suppose f:[a,b]-> real numbers is continuos and for all continuous functions g:[a,b] -> real numbers, then intergrate from b to a of (f.g)=0. show that f is identically 0.
how do i go about solving this?
Just as a remark, this is true more generally in the sense that if $\displaystyle \displaystyle \int_a^b f(x)h(x)\text{ }dx=0$ for every $\displaystyle h \in C^k[a,b]$ (for some arbitrary but fixed $\displaystyle k\geqslant 0$) with $\displaystyle h(a)=h(b)=0$ then $\displaystyle \displaystyle f\equiv 0$. The proof is not much different, merely take $\displaystyle h(x)=f(x)(x-a)(x-b)$ to conclude. I mention this because it shows that $\displaystyle C^k[a,b]$ separates points with respect to the usual weak topology on it. In this context this theorem (as simple as it is) actually has a name. It's the Fundamental Lemma of the Calculus of Variations.