Show that if is a continuous real-valued function on the interval and , then

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- January 30th 2009, 01:29 PMchiph588@Real Analysis
Show that if is a continuous real-valued function on the interval and , then

- January 30th 2009, 01:50 PMOpalg
Let . It should be fairly clear that . Thus (if it exists) is less than or equal to m.

To get the reverse inequality, let . The continuity of f ensures that there is some interval of length on which . Then . Take the n'th root and let to see that . Finally, take the limit as . - January 31st 2009, 08:10 AMchiph588@
- January 31st 2009, 08:41 AMOpalg
- January 31st 2009, 03:29 PMchiph588@
This is one of my favorite theorems.

- January 31st 2009, 04:06 PMNonCommAlg
it's not bad to know, if you haven't noticed yet, that the limit is some sort of "continuous version" of this: where are postive numbers.

- January 31st 2009, 04:10 PMchiph588@
Ah yes, because an integral is just a Riemann sum.

- January 31st 2009, 05:15 PMThePerfectHacker
I do not want to argue against

**Opalg**because I am a noob compared to him in analysis (Blush), but I offer a suggestion.

I think this problem is better solved using the concepts of limsup's and liminf's.

Let be the length of the interval. Let .

Opalg hath shown us that .

Then he says by .

But we need to know the limit exists. (Thinking)

So a way around this problem is to perhaps say first take and then .

But the problem is depends on and so how we know that as ? (Thinking)

Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because and so . However, . Thus, we have shown but . This means that is a convergent sequence with .

--- - February 1st 2009, 02:00 AMOpalg
That is quite correct. The efficient way to explain the solution is to use limsup in connection with the upper estimate, and liminf for the lower estimate.

I didn't do that, because I didn't want to deter anyone who might be unfamiliar with limsup and liminf. So I just included a parenthetical comment to the effect that you need to establish that the limit exists. - February 17th 2009, 03:13 PMchiph588@
- February 17th 2009, 09:06 PMThePerfectHacker
- February 18th 2009, 09:05 AMheathrowjohnny
- February 18th 2009, 06:32 PMThePerfectHacker