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Let $f,g:R\to R\setminus\{0\}$ and $\forall x,y\in R$,such $$\color{crimson}{f(x-y)=f(x)g(y)-f(y)g(x)}$$ I have prove the function $\color{crimson}f$ odd function.

because let $y=0$ we have $$f(x)=f(x)g(0)-f(0)g(x)\tag{1}$$ Let $x=0,y=x$ we have $$f(-x)=f(0)g(x)-f(x)g(0)\tag{2}$$ by $(1),(2)$ ,then $$f(x)=-f(-x)$$

Conjecture： $\color{crimson}{g(x)}$ is even function

For this function $g(x)$ is even problem ,I don't have any idea to prove it.But I think is right,because
$$\color{blue}{\sin{(x-y)}=\sin{x}\cos{y}-\sin{y}\cos{x}}$$ $$\color{crimson}{\sinh{(x-y)}=\sinh{x}\cosh{y}-\sinh{y}\cosh{x}}$$

math110
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  • First, necessarily $f(0)=0$ as you showed, otherwise there is no such solutions. More generally, necessarily $f$ is odd. But $g$ cannot be any even function: $f(x-0)=f(x)g(0)=f(x)$ so if $f(x)\neq 0$, then $g(0)=1$. The open question is, can $g$ be any even function provided $g(0)=1$? – anderstood May 13 '16 at 14:03
  • Note that it's possible for $g$ to be odd and not even. Let $f(x) = 0$ and $g(x) = \sin(x)$. Then $f(x-y)=f(x)g(y) - f(y)g(x)$, as desired. This isn't a very interesting case, though. – diracdeltafunk May 16 '16 at 05:08

2 Answers2

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The conjecture is false.

Consider $f(x) = x$ and $g(x) = 1+x$. Now, for all $x,y \in \mathbb{R}$, $$f(x-y) = x-y$$ and $$f(x)g(y) - f(y)g(x) = x(1+y) - y(1+x) = x + xy - y - xy = x-y,$$ so $f(x-y) = f(x)g(y) - f(y)g(x)$. However, $g$ is not even.

diracdeltafunk
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You're in the right direction when you got $Eq-2$ by putting $x=0$ and $y=x$. From that you get $g(x) = \frac{f(-x) + f(x)g(0)}{f(0)}$. Similarly, putting $x=0$ and $y=-x$ can give you $g(-x)$ which is same as $g(x)$. Thus proving $g(x)$ is even. Note that dividing by $f(0)$ is safe because the range of $f$ does not include 0.

BTW, the function range for odd functions usually include 0, unless you don't want to define your function to be defined at 0. You could argue that $\lim_{x \rightarrow 0} f(x) = 0$ but then I am talking things that I am not sure about :D

wadkar
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  • No, Putting $x=0,y=-x$ then we have $$g(-x)=\dfrac{f(x)+f(-x)g(0)}{f(0)}$$ it doesn't usefull – math110 May 13 '16 at 04:49
  • Yep, you're right, I was trying to give you directions of getting $g(x)$ and $g(-x)$ and assuming that would lead you to the answer. I should've checked myself if it's useful. Sorry! – wadkar May 13 '16 at 04:57