If $\int_{-1}^1 fg = 0$ for all even functions $f$, is $g$ necessarily odd?

Question

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!

Answer 1

For the edited question: The answer is yes. Indeed, let $f$ be an arbitrary continuous function on $[0,1]$ and extend it to a continuous even function

$$ \tilde{f}(x) = \begin{cases} f(x), & x \geq 0; \\ f(-x), & x < 0; \end{cases} $$

on $[-1, 1]$. (Check that $\tilde{f}$ is indeed continuous!) Then by the assumption,

$$ 0 = \int_{-1}^{1} \tilde{f}(x)g(x) \, \mathrm{d}x = \int_{0}^{1} f(x)(g(x)+g(-x)) \, \mathrm{d}x. $$

Now pick $f(x) = g(x)+g(-x)$ and note that

$$ \int_{0}^{1} (g(x)+g(-x))^2 \, \mathrm{d}x = 0. $$

Together with the continuity of $g$, this implies that $g(x)+g(-x) = 0$ for all $x \in [0, 1]$, which then implies that $g$ is odd.

Answer 2

Simple counterexample: let $f=0$, then $g$ can be any function. For a non-zero $f$ this still does not hold. Can you think of a counterexample?