# prove or disprove: Multiplication of infinite number of even functions is also an even function

I know that if $$f_1(x)$$,$$f_2(x)$$,...,$$f_n(x)$$ are even functions and $$c_1$$,$$c_2,...,c_n$$ are fixed real numbers then
$$c_1f_1(x)$$ $$c_2f_2(x)$$ ... $$c_nf_n(x)$$ is also an even function.

But how for infinite Multiplication ((Multiplication of ($$f_i$$ , i is member of $$I$$))?!

exactly i want to prove or disprove that "Multiplication of infinite number of even functions is also an even function".

i think its wrong but not sure !!! i have an example in my mind but i dont know its correct or not!!

im thinking about multiplication of a divergent series,like this :

let $$f_1(x)=1$$ , $$f_2(x)=-1$$ , $$f_3(x)=1$$ , f$$_4(x)=-1$$ ,....

all of $$f_i(x)$$ are even. but is $$f_1(x)f_2(x)f_3(x)f_4(x)..... = (1)(-1)(1)(-1)....$$ also even function?

i dont know this counter example is correct or not. (if correct how to continue my argument and finish it,and if not please prove its also even function.)

• An infinite product (if it converges) is a limit of partial, finite products. The latter are even functions. Does it give you a hint? Besides that, welcome to MSE and please write your formulas in MathJax: same syntax as LaTeX
– Ilya
Mar 20 at 12:32
• @llya so you said that it depend if product is converges or not?!! Mar 20 at 12:38
• This is very hard to read. In your example, are we only looking at constant functions? But the infinite product you appear to write down certainly doesn't converge.
– lulu
Mar 20 at 12:52
– lulu
Mar 20 at 12:53
• Well, a function defined by an infinite product should have this product converging in every point of its definition domain
– Ilya
Mar 20 at 12:55

Suppose that $$E\subseteq\Bbb R$$ such that for all $$x\in E,$$ $$-x\in E.$$ Similarly, suppose for all $$n\in\Bbb N$$ that $$E_n\subseteq\Bbb R$$ such that for all $$x\in E_n,$$ $$-x\in E_n.$$ Suppose that for each $$n\in\Bbb N,$$ $$f_n:E_n\to\Bbb R$$ such that for all $$x\in E_n,$$ $$f_n(-x)=f_n(x).$$ Finally, suppose $$E\subseteq E_n$$ for all $$n\in\Bbb N$$ and that for all $$x\in E$$ we have $$f(x)=\prod_{n=1}^\infty f_n(x).$$ Can we conclude that for all $$x\in E,$$ $$f(-x)=f(x)$$?
The answer to this question is yes. This is because for each $$x\in E,$$ we have $$f(-x)=\lim_{k\to\infty}\prod_{n=1}^kf_n(-x)=\lim_{k\to\infty}\prod_{n=1}^kf_n(x)=f(x).$$
Note in particular that the function in your example is technically even, since for all $$x\in E,$$ we have $$f(-x)=f(x).$$ That just isn't very interesting, since $$E=\emptyset,$$ even though $$E_n=\Bbb R$$ for all $$n\in\Bbb N.$$