Suppose we have a functional equation in the form $$f(1-x)=\chi (x) f(x)$$ with given function $\chi (x)$. What is the condition on the function $\chi (x)$ so that we can write this reflection relation in a symmetric form $$\xi(1-x)=\xi(x) ?$$ I have the only simple answer: the $\chi (x)$ function should be of the type:$$\chi (x)= \phi(1-x)/\phi(x)$$ then $\xi(x)=f(x)/\phi(x)$. Is it correct?

Edit I want to illustrate my question with an example.

Let first $\chi (x)$ be $$ \chi (x)= \frac {\pi}{\sin\pi x (\Gamma (x))^2} $$ And we have functional equation $\Gamma (x) \Gamma (1-x)=\frac {\pi}{\sin\pi x}$ without symmetric form.

Let then $\chi (x)$ be $$ \chi (x)= \frac { \pi^{-\frac {x}{2}}\Gamma(\frac{x}{2})} { \pi^{-\frac {1-x}{2}} \Gamma(\frac{1-x}{2})} $$ Here $\phi(x)= ({ \pi^{-\frac {x}{2}}\Gamma(\frac{x}{2})})^{-1} $ and we have symmetric form with $\xi(x)=\zeta(x)/\phi(x)$

  • $\begingroup$ I have the feeling that you can always write $f(1-x)=\chi (x) f(x)$ for an appropriate $\chi$ and set $\xi(x) = f(x)f(1-x)$ or some other symmetric form. Is there something I'm missing? $\endgroup$ – G.Carugno Apr 10 at 10:41
  • $\begingroup$ @G.Carugno, I mean the case when we don't know the expliсit expression for $f$ and do have an explicit expression for $ \chi$ and if it is in the form $\chi (x)= \phi(1-x)/\phi(x)$ we can get symmetrical (but also unknown) function $ \xi (x) = f(x)/ \phi(x)$ $\endgroup$ – Aleksey Druggist Apr 10 at 12:52
  • $\begingroup$ Taking the log derivative we have $F(x)-F(-x) = G(x)$ where $F(x) = f'/f(1/2+x), G(x) = \chi'/\chi(1/2+x)$ so $G$ is odd so $G(x) = H(x)-H(-x)$ and $F(x)+H(x)$ is odd ie. $f(1/2+x) h(1/2+x)$ is even where $h(x) = \exp(\int H(1/2- x)$. With $f(x) = \zeta(x)$ then $h(x) = \pi^{-x/2} \Gamma(x/2)$ $\endgroup$ – reuns Apr 11 at 11:08
  • $\begingroup$ I still want to understand the answer to my question As far as I can see the function $\chi$ must be the kind of $ \chi (x)= \phi(1-x)/\phi(x) $ to satisfy the original functional equation $f(1-x)=\chi (x) f(x)$ hence the additional condition for the existence of a symmetric form is not required $\endgroup$ – Aleksey Druggist Apr 12 at 9:46

There is not much going on here. Just to make that clear, without loss of essential generality, suppose that we have any function $\ f(x)\ $ and define $\ g(x) := f(x) - f(-x).\ $ Now we have $\ g(-x) = f(-x) - f(-(-x)) = f(-x) - f(x) = -g(x).\ $ Thus, $\ g(x)\ $ is an odd function as in Even and odd functions. In fact, it is twice the odd part of $\ f(x).\ $ Further, it follows that by adding any even function to $\ f(x)\ $ we can recover the same odd function $\ g(x).\ $

In your case, addition and subtraction is replaced by multiplication and division, and also the involution $\ x \to -x\ $ is replaced by $\ x \to 1-x.$


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