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given any even function $ g(x)=g(-x) $ is it always possible to write it as the product of two functions ? i mean

$ g(x) = f(x)f(-x) $ so $ g(x) $ is always an even function even though $ f(x) $ it isn't

for example given the Riemann Xi function $ \xi(1/2+s)= \xi(1/2-s) $ can we represent it as the product of two functions ? $ \xi(1/2+s)=f(s)f(-s) $

in this case and using the Hadamard product $ f(x)= \prod _{n}(1- \frac{ix}{n}) $

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$g(0)=f(0)^2 \geq 0$ would be a necessary condition. –  Siminore Jun 27 '12 at 9:44
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3 Answers 3

Consider $g(x)=-5+x^2$ then $g$ is even and $g(0)=-5<0$. Now if $g(x)=f(x)f(-x)$ for some $f$ then $g(0)=f(0)^2\geq 0$, a contradiction.

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The natural question now is: can the representation hold true for positive functions? –  Siminore Jun 27 '12 at 9:57
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@Siminore If $f$ is positive, we can let $g = \sqrt{f}$ ... then $g$ is even and $f = g^2$. –  martini Jun 27 '12 at 10:04
    
@martini this is true for complex functions too. –  pritam Jun 30 '12 at 9:58
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There are many irreducible even functions over real numbers,which is a whole class of function that contradicts your statement.

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But this $f$ is of the form $f(x) = g(x)g(-x)$ with $g(x) = \sqrt{5 + x^2}$ ... –  martini Jun 27 '12 at 10:00
    
but it is true if $f(x)$ is negative for some $x$. –  Aang Jun 27 '12 at 10:03
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As already mentioned by @pritam $g(0) < 0$ will lead to a contradiction. If we have $g(0) \ge 0$ the function

$$f(x)=\begin{cases}g(x) &; x > 0 \\ \sqrt{g(0)} &; x=0 \\ 1 &;x < 0 \end{cases}$$

would be a simple function $f(x)$ with $g(x)=f(x)f(-x)$.

If we allow complex values for $f$ then $f=\sqrt{g}$ would be a solution where $\sqrt{\cdot}$ is defined as one steam of the complex square root.

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