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$f(x)$ is a nonincreasing function of $|x|$, does this automatically imply that $f$ is symmetric or even? I haven't seen it written this way. Sorry for asking such basic questions.

amanda zhang
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    I'm unsure of your meaning. If $f(x)$ can be considered a function of $|x|$, then it is an even function (since $|x|=|-x|$ will imply that $f(x)=f(-x)$. This doesn't rely on any "nondecreasing" behavior of $f(x)$. – hardmath Nov 28 '13 at 20:26
  • @TrevorWilson Yes, I assume I confused myself too. Sorry. – LeeNeverGup Nov 28 '13 at 20:36

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Any function of $|x|$ is even (that is, symmetric) when considered as a function of $x$.

To see this, suppose that we can write $f(x) = g(|x|)$. (Any function $f$ that only depends on the magnitude of its argument and not on the sign of its argument can be written in this way for an appropriate function $g$.)

Then the function $f$ is even: $$f(-x) = g(\left|\mathop{-}x\right|) = g(|x|) = f(x).$$ (Because $\left|\mathop{-}x\right| = |x|$.) Note that it doesn't matter whether or not the function $f$ is non-increasing.

Trevor Wilson
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