What are the consequences of $P \subseteq L/poly$? A language is in $L/poly$ if there exists a logspace Turing machine that decides the language with polynomial amount of advice.

See here for more info: https://en.wikipedia.org/wiki/L/poly


  Question
  
  What are the consequences of $P \subseteq L/poly$?
One simple consequence is $\mathbf{P}/\text{poly} = \mathbf{L}/\text{poly}$. Proof: For any language $A \in \mathbf{P}/\text{poly}$, there is a language $B \in \mathbf{P}$ and a sequence of polynomial-length advice strings $y_1, y_2, y_3, \dots$ such that $x \in A \iff (x, y_{|x|}) \in B$. By assumption, there is a language $C \in \mathbf{L}$ and a sequence of polynomial-length advice strings $z_1, z_2, z_3, \dots$ such that $(x, y) \in B \iff (x, y, z_{|(x, y)|}) \in C$. This implies $A \in \mathbf{L}/\text{poly}$; the advice string for $x$ is $(y_{|x|}, z_{|(x, y_{|x|})|})$.

(A concise version of the proof: $\mathbf{P} \subseteq \mathbf{L}/\text{poly} \implies \mathbf{P}/\text{poly} \subseteq (\mathbf{L}/\text{poly})/\text{poly} = \mathbf{L}/\text{poly}$.)