Uniqueness of functional equation amongst smooth functions?

Question

Conjecture: the functional equation $x \cdot F(x) - (1-x) \cdot F(1-x) \equiv 0$

has a unique "family" of solutions $x \mapsto \frac{K}{x}$ (indexed by the real number $K$) if we restrict ourselves to smooth functions on $(\epsilon, 1-\epsilon)$ for some small $\epsilon$.

Is this true?

If we do not restrict ourselves to smooth functions uniqueness does clearly not hold. You can take some function $G$ and let $F = \mathcal{1}_{\{x > 0.5\}} \cdot G(x) + \mathcal{1}_{\{x \leq 0.5\}} \cdot \frac{1-x}{x} \cdot G(1-x) $. This will even be continuous if $G$ is continuous.

Answer 1

From the relation that $F$ has to satisfy we get \begin{equation} F(x)=\frac{1-x}{x}F(1-x), \end{equation} therefore we only need to define $F$ in the segment $(\epsilon, \frac{1}{2}]$. Take a smooth function $\phi$ with support in $[\frac{1}{6}, \frac{1}{3}]$ and define $F(x)=\phi(x)$ for $x$ in $(\epsilon, \frac{1}{2}]$. Now you can define $F$ by symmetry in the remaining region.