2 expand on recursion theorem

Note: The previous paragraph requires more than the usual statement of the recursion theorem, it requires some knowledge of the proof to show that $\phi_k$ is polynomial-time. Here is the construction I need.

Let $s(j,k)$ be the usual polynomial-time function such that $\phi_{s(j,k)}(n) \simeq \phi_j(k,n)$; the key point we need is that the running time of $\phi_{s(j,k)}(n)$ is polynomially bounded if the running time of $\phi_j(k,n)$ is polynomially bounded, and the first of these is not smaller than the second. This can be checked by examining the construction of $s$ in the chosen model of computation.

Now let $d$ be the index for the computable function $\phi_d(j,n) = B(s(j,j),n)$ obtained by simple composition. Let $k = s(d,d)$. Then $\phi_k(n) = \phi_d(d,n) = B(k,n)$ as desired; this is the proof of the recursion theorem. Moreover, the implementation of these functions ensures that $\phi_k(n)$ runs in polynomial time but not faster than $B(k,n)$, because each computation of $\phi_k(n)$ consists of some polynomial-time-in-$n$ invocations of $s$ functions followed by the literal execution of the program for $B(k,n)$.

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You can also diagonalize directly against a purported bound-producing algorithm. Say that $R(j)$ returns a polynomial when run with any index $j$ of a polynomial time function as input.

Define a function $B(j,n)$ as follows. On input $n$, run $R(j)$ for $n$ steps. If this doesn't halt, return $0$ immediately. Otherwise, if $R(j)$ does not return a polynomial when it halts, return $0$ immediately. Otherwise, if the polynomial is $p(x)$, waste at least $(n+p(n))^2$ steps and then return $0$.

Note that for any $j$, the function $C_j(n) = \lambda n . B(j,n)$ is total and runs in polynomial time, and if $R(j)$ returns a polynomial then this is not a bound on the running time of $C_j(n)$.

Now the function that takes a number $j$ and returns an index for the $C_j$ is a total computable function. So we can use the recursion theorem to produce an index $k$ such that $\phi_k(n) = C_k(n)$. Then $\phi_k$ will be a total polynomial-time function, but if $R(k)$ returns a polynomial then this is not an upper bound for $\phi_k$.