# even function representation

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

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
@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

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
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.