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Let $f$ be an even function, and $g$ odd.

Let $h$ be some arbitrary function.

Is it the case that $f(x) + h(x),\ fh(x),\ hf(x),\text{ and }f(x)h(x)$ are each even or odd according to $h$, and that $g(x) + h(x),\ gh(x),\ hg(x),\text{ and }g(x)h(x)$ are each odd irrespective of $h$?

I thought I noticed this pattern reading some examples in A Course in Pure Mathematics, but I was unsure if it was coincidence or something I could rely on.

NB: I'm not asking for proofs of each case, just a general "yes, you're right, see here", "no, you're way off", or "almost - but not in these cases, see here" would suffice - it's not homework!

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By $fh(x)$, do you mean $f(x)\cdot h(x)$ or $f(h(x))$? –  5xum Apr 15 at 13:53
    
Obviously the latter, since I listed the former separately.. –  Ollie Ford Apr 15 at 14:17
    
I see. Then, for a general function $h$, see my answer and for special cases when $h$ is even or odd, see the answer by @GitGud –  5xum Apr 15 at 18:07

2 Answers 2

up vote 2 down vote accepted

I assume $f,g$ and $h$ are defined on $\mathbb R$.

If $h$ is even:

  1. $f+h$ is even. True.
  2. $fh$ is even. True.
  3. $g+h$ is even. False. Take $g$ the identity function and $h$ constantly $1$.
  4. $g+h$ is odd. False. Same as 3..
  5. $gh$ is odd. True.

If $h$ is odd:

  1. $f+h$ is even. False. See 3. from the even $h$ case.
  2. $f+h$ is odd. False. Same as above.
  3. $fh$ is odd. True.
  4. $g+h$ is odd. True.
  5. $gh$ is even. True.
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Thank you! Must just have been a coincidental collection of examples that threw me. Mostly trigonometrical. –  Ollie Ford Apr 15 at 14:21

For most of these cases, simple counterexamples can be found.

  • For an even function $f(x)=1$ and the function h(x)=x, the function $f(x)+h(x) = 1+x$ is neither odd nor even
  • For an even function $f(x) = |x|$ and the function $h(x) = x+1$, the function $f(h(x))=|x+1|$ is neither odd nor even
  • For the even function $f(x) = 1$ and the function $h(x) = x+1$, the function $f(x)\cdot h(x)$ is neither odd nor even.

The only case that can be proven is that if $f(x)$ is even, then for any function $h(x)$, you have $h(f(-x)) = h(f(x))$ meaning the function is even.

For the second case, again, counterexamples are easy to find:

  • For an odd function $g(x) = 0$ and the function $h(x) = x+1$, the function $g(x)+h(x)=x+1$ is neither odd nor even.
  • For an odd function $g(x) = x$ and the function $h(x) = x+1$, neigher of the functions $g(h(x)) = x+1$ nor $h(g(x)) = x+1$ is either even or odd.
  • For an odd function $g(x) = x$ and the function $h(x) = x+1$, the function $g(x)h(x) = x^2+1$ is neither even nor odd.
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