# Is P equal to the intersection of all superpolynomial time classes?

Let us call a function $f(n)$ superpolynomial if $\lim_{n\rightarrow\infty} n^c/f(n)=0$ holds for every $c>0$.

It is clear that for any language $L\in {\mathsf P}$ it holds that $L\in {\mathsf {DTIME}}(f(n))$ for every superpolynomial time bound $f(n)$. I wonder, wether the converse of this statement is also true? That is, if we know $L\in {\mathsf {DTIME}}(f(n))$ for every superpolynomial time bound $f(n)$, does it imply $L\in {\mathsf P}$? In other words, is it true that $${\mathsf P} = \cap_f {\mathsf {DTIME}}(f(n))$$ where the intersection is taken over every superpolynomial $f(n)$.

• A general advice about writing questions is that you should make your question (stated in the easiest way to understand) your title. – Kaveh Sep 18 '14 at 3:12

In fact, by the McCreight-Meyer Union Theorem (Theorem 5.5 of McCreight and Meyer, 1969, free version here) a result of that I believe is due to Manuel Blum, there is a single function $f$ such that $\mathsf{P} = \mathsf{DTIME}(f(n))$. This function is necessarily superpolynomial, but "just barely."
The theorem applies more generally to any Blum complexity measure $\Phi$ and any union class $\bigcup_{f \in S} \mathsf{BLUM}_{\Phi}(f(n))$ where $S$ is a c.e., self bounded set of total computable functions. (A set of functions $S$ is c.e. if there is a single partial computable function $F(i,\vec{x})$ such that $S = \{f_i(\vec{x}) | i \in \mathbb{N}\}$ where $f_i(\vec{x}) := F(i,\vec{x})$. Self-bounded means that for every finite subset $S_0 \subset S$, there is a function in $S$ that dominates all $g \in S_0$ almost everywhere. "$\mathsf{BLUM}_{\Phi}$" is a notation I haven't seen before, but I like it :) - I'm using it for the $\Phi$-bounded analog of a time-bounded complexity class.)
• I think the catch is that $f$ is not time-constructible. – Sasho Nikolov Sep 18 '14 at 3:52