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.)

  • 2
    $\begingroup$ 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 $\endgroup$
    – Ilya
    Mar 20 at 12:32
  • $\begingroup$ @llya so you said that it depend if product is converges or not?!! $\endgroup$ Mar 20 at 12:38
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Mar 20 at 12:52
  • 1
    $\begingroup$ Please edit your post for clarity. $\endgroup$
    – lulu
    Mar 20 at 12:53
  • $\begingroup$ Well, a function defined by an infinite product should have this product converging in every point of its definition domain $\endgroup$
    – Ilya
    Mar 20 at 12:55

1 Answer 1


It sounds like you are asking the following:

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 $n\in\Bbb N.$

  • $\begingroup$ thank you so much :) $\endgroup$ Mar 20 at 19:13

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