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Let $u:\mathbb{R}^d\rightarrow\mathbb{R}$ be a continuous function that is not identically equal to zero. Suppose further that $u$ is an odd function (ie. $u(\mathbf{x})=-u(-\mathbf{x})$). Let $g:\mathbb{R}^d\rightarrow\mathbb{R}$ be continuous. I'd like to prove the following:

If

$$ \int_{\mathbb{R}^d}u(\mathbf{x})f(g(\mathbf{x}))d\mathbf{x}=0 \qquad \forall f\in C_b(\mathbb{R})$$

then $g(\mathbf{x})=g(-\mathbf{x})$ for all $\mathbf{x}\in\mathbb{R}^d$, that is $g$ is an even function.

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What is $C_b(\mathbb{R})$? –  user7530 Jan 23 '13 at 18:51
    
I think it is the space of functions with compact support, @user7530 –  Tomás Jan 23 '13 at 19:19

1 Answer 1

up vote 1 down vote accepted

Isn't the statement false? Let $B:\mathbb{R}\to\mathbb{R}$ be an even bump function centered at $0$ of radius 1. Then if $$u(x) = B(x+1)-B(x-1)$$ and $$g(x) = B(x-k)$$ for $|k|>3$, then your integral converges and is zero for any $f$, but $g$ is not an even function.

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That's a nice counterxample. Thanks! –  red271 Jan 23 '13 at 20:08

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