# Proving that this function must be even (II)

Suppose $g:\mathbb{R}^d\rightarrow\mathbb{R}$ is continuous. Also let $\mathbf{x}=(x_1,\ldots,x_d)\in\mathbb{R}^d$.

I'd like to prove the following:

If

$$\int_{\mathbb{R}^d}x_k\exp(-\mathbf{x}^T\mathbf{x})f(g(x))\,d\mathbf{x}=0$$

for all bounded continuous functions $f$, 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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Something is missing here, I think. Right now you are integrating a vector-valued function - perhaps your integrand is supposed to be $\vert\textbf{x}\vert\exp(-\textbf{x}^\intercal\textbf{x})f(g(\textbf{x}))$? – icurays1 Jan 24 '13 at 19:06
Presumably the integral is the vector valued integral, ie, $[\int_{\mathbb{R}^d}\mathbf{x}\exp(-\mathbf{x}^T\mathbf{x})f(g(x))\,d\mathbf{x}‌​]_k = \int_{\mathbb{R}^d}\mathbf{x_k}\exp(-\mathbf{x}^T\mathbf{x})f(g(x))\,d\mathbf{x}‌​$. – copper.hat Jan 24 '13 at 19:31
Sorry I meant what copper.hat has written. Thanks for spotting the mistake! – red271 Jan 25 '13 at 8:24