Suppose for a fixed continuous function $g$, all even continuous real-valued functions $f$ satisfy $\int_{-1}^1 fg = 0$, is it true that $g$ is odd on $[-1,1]$?
My intuition is telling me that this is correct, as I have not found any counterexamples. I've tried proving this by generating a contradiction by assuming $g$ is not odd and so there is a point such that $g(-x) \neq -g(x)$ on $[-1,1]$, but this doesn't yield any information that I think I can readily use to prove the claim.
Any help would be very much appreciated!