# prove...

• Feb 28th 2010, 01:11 AM
flower3
prove...
prove that if f is continuous on [a,b] with $\displaystyle f(x) \geq 0 \ , \forall x \in [a,b]$ prove that if g is strictly increasing on [a,b]
with $\displaystyle \int_a^b f . dg =0$ then f(x)=0 $\displaystyle \forall x \in [a,b]$
• Feb 28th 2010, 03:25 AM
Laurent
Quote:

Originally Posted by flower3
prove that if f is continuous on [a,b] with $\displaystyle f(x) \geq 0 \ , \forall x \in [a,b]$ prove that if g is strictly increasing on [a,b]
with $\displaystyle \int_a^b f . dg =0$ then f(x)=0 $\displaystyle \forall x \in [a,b]$

The proof is the same as for usual Riemann integral. Procede by contradiction: assume that $\displaystyle f$ is not identically zero on $\displaystyle [a,b]$. Using continuity, prove that there exists $\displaystyle \epsilon>0$ and $\displaystyle a\leq u<v\leq b$ such that $\displaystyle f(x)\geq \epsilon$ when $\displaystyle x\in[u,v]$. Then, justify the following: $\displaystyle \int_a^b f\, dg\geq \int_u^v f\, dg\geq \epsilon \int_u^v dg=\epsilon(g(v)-g(u))>0$. And conclude.