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Let $f:ℂ→ℂ$ be a function.

(a) How I can solve this functional equation:

$f(b-s)+f(s)=0,b∈ℝ$ with respect to $f$ and $b=1$.

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In part (a), what do you mean by "and $b=1$". Should the equation simply read $f(b-s) + f(s) = 0$? – Calvin Lin Dec 30 '12 at 16:38
No, I mean the case: $f(1-s)+f(s)=0$ – ZE1 Dec 30 '12 at 16:40
up vote 1 down vote accepted

(a) For $s=\frac {1}{2}$, we have $f(\frac {1}{2}) +f(\frac {1}{2}) = 0$ , so we must have $f( \frac {1}{2}) =0$.

Consider $S = (\frac {1}{2}, \infty) \times (-\infty, \infty) \cup \frac {1}{2} \times (0, \infty)$.

Define $f(s): S \rightarrow \mathbb{C}$ however you wish. It does not even need to be continuous.

Consider $ T = \mathbb{C} \setminus S \setminus (\frac {1}{2},0)$

For $t \in T, 1-t \in S$, and so we define $ f(t) = -f(1-t)$. This then clearly satisfies your condition.

For (b), do the same with similar sets.

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@ Calvin Lin : Thank you very much for this answer. As far I a can see the only property of $f$ is to be an odd function. You determined only the set where the equation holds. – ZE1 Dec 30 '12 at 17:01
By setting $g(s)=f((1/2)+s)$ – ZE1 Dec 30 '12 at 17:25
@ Calvin Lin: By the set $ T = \mathbb{C} \setminus S \setminus (0, \frac {1}{2})$ you mean in inside: $S$ minus the two points not the interval $(0, \frac {1}{2})$. – ZE1 Dec 30 '12 at 18:31
@RH1 Think about the symmetry, which you stated as "$g$ is an odd function. You could create the sets $S$ and $T$ in a variety of ways. – Calvin Lin Dec 31 '12 at 4:55
@ Calvin Lin: In the set $S = (\frac {1}{2}, \infty) \times (-\infty, \infty) \cup \frac {1}{2} \times (0, \infty)$. Is the part $(1/2)×(0,∞)$ is an open set (not containing $(1/2))$. Should'nt $S$ be $S=((1/2)+i0)∪(((1/2),∞)×ℝ)∪((1/2)×(0,∞))$. – ZE1 Jan 1 '13 at 11:47

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