I'm looking for two injective functions $f, g:\mathbb R\to\mathbb R$ with $f(x)+g(x)=x$ for all $x\in\mathbb R$ and $\operatorname{Im} f\cap\operatorname{Im} g=\emptyset$.
I've tried nothing and I'm all out of ideas.
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$f(x)=e^x$ $g(x) = x-e^x$, $\text{Im} f = (0, \infty)$, $\text{Im} g = (-\infty, -1)$ – Apr 06 '14 at 05:11
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@Mike $x-e^x$ is not injective. – Amateur Apr 06 '14 at 05:13
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@Amateur Ah, you're right. How obvious. Sorry. – Apr 06 '14 at 05:13
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I think $f(x) := e^x+2$ and $g(x) := -e^x-1$ work. – Amateur Apr 06 '14 at 05:22
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@Amateur OP wants $f(x)+g(x)=x$, not $f(x)+g(x)=1$. If it were the latter, that would work. – Apr 06 '14 at 05:23
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@Mike Okay, my bad, it is late here I think I should go to sleep :). – Amateur Apr 06 '14 at 05:24
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@Amateur As a fellow mistake maker, I agree! – Apr 06 '14 at 05:28
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Hint: look for something of the form $$\eqalign{f(x) &= \cases{h(x) & if $x \le 0$\cr h(x) + x & if $x > 0$\cr}\cr g(x) &= \cases{x-h(x) & if $x \le 0$\cr - h(x) & if $x > 0$\cr}}$$
for suitable "small" $h$.
Robert Israel
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What would be the requirements on $h$ for $f$ and $g$ to be increasing? – Robert Israel Apr 06 '14 at 18:29