# analysis integration continuity question

I'm trying to show that if f is continuous real-valued function on [a,b] satisfying integral from 1 to b (f(x)*g(x)) dx = 0 for every confinuous function g on [a,b] then f(x) = 0 for all x an element in [a,b]

What I tried is assuming that f(x) >= 0 for all x an element in [a,b] but I need help in breaking the cases and how many cases are there?

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Hint:

What about using $g(x)=f(x)$ ...

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This is ONE case, but what if f <0, f > 0, etc? – mary Apr 29 '11 at 5:15
Using $g=f$ works for every function $f$. User: you seem to be under the impression that either $f\ge0$ on $[a,b]$ or $f\le0$ on $[a,b]$. This is not true, think about $f$ defined by $f(x)=\sin(\omega x)$ for every $x$, for a large $\omega$. – Did Apr 29 '11 at 5:27
Fabian means that the problem is really not that hard. However, more general statements are true $f$ may be chosen from larger classes of functions than that of continuous ones (e.g. $f\in L^1$) and in the same manner the annihilating family, the $g$'s, may be chosen more restrictive (e.g. trigonometric polynomials $g$). – AD. Apr 29 '11 at 6:40

If you want to prove this by contraposition or contradiction, you don't want to assume that $f(x)\geq 0$ for all $x\in [a,b]$.

The negation of the conclusion of the stated theorem is that $f(x) \neq 0$ for at least one $x\in [a,b]$.

Assuming this, there is has to be some $c\neq 0$ and $x_c \in [a,b]$ such that $f(x_c)=c$. Now, a quick argument using continuity of $f$ lets us conclude that $f$ is actually bounded away from $0$ on some interval around $x_c$.

Once you have that, you can think about how to construct a continuous function $g$ such that $\int fg$ is nonzero.

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