I've seen several instances where an undecidability/uncomputability result can be used to produce a lower bound complexity result, and I am interested in when the process can go the other way. In particular: suppose you have a computable subset $L\subset\mathbb{N}\times\mathbb{N}$ (computable in the sense that membership of a given pair can be determined by some algorithm.) Then create the decision problem $P$ as follows: on input $n$ decide whether there exists $m$ so that $(n, m)\in L$.
Observation: suppose $P$ is undecidable, then there is a sequence $n_i\in P$ where the minimal solutions $m_i$ (that is, the smallest $m_i$ so that $(n_i,m_i)\in L$) grow faster than any computable bound relative to $n_i$. Indeed, if it were not so, then you could get an algorithm for $P$ by checking all $m_i$ up to this computable bound.
Question: Is there a way to get a converse result? To clarify (in response to a comment) a converse would be something like: if there exist a sequence $n_i$ so that each $n_i$ is in $P$ but the sequence $m_i$ defined on each term as the smallest $m$ such that $(n_i,m_i)\in L$ grows faster than any computable bound relative to $n_i$, then P is undecidable. Equivalently, if there is a sequence $n_i\in P$ such that any sequence $m_i$ satisfying $(n_i,m_i)\in L$ satisfies $m_i>f(n_i)$ for infinitely many $i$ and any computable $f$, then $P$ is uncomputable. At first I thought obviously no: the $L$ defined by $(n, BB(n))$ for all $n$ has such a sequence and gives trivial $P$, but this isn't a computable subset of $\mathbb{N}\times\mathbb{N}$
A motivation here is the problem of diophantine equations. The fact that the existence of a solution to diophantine equations is undecidable means that there are sequences of equations with solutions that grow faster than any bound computable from the coefficients. Suppose though that one wanted to consider a subset of diophantine equations, but one where membership in the subset is computable. Then one might wonder whether containing such sequences meant that even restricting to this subset led to an undecidable problem.
