Lower and Upper Darboux Sums

seniorcalculus · #1 $Old$ December 13th, 2008, 04:11 PM

Let $f(x) : [a,b] -> R$ be a bounded function such that $U(f,P) = L(f,P)$ for any partition P of [a,b].
Prove that there exists a constant c such that $f(x) = c$ for any x in [a,b] (by contradiction) .
Appreciate if someone could help on this one. Thanks...

NonCommAlg · #2 $Old$ December 13th, 2008, 05:25 PM

Quote:

Originally Posted by seniorcalculus $View Post$

Let $f(x) : [a,b] -> R$ be a bounded function such that $U(f,P) = L(f,P)$ for any partition P of [a,b].
Prove that there exists a constant c such that $f(x) = c$ for any x in [a,b] (by contradiction) .
Appreciate if someone could help on this one. Thanks...

it's really trivial! let $P=\{a,b\}.$ let $M=\sup_{x \in [a,b]}f(x), \ m=\inf_{x \in [a,b]}f(x).$ then: $0=U(f,P)-L(f,P)=(M-m)(b-a).$ thus $M=m,$ which means $f$ is constant.

seniorcalculus · #3 $Old$ December 13th, 2008, 05:46 PM

Okay, thanks. But that's a direct proof, right? I kinda need help for proof by contradiction though.

NonCommAlg · #4 $Old$ December 13th, 2008, 06:09 PM

Quote:

Originally Posted by seniorcalculus $View Post$

Okay, thanks. But that's a direct proof, right? I kinda need help for proof by contradiction though.

it's the same proof! let $P,M, m$ be as above. suppose $f$ is not constant. then $M \neq m.$ but $0=U(f,P)-L(f,P)=(M-m)(b-a).$ thus $b=a.$ contradiction!

seniorcalculus · #5 $Old$ December 13th, 2008, 06:28 PM

Okay, I understand it now. Thanks, NonCommAlg

But, I realized that you have picked an instance of a partition in this case, which is P={a,b}. But in the proof we have to show that for any partition interval. Shouldn't we have to assume for any partition P?