 Solution

The algebraic dual of a direct sum is a product, rather than another direct sum. The mapis injective, since the summands span $V$. Given any family $({\ell }_i{)}_i$, define $\ell ((v_i{)}_i)={\sum }_i{\ell }_i(v_i)$. This sum is finite for each vector, regardless of whether the family of functionals has finite support. It defines a linear functional and is inverse to $R$. Hence this is a natural vector-space isomorphism.

The algebraic contragredient representation is $({\pi }^{\ast }(g)\ell )(v)=\ell (\pi (g^{-1})v)$. On the $i$th coordinate,Thus $R$ intertwines the actions and givesNo smoothness restriction has yet been imposed on that product.