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

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!

• If $g$ is continuous and the statement holds for any even continuous function $f$, then the conclusion is true. But if it holds only for a given function $f$, then it need not be true. We have so much room for cooking up counter-examples. For instance, fix $f \equiv 1$ and try to determine a non-trivial $g(x)=ax^2+b$. Jul 5, 2020 at 22:59
• For a simple family of counterexamples, let $f$ be $0$ on $[-1/2, 1/2]$. Outside this interval, define $f$ to be any even function you want. Then you can define $g$ to be anything you want on $[-1/2, 1/2]$ without affecting the integral of the product. Jul 5, 2020 at 23:03
• @SangchulLee Sorry! I totally forgot to include the $\forall$ quantifier on $f$ in the question description. Given that this holds for all even functions $f$, you mentioned that the statement is true? Jul 5, 2020 at 23:04

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.

• Thanks for the answer! I'm wondering how the last integral implies that $g(x) + g(-x) = 0$? Jul 5, 2020 at 23:30
• @learning_linalg, Note that $(g(x)+g(-x))^2\geq0$ and its integral is zero. This and the continuity together implies that $(g(x)+g(-x))^2=0$. For more details, this posting might be helpful. Jul 5, 2020 at 23:34

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?

• Edited the question! I'm sorry. I totally missed saying that the statement holds for all even functions. Careless on my part, sorry. Jul 5, 2020 at 23:03