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Does the equation $f(x+y)=f(x)+f(y)$ (with $x \in\mathbb{R^+}$ and $y \in\mathbb{R^-}$) ($0$ in included in $\mathbb{R^+}$ and $\mathbb{R^-}$) have the same solution of the normal equation knowing that we have that $f$ is bijective and increasing

Xander Henderson
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Math player
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    What is the normal equation? – jet457 Aug 01 '23 at 18:38
  • You get $f(0)=0$, by putting both $x=y=0$. You get $f(-a)=-f(a)$, for $a\geq0$, by putting $x=a$, $y=-a$. For $a<0$, put $x=-a$ and $y=a$. Now, for $r,s>0$, assume that $r>s$. Put $x=r+s$ and $y=-r$. Then $f(s)=f((r+s)+(-r))=f(r+s)+f(-r)=f(r+s)-f(r)$. So, $f(r+s)=f(r)+f(s)$. So, the equation also holds for both $x,y\geq0$. Using $f(-x)=-f(x)$ from before, we also get that it holds for both negative. So, $f$ satisfies Cauchy's functional equation. – NDB Aug 01 '23 at 18:46
  • How is this not a duplicate of https://math.stackexchange.com/q/423492/ ? Or https://math.stackexchange.com/q/1656500/ ? – Xander Henderson Aug 01 '23 at 18:51
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    I think it's slightly different because we've put different requirements on $x$ and $y$. – jet457 Aug 01 '23 at 18:52
  • @jet457 It is slightly different, but not in a way that significantly, materially changes the problem. – Xander Henderson Aug 01 '23 at 18:54

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Yes. For $x, y \in \mathbb R_+$,

$$f(x) = f(x+y-y) = f(x+y) + f(-y) = f(x+y) + f(0) - f(y) = f(x+y) - f(y)$$

Do the same for $x, y\in \mathbb R_-$ and you will have the "normal equation".

Kroki
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