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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.

Let $\ell =({\ell }_i{)}_i$ be a smooth vector in the product description of the algebraic contragredient representation. Its open stabilizer contains $U_m$ for some $m$. For every $i\ge m$, choose $u_i\in U_i\subseteq U_m$ with ${\chi }_i(u_i)\ne 1$. Invariance of $\ell$ at its $i$th coordinate givesforcing ${\ell }_i=0$. Hence a smooth functional has finite support, contained among $i<m$.

Conversely a finitely supported family has indices bounded by some $N$. Each inverse linear character has the same kernel as its original linear character, and all those kernels contain $U_{N+1}$. Thus this subgroup fixes the family. The smooth dual is thereforeThis is the principle that conductors force finite support in a smooth dual. For instance a product vector with every coordinate nonzero is a perfectly valid algebraic functional on the original direct sum, but is not smooth.

The smooth dual consists of the smooth vectors of the algebraic contragredient representation. Their union is a vector subspace because finite intersections of open stabilizers remain open; it is invariant because translation conjugates stabilizers.