This is just a curious thought of mine.
Suppose $f$ and $g$ are real continuous functions on closed interval $[a,b]$ and $\int_{a}^{b} f = 0$, does that therefore imply that $\int_{a}^{b} fg=0$? If so could someone please construct a proof for me please?
If I remove the condition that both functions are continuous, then will this affect the conclusion?
Thanks very much in advance!
On
Answer when you add the condition that $f$ is non-negative and continuous (following your comment to the previous answer): in this case let us show that $f(x)=0$ for all $x$ which makes it obvious that $\int fg=0$. Suppose, if possible, $f(x) \neq 0$ for s0me $x$. By continuity there exists $\delta >0$ such that $|f(y)-f(x)| <\frac {f(x)} 2$ for $|y-x| <\delta$. Now $f(y) >f(x)-\frac {f(x)} 2=\frac {f(x)} 2$ for $y \in (x-\delta,x+\delta)$. Hence $\int_a^{b} f(y)\, dy \geq \int_{x-\delta}^{x+\delta} f(y)\, dy > 2\delta\frac {f(x)} 2 >0$ contradicting the hypothesis. Hence $f \equiv 0$.
Nobody can prove it, since it is false. Take $a=-1$, $b=1$, and $f(x)=g(x)=x$, for instance. The $\int_a^bf(x)\,\mathrm dx=0$, but $\int_a^bf(x)g(x)\,\mathrm dx=\frac23$.