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I am now investigate some behavior of n-dimensional even functions on $\Bbb R^n$.

For 1-dimensional even functions, because $f(x) = f(-x)$ for all $x \in \Bbb R$, so we only have to investigate for $x \geq 0$.

For 2-dimensional even functions, because $f(x_1, x_2) = f(-x_1, -x_2)$ for all $x = (x_1, x_2) \in \Bbb R^2$, we only have to investigate for the region $R = \{ (x_1, x_2) \in \Bbb R^2 \; | \;x_2 \geq -x_1 \}. $

Then how can I generalize this for $n$ dimensional even functions? (I mean how can I genralize the region $R$ above)

Please help!

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Welcome to Math.SE! Since "$n$-dimensional even function" is not a standard term, it is best to add a definition of "$n$-dimensional even" to your question. For example, some people might think it means $f(\pm x_1,\pm x_2)=f(x_1,x_2)$ for all choices of $\pm$. –  user53153 Jan 19 '13 at 21:23
    
@5PM Thank you, I mean the function which satisfies $f(x_1, \cdots , x_n) = f(-x_1, \cdots, -x_n)$. –  Konsta Jan 19 '13 at 21:24
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1 Answer 1

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I interpret your question as asking for a minimal region which uniquely defines an even function. In that case, your region for $n=2$ contains points it shouldn't: it contains both $(-1,1)$ and $(1,-1)$, but these map to the same value for even functions. For general $n$, take the set of $x=(x_1,\ldots,x_n)$ such that the first nonzero coordinate of $x$ is positive. For any nonzero $x\in\mathbb R^n$, this region contains either $x$ or $-x$, but not both. For $n=2$ this gives you the union of a half-plane and a half-line: $\{(x_1,x_2)\in\mathbb R^2: x_1>0$ or $(x_1=0$ and $x_2>0)$ or $(x_1=0$ and $x_2=0)\}$.

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Thank you very much. Then for 3 dimensional case, is $R = \{ (x_1, x_2, x_3) : x_1 >0 \; or \; (x_1 = 0 \; and \; x_2 > 0 ) \; or \; (x_1 = 0, x_2 = 0, x_3 >0) \; or (x_1 = x_2 = x_3 = 0) \}$ right? –  Konsta Jan 19 '13 at 21:46
    
Then does this also work? For example, $R = \{ (x_1, x_2, x_3) : x_1 < -x_2 \; or \; (x_1 = -x_2 \; and \; x_2 < -x_3) \; or \; (x_1 = -x_2, x_2 = -x_3, x_3 >0) \; or \; (x_1 = x_2 =x_3 =0) \}$ –  Konsta Jan 19 '13 at 21:57
    
@Konsta: Yes to both. –  Samuel Jan 20 '13 at 5:37
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