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$f(x)$ is defined on $\mathbb{R}$, not constant and even. $g(x)$ is defined on $\mathbb{R}$, not constant and periodic.

Why is $f(x) + g(x) + x$ then not even?

It is known that a sum of an even and an odd functions ($f(x) + x$) is neither odd nor even.

But why does adding a periodic function definitely not turn the result into an even function?

alexpetnet
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  • Why would you expect otherwise? – anomaly Aug 10 '16 at 18:33
  • I don't know, to be honest. What should've led me to treat this idea as unreasonable right away? – alexpetnet Aug 10 '16 at 18:36
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    Periodicity has nothing to do with odd- or evenness. Also, in broad terms, even and odd functions are eigenfunctions of the operator $(Tf)(x) = f(-x)$, and periodic functions are eigenfunctions (with eigenvalue $1$) of the operator $(Pf)(x) = f(x + t)$ for some $t$. Since $T$ and $P$ don't commute, their eigenspaces do not coincide, and the sum of their respective eigenfunctions isn't an eigenfunction of either one. – anomaly Aug 10 '16 at 18:46

2 Answers2

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Suppose $f(x)+g(x)+x$ was even. Since $f(x)$ is even, this implies $g(x)+x$ is even, i.e., for all $x$, $$g(x)+x=g(-x)-x$$ or equivalently $$g(x)-g(-x)=-2x$$ Now let $p>0$ be the period of $g$. Then $g(-p/2)=g(p/2)$, so setting $x=p/2$ above yields $$p=0$$ a contradiction (note that we did not use the fact that $f$ and $g$ are not constant).

Luiz Cordeiro
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  • Thank you! So the contradiction method with an observation that the sum of two even functions is definitely even should be used. – alexpetnet Aug 10 '16 at 18:32
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You have it wrong: the sum of an even, an odd, and a periodic function can be even. You can easily see this by choosing $\sin x$ as your odd function and $-\sin x$ as your periodic function.

TonyK
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