Proposition 49

Let A be a complete normable unital associative algebra. We identify A with $L(A^*)$. Then, if $g \in A^*$ and $y \in A$, $(\mathrm{Ad}\, g)y = gyg^{-1}$.

Recall that $\mathrm{Ad}\, g = T_1(\mathrm{Int}\, g)$. Let $u_g$ be the mapping $x \mapsto gxg^{-1}$ of A into A. The identity chart of $A^*$ into A transforms $\mathrm{Int}\, g$ into $u_g \mid A^*$. The tangent mapping at each point of $A^*$ to this mapping is equal to $u_g$, whence the proposition.