Fuzzy's Involution Generalization As a consequence from my personal attempt to generalize the Fuzzy Logic's negation, I've got the following functional equation: \begin{align*} &f(x) + f(\alpha + \beta - x) = 2*f\left(\frac{\alpha+\beta}{2}\right),\,\text{where}\,\,\alpha < \beta,\\\ &f(\alpha) = \beta,\,\,f(\beta) = \alpha\,\,\text{and}\,\,f:[\alpha,\beta]\rightarrow[\alpha,\beta]\,\,\text{is continuous}. \end{align*} Thence it comes the question: could anyone please help me to solve it? Thank you in advance.

If you're looking for one solution, then here's one:

Let $f(x) = \alpha+\beta -x$. Then $f(\alpha)=\beta$, $f(\beta)=\alpha$, $f(x)+f(\alpha+\beta-x)=\alpha+\beta-x +\alpha+\beta -(\alpha+\beta -x)=\alpha+\beta$. Since this is true for all $x$, then it is in particular true when $x=(\alpha+\beta)/2$. Continuity is obvious.

But really by the second equation combined with the first equation, we have when $x=\alpha$, $f(\alpha)+f(\beta)=\beta+\alpha = 2f((\alpha+\beta)/2)$.

Now choose any continuous function $g:[\alpha,(\alpha+\beta)/2]$ satisfying $g(\alpha)=\beta$, and $g((\alpha+\beta)/2)=(\alpha+\beta)/2$. Define

$$ f(x)=\left\\{ \begin{array}{ll} g(x)&x\le (\alpha+\beta)/2\\\ \alpha+\beta - g(\alpha+\beta-x) & x \ge (\alpha+\beta) /2 \end{array}\right.$$

It should be easy to check that this has the required properties, and any solution has this form.