$f : X \to Y$ is a deterministic polynomial-time algorithm for problem inputs $x \in X$ and problem outputs $f(x) = y \in Y \iff $there exists a polynomial $P_f \in \Bbb{Z}[x_1]$ such that $C\cdot P_f(|x|) = $ the running time on input $x, |x| = $ size of $x, C \in \Bbb{R}$.
Conjecture
Let $f: X \to Y$ specifiy a problem.
Then there is a det. poly-time algorithm for $f$ that uses at most $k$ words of memory each of size $n$ $\iff$ there exists $O(1)$ algorithms $\phi : X \to (\Bbb{Z_n})^k, \ \psi: (\Bbb{Z_n})^k \to Y$ and $k$ polynomials in $P_i \in \Bbb{Z_n}[z_1, \dots, z_k]$ such that $f = \psi \circ (P_1, \dots, P_k) \circ \phi$.
I think I can prove that too. Has anyone looked into this yet?
$\phi, \psi$ are an encoder and decoder. You have to represent sufficiently your inputs and outputs in your typical computer storage before you can solve the problem, right? So it assumes all data needed to solve a problem has to be calculated when you run the algorithm. No I think the precomputed data used to solve a problem is held in the constants of the polynomial or equivalently in the code specifying the algorithm.
This might have important implications such as the set of all poly-time algorithms is finite for finite data memory (or something...) and a natural ring structure on the set of algorithms.
I believe that you are asking the question does LOGSPACE = P, which remains an open question similar to the P vs NP problem.
It is known that the use of constant space $\subset$ LOGSPACE $\subset$ PSPACE, where PSPACE uses polynomial space, and it known that LOGSPACE is contained within P. It is generally believed that LOGSPACE $\subset$ P $\subset$ NP $\subset$ PSPACE, but only the first and last containments are known to be strict.