Let $S=\{g(x)|g(x) $ is integrable in $[a,b]$ and $\int_a^bg(x)dx=0\}$,$f(x)$ is continuous on $[a,b]$.if $\int_a^bf(x)g(x)dx=0$ for every $g(x) \in S$，then $f(x)$ is constant
I tried in several direction but seemingly can't solve it genuinely. I want to approximate f(x) with polygon thus thinking $x^n$ first, or handle alternately $f(x)^n$,it doesn't seem to work. If I see S as a vector space(is it infinite dimensional),then let $g(x)$ be all basis of S,we can conclude $f(x)=0$,or similarly take $g(x)$ as $e^{\frac{2\pi in}{b-a}}$ then $f(x)=0$ since all the fourier coefficient of $f(x)$ is $0$,but this seemingly only half solve it since $f(x)$ can be any constant. Is there anything else elegant method that solve it completely
Lemma: we find that if $q(x) = f(x) - \frac{1}{b-a} \int_a^b f(x) dx $ then $q(x) \in \mathbb{S}$ for any integrable $f \in \mathbb{S}$
Proof:
$$ \int_a^b q(x) dx = \int_a^b f(x) dx - \frac{1}{b-a} \int_a^b f dx \left[b-a \right] =0$$
Returning to the problem:
Now, multiply both side by $g(x)$ in the equation for $q(x)$:
$$ q(x) g(x) = f(x) g(x) -\left[ \frac{1}{b-a} \int_a^b f(x) dx \right]g(x)$$
– tryst with freedom Jul 22 '21 at 06:30$$ \int_a^b q(x) g(x) dx = \int_a^b f(x) g(x) dx - \left[ \frac{1}{b-a} \int_a^b f(x) dx \right] \int_a^b g(x) dx$$
The second term is zero, since it is a member of $S$, now the problem has reduced to proving that two members of $S$ when producted together give zero and also the first term is zero due to result in question. Hence,
$$ \int_a^b q(x) g(x) dx = 0$$
– tryst with freedom Jul 22 '21 at 06:30