0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.