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 $b=1$.
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 $b=1$.
(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.