# Lower and Upper Darboux Sums

• Dec 13th 2008, 03:11 PM
seniorcalculus
Lower and Upper Darboux Sums
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...
• Dec 13th 2008, 04:25 PM
NonCommAlg
Quote:

Originally Posted by seniorcalculus
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.
• Dec 13th 2008, 04:46 PM
seniorcalculus
Okay, thanks. But that's a direct proof, right? I kinda need help for proof by contradiction though.
• Dec 13th 2008, 05:09 PM
NonCommAlg
Quote:

Originally Posted by seniorcalculus
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!
• Dec 13th 2008, 05:28 PM
seniorcalculus
Okay, I understand it now. Thanks, NonCommAlg (Wink)
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?