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An even function satisfies $$ f_e(x) = f_e(-x) $$ and a odd function $$ f_o(x) = -f_o(-x) $$ Every function can be split into an even and an odd part $$ f(x) = f_e(x) + f_o(x) = \frac{1}{2}(f(x)+f(-x)) + \frac{1}{2}(f(x)-f(-x)) $$ Now even and odd functions are basically related to the operation of addition and the inverse of addition. I wonder if it would make sense to define even and odd functions related to multiplication and the inverse of multiplication. So I imagine a multiplicative even function would satisfy $$ f_{me}(x)=f_{me}\left(\frac{1}{x}\right) $$ and a multiplicative odd function $$ f_{mo}(x)=\frac{1}{f_{mo}\left(\frac{1}{x}\right)} $$ Are these functional equations the correct generalizations of even and off functions to the operation of multiplication? If so, what functions can fulfill these equations? Is it possible to split each function into a multiplicative even and odd part?

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  • $\begingroup$ Do you know what a group is? $\endgroup$ – dfsfljn Mar 6 '16 at 20:12
  • $\begingroup$ For functions $\mathbb{R}^*_+ \rightarrow \mathbb{R}^*_+$ it makes sense, and you would get $f_{me}(x) = \sqrt{f(x)f(1/x)}$ and $f_{mo}(x) = \sqrt{f(x)/f(1/x)}$, but I have never seen anyone use such notions. $\endgroup$ – Captain Lama Mar 6 '16 at 20:13
  • $\begingroup$ I haven't gone into the details but i would think you can split a function into the product of m-odd and m-even functions, where the m-even function is the geometric mean if f(x) and f(1/x), and the m-odd is the GM of f(x) and 1/f(1/x). How usefil this would be, i don't know, and certainly there would be several divide-by-zero gotchas to look out for. $\endgroup$ – IanF1 Mar 6 '16 at 20:15
  • $\begingroup$ Hang on, does this mean a m-even function is one whose logarithm is additively-even? $\endgroup$ – IanF1 Mar 6 '16 at 20:27
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You can generalize even and odd functions $\mathbb{R}\to\mathbb{R}$ in a natural way in the setting of functions $G_1\to G_2$ between groups. If $f:G_1\to G_2$, then $f$ would be "even" if $f(x^{-1})=f(x)$ for all $x\in G_1$ and $f$ would be "odd" if $f(x^{-1})=f(x)^{-1}$ for all $x\in G_1$.

Your example is precisely this construction with $G_1=G_2=\mathbb{R}^\times$, the group of nonzero real numbers under multiplication, or $G_1=G_2=(0,\infty)$, the multiplicative group of positive real numbers. Note that this avoids the problem of potentially dividing by zero.

I think that in this generality, it is probably not possible to split every function into a product (or sum, if you like, if $G_2$ is additive) of "even" and "odd" parts. In order to do this for functions $\mathbb{R}\to\mathbb{R}$, we have to take advantage of some extra multiplicative structure to divide $f(x)+f(-x)$ and $f(x)-f(-x)$ by $2$. The most obvious way to symmetrize $f$, $f(x)f(x^{-1})$, does not even work if $G_2$ is not abelian.

In analogy with the $G_1=G_2=\mathbb{R}$ example, if we assume that $G_2$ is abelian, then $E(x):=f(x)f(x^{-1})$ is an "even" function and $O(x)=f(x)f(x^{-1})^{-1}$ is an "odd" function. We have $E(x)O(x)=f(x)^2$. So in your $\mathbb{R}^\times$ example, you can at least write the square of any function as the product of an even and odd function. This further implies that if $f:(0,\infty)\to(0,\infty)$, then $f$ can be written as the product of an "even" and an "odd" function by setting $g=\sqrt{f}$, $E(x)=g(x)g(x^{-1})$, and $O(x)=g(x)g(x^{-1})^{-1}$, so that $f(x)=g(x)^2=E(x)O(x)$.

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