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I have a problem like this:

Let $f:[-a,a]\to\mathbb R$ be a continuous function where $a>0$. If $f$ satisfies that $$\int_{-a}^a f(x)g(x)dx=0$$ for every integrable even function $g:[-a,a]\to\mathbb R,$ show that $f$ is an odd function.

My attempt:
Since $f(x)g(x)$ is integrable on $[-a,a]$ then the integral can be written as $$\int_{-a}^0 f(x)g(x)dx+\int_0^a f(x)g(x)dx=0$$Since $g(x)$ is even, then we have $$\int_{-a}^0f(-x)g(x)dx+\int_0^af(x)g(x)dx=0.$$ This is where I get stuck. There was a hint that says obtain the integral equation $f(x)+f(-x)$ but I'm not sure what to do next. Any help is appreciated. Thanks.

homegrown
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  • Doesn't this prove much more, namely that $f = 0$? If $f(x) \neq 0$ for some $x$, pick a bump function $g$ that is non-zero only on a small enough interval around $x$, and the integral will be non-zero, no? – fgp Apr 01 '14 at 18:18
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    $g$ would have to have a corresponding bump at $-x$ though – Rookatu Apr 01 '14 at 18:29
  • Hint: Use the change of variables $y=-x$ in the integral over $[-a,0]$. Note that the second sum that you have written is incorrect as you are using the assumption that $f$ is an odd function, which you are trying to prove – Keeran Brabazon Apr 01 '14 at 18:32

1 Answers1

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Put $g(x):=\frac{f(x)+f(-x)}{2}$, which is even.

Notice that $f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$.

Then,

$$\begin{align}0&=\int_{-a}^{a}f(x)g(x)\\&=\int_{-a}^{a}\left(\frac{f(x)+f(-x)}{2}\right)^2+\int_{-a}^{a}\left(\frac{f(x)-f(-x)}{2}\right)\left(\frac{f(x)+f(-x)}{2}\right)\\&=\int_{-a}^{a}\left(\frac{f(x)+f(-x)}{2}\right)^2\end{align}$$

But $\left(\frac{f(x)+f(-x)}{2}\right)^2$ is non-negative and continuous.

If its integral is zero it means that $\frac{f(x)+f(-x)}{2}=0$.

Therefore $f(x)=\frac{f(x)-f(-x)}{2}$, which is odd.

  • What happened to $\int_{-a}^a\left(\frac{f(x)-f(-x)}{2}\right)\left(\frac{f(x)+f(-x)}{2}\right)$? – homegrown Apr 01 '14 at 19:10
  • @jnh The function inside the integral there is odd (notice the first factor is odd and the second is even), and the interval is symmetric with respect to the $0$. Therefore that integral is zero, no matter what $f$ is. –  Apr 01 '14 at 19:12
  • Awesome! Thanks. – homegrown Apr 01 '14 at 19:17