Technically, the answer is yes $u(s)=1$.
If $\overline{f}(a-s)=f(s)$, \overline{f(a-s)}=f(s)$, then$2 \ \Re{f}(s)$would be holomorphic in that region, as it is equal to$f(s)+\overline{f}(s)= f(s)+\overline{f(s)}= f(s) + f(a-s)$a sum of holomorphic functions. Yet, then as a real-valued holomorphic function$\Re{f}(s)$is constant. And, so$f(s)$is constant. Thus, the only functions$f$fulfilling your assumptions are constant functions, for which what you ask about is clear. 1 Technically, the answer is yes$u(s)=1$. However, for a quite boring reason (as hinted at in the comment of Xogn Ambandl): If$\overline{f}(a-s)=f(s)$, then$2 \ \Re{f}(s)$would be holomorphic in that region, as it is equal to$f(s)+\overline{f}(s)= f(s) + f(a-s)$a sum of holomorphic functions. Yet, then as a real-valued holomorphic function$\Re{f}(s)$is constant. And, so$f(s)$is constant. Thus, the only functions$f\$ fulfilling your assumptions are constant functions, for which what you ask about is clear.