Let the linear mapping ${P} : \mathbb{V}\rightarrow \mathbb{V}$ where $\mathbb{V}$ is the space of all functions real valueted of real variable, namely x, we set the image under ${P}$ of a function f(x) to be $$(Pf)(x)= \frac{f(x) + f(-x)}{2},$$ what can i say about the image of $\mathbb{V}$ under ${P}$ ?

My initial hypothesis was that the direct image on the whole domain would be the set of all function but the odd functions, i proved if $\mathbb{V}$ is composed only by either odd function or even functions, but trying to be more general, stating $\mathbb{V}$ to be the set defined on the first paragraph i could not prove. Someone know a way to do this?

  • $\begingroup$ i don't understand how the image can contain odd functions. If $g(x) = \frac{f(x)+f(-x)}{2}$ then $g(-x) = g(x)$ so the image contains only even functions. $\endgroup$
    – gt6989b
    Jun 7, 2016 at 15:07
  • 1
    $\begingroup$ You're not on a bad track. Any real function decomposes uniquely as sum of an odd and an even function. $\endgroup$
    – Pedro
    Jun 7, 2016 at 15:08
  • $\begingroup$ @gt6989b i agree that the image can not contain odd function, but this set could also contain a function whose is neither odd nor even. $\endgroup$ Jun 7, 2016 at 16:20
  • $\begingroup$ @PedroTamaroff there is a name for this statement? i mean where can i find more informations about this? $\endgroup$ Jun 7, 2016 at 16:21
  • $\begingroup$ @YassinRany no it cannot... see Ashwin's answer $\endgroup$
    – gt6989b
    Jun 7, 2016 at 16:22

1 Answer 1


The image would be the set of all even functions. It's clear the image consists only of even functions because for every function $f(x)$, $\frac{f(x)+f(-x)}{2}$ is an even function. Also, the image consists of all even functions because if $g(x)$ is any even function, then $g(x)$ is in the preimage of $g(x)$. So $P$ is a map from the set of all functions onto the set of all even functions.


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