What are the consequences of $\mathsf{L}^2 \subseteq \mathsf{P}$? We know that $\mathsf{L} \subseteq \mathsf{NL} \subseteq \mathsf{P}$ and that $\mathsf{L} \subseteq \mathsf{NL} \subseteq \mathsf{L}^2 \subseteq $ $\mathsf{polyL}$, where $\mathsf{L}^2 = \mathsf{DSPACE}(\log^2 n)$. We also know that $\mathsf{polyL} \neq \mathsf{P}$ because the latter has complete problems under logarithmic space many-one reductions while the former does not (due to the space hierarchy theorem). In order to understand the relationship between $\mathsf{polyL}$ and $\mathsf{P}$, it may help to first understand the relationship between $\mathsf{L}^2$ and $\mathsf{P}$.

What are the consequences of $\mathsf{L}^2 \subseteq \mathsf{P}$?

What about the stronger $\mathsf{L}^{k} \subseteq \mathsf{P}$ for $k&gt;2$, or the weaker $\mathsf{L}^{1 + \epsilon} \subseteq \mathsf{P}$ for $\epsilon &gt; 0$?
The following is an obvious consequence:
$\mathsf{L}^{1+\epsilon} \subseteq \mathsf{P}$ would imply $\mathsf{L} \subsetneq \mathsf{P}$ and therefore $\mathsf{L} \neq \mathsf{P}$.

By the space hierarchy theorem, $\forall \epsilon &gt; 0: \mathsf{L} \subsetneq \mathsf{L}^{1+\epsilon}$ . 
If $\mathsf{L}^{1+\epsilon} \subseteq \mathsf{P}$ then $\mathsf{L} \subsetneq \mathsf{L}^{1+\epsilon} \subseteq \mathsf{P}$.