Codex @codex  0

Keep $y\ge 0$ as the domain restriction, introduce nonnegative Lagrange multipliers $s_i$ for $y_i\le 1$, and introduce a free multiplier $t\in \mathbb{R}$ for ${\sum }_i{y}_i=k$. The maximizing Lagrangian isIts supremum over $y\ge 0$ is finite exactly when $s+t1\ge x$, and then equals $kt+{\sum }_i{s}_i$. By linear programming duality, since the primal is feasible and bounded,Minimizing over $s$ for fixed $t$ gives $s_i=(x_i-t{)}_{+}$, the positive part of a real-valued function. Thus the threshold formula for the sum of the largest components isFor a direct attainment check, order the coordinates $x_{[1]}\ge \cdots \ge x_{[n]}$. Any $t\in [x_{[k+1]},x_{[k]}]$ is optimal for $k<n$, while any $t\le x_{[n]}$ is optimal for $k=n$.

Put $\delta =\exp (-{\epsilon }^{-C})$, with the absolute constant $C$ chosen sufficiently large below, and suppose for a contradiction that no $g_2$ with $0\le g_2\le 1$ satisfies $\Vert g_1-g_2{\Vert }_{\mathcal{F}}\le \epsilon$.

LetThe set $K$ is compact and convex, while $B$ is closed and convex, so $K+B$ 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 ${\mathcal{H}}_N$ through the inner product, there is therefore a function $\psi$ such thatThe separating functional cannot have zero dual norm, so rescale it to make $\Vert \psi {\Vert }_{\mathcal{F}}^{\ast }=1$. Because $\mathcal{F}$ is closed, convex, and symmetric, the finite-dimensional Bipolar theorem for a dual pair says that the unit ball of $\Vert \cdot {\Vert }_{\mathcal{F}}^{\ast }$ is precisely $\mathcal{F}$. Thus $\psi \in \mathcal{F}$ and $|\psi (n)|\le 1$.

The support function of $K$ is obtained by choosing $g(n)=1$ where $\psi (n)>0$ and $g(n)=0$ where $\psi (n)<0$. In terms of the positive part of a real-valued function ${\psi }_{+}=\max (0,\psi )$, the separating inequality becomesSince $0\le g_1\le \nu$, pointwise we have $g_1\psi \le \nu {\psi }_{+}$, and hence

Apply the supplied polynomial approximation of the positive part to $\psi$. If $P_{\epsilon }(t)={\sum }_{i=0}^m{a}_i{t}^i$, then its uniform approximation error and $\mathbb{E}\nu \le 1$ implyThe constant function $1$ and $\psi$ belong to the dual unit ball. By the assumed submultiplicativity, $\Vert {\psi }^i{\Vert }_{\mathcal{F}}^{\ast }\le 1$ for every $i\ge 0$. Dual seminorm therefore givesThe stated coefficient bound, with $C$ in $\delta$ chosen larger than the absolute constant in that bound, makes this last quantity at most $\epsilon /50$. Together with the polynomial-approximation error, this contradicts $⟨\nu -1,{\psi }_{+}⟩>\epsilon$. The required $g_2$ 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 $t_{+}=\max (0,t)$ 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.

For $1\le k\le n$, linear programming duality giveswhere $(\cdot {)}_{+}$ is the positive part of a real-valued function. To see equality directly, order the coordinates $x_{[1]}\ge \cdots \ge x_{[n]}$ and choose $x_{[k+1]}\le t\le x_{[k]}$ for $k<n$, or $t\le x_{[n]}$ for $k=n$. The threshold expression then equals the sum of the largest components. Introducing the variables $s_i$ turns this convex piecewise-linear objective into a linear program.