I'm trying to prove that $\operatorname{bb}(x)= \max \{U (e, 0) \mid e \leq x \text{ and } (e, 0) \in \operatorname{Dom} (U)\}$ is not recursive.
($U$ is the universal recursive function, i.e. $U$ is recursive s.t. for any recursive function $f$ there is a code $e(f)$ s.t. $U(e(f),x)=f(x)$ for every $x$ in $\operatorname{Dom}(f)$.)
But I am required to use Rogers's fixed point thm and I don't know on which function I should apply it.
Thank you very much !
Nobody answered and I found a solution so Ill leave it here in case anyone needs it.
Assume by contradiction it is recursive.
WLOG we can suppose $bb$ is a fully defined function (its not so clear what does $U(0,x)$ is but we can for example define a new function $bb'(x) = bb(x+N)$ where N is the code of the zero function).
Now we define $$\sigma(x) = Code(n\rightarrow bb(x)+1) $$ Since $bb$ is recursive and fully defined, so is $\sigma$. Then by Rogers theorem, there is some $x$ s.t $U(x, 0)=U(\sigma(x),0)=bb(x)+1$ which is a contradiction to the definition