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Solve this functional equation: $$f(s,t)=4f(s,u)f(u,t)-f(s,u)-f(u,t)+\frac{1}{2}, \ \ \mbox{for any} \ \ 0 \leq s<u<t.$$

I have found only one constant solution that is $f(s,t)=\frac{1}{4}$, and then got stuck. Thanks in advance.

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  • $\begingroup$ What function space do you work in? $\endgroup$
    – Emil
    Mar 23 '20 at 12:36
  • $\begingroup$ Could you tell the context of this issue ? $\endgroup$
    – Jean Marie
    Mar 23 '20 at 12:38
  • $\begingroup$ @Emil The function can be any function from $[0,+\infty)^2$ to the set of real numbers. $\endgroup$
    – VIVID
    Mar 23 '20 at 13:04
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The solution you found suggests a change of variables. Define $g(s, t) = 4f(s, t) - 1$. Then the functional equation is just $$g(s, t) = g(s, u) g(u, t) \qquad (0 \leq s < u < t).$$ There are rather a lot of solutions to this.

Focusing on the case $g > 0$, we may write $g(s, t) = \exp m(s, t)$, and then the functional equation becomes $$m(s,t) = m(s,u) + m(u, t),$$ which is more or less just the definition of a finitely additive signed Borel measure on $[0, \infty)$. For example, for any $f \in L^1$ there is a solution defined by $m(s, t) = \int_s^t f$. There are plenty of more exotic solutions.

In general, the sign and magnitude parts of the problem for $g$ separate. Both parts of the problem have lots of exotic solutions.

Are you sure you don't have any further hypotheses?

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  • $\begingroup$ Your solution contains the step that I was looking for! Thanks! $\endgroup$
    – VIVID
    Mar 23 '20 at 18:12

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