Put , with the absolute constant chosen sufficiently large below, and suppose for a contradiction that no with satisfies .
Let
The set is compact and convex, while is closed and convex, so is closed and convex. We use the following finite-dimensional form of the Hahn-Banach separation theorem: if a point lies outside a nonempty closed convex set, there is a linear functional whose value at the point is strictly greater than its supremum over that set. Identifying linear functionals on through the inner product, there is therefore a function such that
The separating functional cannot have zero dual norm, so rescale it to make . Because is closed, convex, and symmetric, the finite-dimensional Bipolar theorem for a dual pair says that the unit ball of is precisely . Thus and .
The support function of is obtained by choosing where and where . In terms of the positive part of a real-valued function , the separating inequality becomes
Since , pointwise we have , and hence
Apply the supplied polynomial approximation of the positive part to . If , then its uniform approximation error and imply
The constant function and belong to the dual unit ball. By the assumed submultiplicativity, for every . Dual seminorm therefore gives
The stated coefficient bound, with in chosen larger than the absolute constant in that bound, makes this last quantity at most . Together with the polynomial-approximation error, this contradicts . The required consequently exists. This proves the dense model theorem for a multiplicative test family.
On any fixed compact interval, the positive part of a real-valued function can be approximated uniformly by a real polynomial. Quantitative dense-model arguments use a version whose degree and coefficient growth are explicitly controlled in terms of the approximation error.