A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps.
It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$
are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.
My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?
Best Answer
It is proven in
Ellis, Robert, Locally compact transformation groups, Duke Math. J. 24, 119-125 (1957). ZBL0079.16602.
that a locally compact Hausdorff topological space equipped with a group structure where both left and right multiplications are continuous, is actually a topological group, i.e. the multiplication is jointly continuous. Now, you can use the result that a topological manifold which is also a topological group admits a unique Lie group structure.