I asked the question if PA can prove any function growing faster than $f_{\epsilon_0}(n)$ to be total. The answer was no.
What about the converse : Can prove PA every function growing slower than $f_{\epsilon_0}(n)$ to be total ? If no, what is the slowest growing function, which cannot be proven total by PA ?
No, and there is no slowest growing function which cannot be proven total by PA.
Let $f$ be any recursive function that cannot be proven total by PA; we will require that $f$ be strictly increasing to avoid some difficulties. Then choose a fast growing function $g$ which can be proven total by PA (faster than $g(n) = n$, of course). Let $h(n) = \sup \lbrace m | g(m) \le n \rbrace$. Then $j(n) = f(h(n))$ is a slower growing function than $f$, and it cannot be proven total by PA.