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If one-dimensional smooth linear characters of a local multiplicative group have conductors tending to infinity, their direct sum can be admissible. Its algebraic dual is the product of inverse linear characters, but a smooth vector in this product has only finitely many nonzero coordinates. A common open stabilizer contains one fixed principal-unit subgroup, on which every sufficiently ramified linear character is nontrivial. Thus the smooth dual is again a direct sum.

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.

A smooth vector in any abstract group representation of a locally profinite group has an open stabilizer, equivalently is fixed by some compact-open subgroup. Smooth vectors form an invariant linear subspace: use intersections for addition and conjugation for translation. A smooth representation is one in which every vector has this property.