose that $f(x) \geq 0$ for all $x \in [a,b]$ and that $f$ is continuous on $[a,b]$. If $f>0$ somewhere in $[a,b]$, prove that $\smallint_{a}^{b} f > 0$.

Proof. If we look at the most extreme case in which $f$ is $0$ everywhere except at $c \in [a,b]$ then $\mathcal{R}(f,P) \geq 0$. E.g. for the "extreme
case" we define $f$ as
$f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ L \ \ \ \ \text{if} \ x = c \end{cases}$

where $c \in [a,b]$ and $L$ is a constant. So $\smallint_{a}^{b} f \neq 0$. Thus $\smallint_{a}^{b} > 0$. $\diamond$

Is this correct?

If $f$ were not continuous then the above statement would be false, right?
• Mar 15th 2009, 03:27 PM
Diamondlance
The way you define f doesn't work since f is assumed to be continuous. Incidentally the integral of your function is 0.

The key idea here is that if f is continuous and f(x)>0 for some x in [a,b], then there exists an open interval (c,d) contained in [a,b] such that x is in (c,d) and f(u)>0 for all u in (c,d).
• Mar 15th 2009, 03:31 PM
manjohn12
Quote:

Originally Posted by Diamondlance
The way you define f doesn't work since f is assumed to be continuous. Incidentally the integral of your function is 0.

The key idea here is that if f is continuous and f(x)>0 for some x in [a,b], then there exists an open interval (c,d) contained in [a,b] such that x is in (c,d) and f(u)>0 for all u in (c,d).

So suppose $a. Let $k$ be any non-zero number. Define $f: \mathbb{R} \to \mathbb{R}$ as follows: $f(x) = \begin{cases} 0 \ \ \ \text{if} \ x < c \\ k \ \ \ \text{if} \ c \leq x \leq d \\ 0 \ \ \ \text{if} \ d < x \end{cases}$
• Mar 15th 2009, 03:34 PM
Diamondlance
Your function still isn't continuous. Actually, you don't need to define an explicit f.

There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?
• Mar 15th 2009, 03:41 PM
manjohn12
Quote:

Originally Posted by Diamondlance
Your function still isn't continuous. Actually, you don't need to define an explicit f.

There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?

$\int_{c}^{d} f \leq \int_{a}^{b} f$ and $\int_{c}^{d} f > 0$.
• Mar 15th 2009, 03:43 PM
Diamondlance
I think you've got it :)!
• Mar 15th 2009, 03:49 PM
Diamondlance
The only technical point we haven't addressed is if x=a or x=b. But this isn't too tough to deal with: we still know f is continuous, so if x=a, there exists d in (a,b] such that f(u)>0 on [a,d), and the same argument then applies. The case where x=b is similar.
• Mar 15th 2009, 03:51 PM
manjohn12
Quote:

Originally Posted by Diamondlance
The only technical point we haven't addressed is if x=a or x=b. But this isn't too tough to deal with: we still know f is continuous, so if x=a, there exists d in (a,b] such that f(u)>0 on [a,d), and the same argument then applies. The case where x=b is similar.

And if $f$ is not continuous then the statement in post #1 does not hold.
• Mar 15th 2009, 03:54 PM
Diamondlance
Right, and you can use the f you came up with in post 1 to illustrate that.
• Mar 27th 2009, 08:41 AM
manjohn12
Quote:

Originally Posted by Diamondlance
Your function still isn't continuous. Actually, you don't need to define an explicit f.

There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?

This is not enough. You have to come up with a piecewise defined function $g(x)$ such that $f(x) \geq g(x)$ aka post #3. E.g. an open interval $(c,d)$ contained in $[a,b]$ such that $f(u) > z_0/2$.