Does the convex cone of monotonic functions on a compact set admit a countable conic generating set?

Question

For any function $f:[0,1]^d\rightarrow\mathbb{R}$ and any $i\in\{1,\dots,d\}$ and $x\in[0,1]^d$, let the functions $f_{i,x}:[0,1]\rightarrow\mathbb{R}$ and $\partial_i f:[0,1]^d\rightarrow\mathbb{R}$ be defined by: $$f_{i,x}(z)=f([x_1,\dots,x_{i-1},z,x_{i+1},\dots,x_d]),\space\forall z\in[0,1]$$ $$\partial_i f(w)=\frac{\partial f(w)}{\partial w_i},\space\forall w\in[0,1]^d.$$

For all $d\in\mathbb{N}$ let: $$R_d^A:=\left\{f\in C([0,1]^d)\middle|\forall i\in\{1,\dots,d\},\space \forall x\in[0,1]^d,\space f_{i,x}\in AC([0,1])\right\},$$ $$R_d^B:=\left\{f\in C([0,1]^d)\middle|\forall i\in\{1,\dots,d\},\space \partial_i f \in C([0,1])\right\},$$ $$S_d^A:=\left\{f\in R_d^A\middle|f\ge 0,\space\forall x,y\in[0,1]^d,\space x\le y \rightarrow f(x)\le f(y)\right\},$$ $$S_d^B:=\left\{f\in R_d^B\middle|f\ge 0,\space\forall i\in\{1,\dots,d\},\space((\partial_i f = 0)\lor(\partial_i f > 0))\right\},$$ where $C(Z)$ is the space of continuous functions on $Z$ and $AC(Z)$ is the space of absolutely continuous functions on $Z$.

Note that $R_d^B\subset R_d^A$ as all continuously differentiable functions are absolutely continuous.

Let $E\in\{A,B\}$. Then $R_d^E$ is a vector space and $S_d^E$ is a convex cone within it. I.e. if $f,g\in S_d^E$, then for all $\alpha,\beta\in \mathbb{R}_{\ge 0}$, $\alpha f + \beta g \in S_d^E$.

Does there exist a countable set $\hat{S}^E_d$ such that for all $f\in S_d^E$, there exists $f_1,f_2,\dots\in \hat{S}^E_d$ and $\alpha_1, \alpha_2, \dots \in \mathbb{R}_{\ge 0}$ such that $\sum_{k=1}^n{\alpha_k f_k(x)}$ converges uniformly to $f(x)$ as $n\rightarrow\infty$?

I.e. does $S_d^E$ admit a countable conic generating set, for $E\in\{A,B\}$?

My conjecture is that it would do to take: $$\hat{S}^A_1:=\{x\mapsto 1\}\cup\{\left\{x\mapsto\int_0^{x_i}{1_{[l_i,r_i]}(z)\space dz}\middle|l,r\in \mathbb{Q}^d\cap[0,1]^d,\space l\ll r\right\}\subset S^A_1,$$ and $$\hat{S}^B_1:=\{x\mapsto 1\}\cup\left\{x\mapsto\int_0^{x_i}{z^{p_i}(1-z)^{q_i}\space dz}\middle|p,q\in \mathbb{N}^d\right\}\subset S^B_1,$$ where $1_Z$ is the indicator function of the set $Z$, with $\hat{S}^E_d$ for $E\in\{A,B\}$ and $d>1$ defined by: $$\hat{S}^E_d:=\left\{x\mapsto\prod_{i=1}^d{f_i(x_i)}\middle|f_1,\dots,f_d\in \hat{S}_1^E\right\}.$$

A possible rough outline of a proof follows. One major claims is unproven (and marked as such). There may also be mistakes in the remainder!

1. First consider the $d=1$ special case.
2. By construction $R_1^B\subset R_1^A=AC([0,1])$. Define: $$L_1^+([0,1]):=\left\{g\in L_1([0,1])\middle|g\ge 0\right\},$$ where $L_1([0,1])$ is the space of Lebesgue integrable functions on $[0,1]$. Then for all $f\in S_1^E$, since $f\in AC([0,1])$, there exists $g\in L_1^+([0,1])$ such that for all $x\in[0,1]$: $$f(x)=f(0)+\int_0^x{g(z)\space dz}.$$ Since $(x\mapsto 1)\in\hat{S}^E_d$ without loss of generality we may assume $f(0)=0$ in the following.
3. CASE A: Suppose $E=A$. By the definition of the Lebesgue integral, there exists $\alpha_1, \alpha_2, \dots \in \mathbb{R}_{\ge 0}$ and $l_1,r_1,l_2,r_2,\dots\in\mathbb{Q}\cap [0,1]$ such that for all $x\in[0,1]$: $$f(x)=\int_0^x{g(z)\space dz}=\int_0^x{\sum_{k=1}^\infty{\alpha_k 1_{[l_k,r_k]}(z)}\space dz}.$$ Thus, by Fubini's theorem: $$f(x)=\sum_{k=1}^\infty{\alpha_k \int_0^x{1_{[l_k,r_k]}(z)\space dz}}.$$ I.e. there exists $f_1,f_2,\dots\in \hat{S}_1^A$ such that for all $x\in[0,1]$, $$f(x)=\sum_{k=1}^\infty{\alpha_k f_k(x)}.$$ Since $f_1,f_2,\dots$ are monotonic increasing, $|f_1|\le f_1(1),\space |f_2|\le f_2(1),\space\dots$, thus since $\sum_{k=1}^\infty{\alpha_k f_k(1)}=f(1)<\infty$, by the Weierstrass M-test, $\sum_{k=1}^\infty{\alpha_k f_k}$ converges uniformly to $f$, as required. Thus $\hat{S}_1^A$ is a countable conic generating set for $S_1^A$.
4. CASE B: Suppose $E=B$. Note $g=\partial_1 f \in C([0,1])$ with $g>0$ without loss of generality. Choose $(p_k)_{k=1}^\infty$ and $(q_k)_{k=1}^\infty$ such that $p_k+q_k$ is weakly increasing in $k$ and such that for any $p,q\in\mathbb{N}$, there exists $k\in\mathbb{N}_{>0}$ such that $p_k=p$ and $q_k=q$. We must then have $p_1+q_1=0$, $p_2+q_2=1$, $p_3+q_3=1$, $p_4+q_4=2$, and so on. For all $n\in\mathbb{N}$ we define $h_n:[0,1]\rightarrow\mathbb{R}_{>0}$ by: $$h_n(z) = g(z)-\sum_{k=1}^{\frac{n(n+1)}{2}}{\alpha_k z^{p_k} (1-z)^{q_k}},$$ where the values of $\alpha_k$ are still to be given. We have that $h_0=g>0$. Now suppose (inductively) that for some $n\in\mathbb{N}$ we have defined $\alpha_k$ for $k\in\left\{1,\dots,\frac{n(n+1)}{2}\right\}$ (i.e. the $\alpha_k$ entering $h_n$ are defined) and we have proven that $h_n>0$. Now by the uniform convergence of Bernstein polynomials, there exists a minimal $N_n\in\mathbb{N}$ such that: $$\max_{z\in[0,1]}{\left| h_n(z) - \sum_{k=\frac{N_n(N_n+1)}{2}+1}^{\frac{(N_n+1)(N_n+2)}{2}}{h_n\left(\frac{p_k}{p_k+q_k}\right) {p_k+q_k \choose p_k} z^{p_k} (1-z)^{q_k}} \right|} < \frac{1}{2}\min_{z\in[0,1]}{h_n(z)}.$$ (Note the $\max$ and $\min$ exist due to the continuity of the objectives and the compactness of the domain.) If $N_n\le n$, then for $k\in\left\{\frac{n(n+1)}{2}+1,\dots,\frac{(n+1)(n+2)}{2}\right\}$ set $\alpha_k:=\frac{1}{2}h_n\left(\frac{p_k}{p_k+q_k}\right) {p_k+q_k \choose p_k}$, else for the same set of $k$ set $\alpha_k:=0$. Since $N_{n+1}=N_n$ in the latter case, eventually we must hit the former case. In the latter case, we have that $h_{n+1}=h_n>0$, and in the former case we have that for all $z\in[0,1]$: $$h_{n+1}(z) - \frac{1}{2}h_n(z)= \frac{1}{2} h_n(z) - \sum_{k=\frac{N_n(N_n+1)}{2}+1}^{\frac{(N_n+1)(N_n+2)}{2}}{\alpha_k z^{p_k} (1-z)^{q_k}}$$ $$= \frac{1}{2} h_n(z) - \frac{1}{2}\sum_{k=\frac{N_n(N_n+1)}{2}+1}^{\frac{(N_n+1)(N_n+2)}{2}}{h_n\left(\frac{p_k}{p_k+q_k}\right) {p_k+q_k \choose p_k} z^{p_k} (1-z)^{q_k}},$$ so: $$\max_{z\in[0,1]}{\left|h_{n+1}(z)-\frac{1}{2}h_n(z)\right|}<\frac{1}{4}\min_{z\in[0,1]}{h_n(z)}.$$ Thus if we define $\underline{z}_{n+1}:=\arg\min_{z\in[0,1]}{h_{n+1}(z)}$, $$\left|h_{n+1}(\underline{z}_{n+1})-\frac{1}{2}h_n(\underline{z}_{n+1})\right|<\frac{1}{4}h_n(\underline{z}_{n+1}),$$ so for all $z\in[0,1]$ $h_{n+1}(z)\ge h_{n+1}(\underline{z}_{n+1})>\frac{1}{4}h_n(\underline{z}_{n+1})>0$ as required for the inductive hypothesis. Having proven this, we know that the procedure must generate infinitely many non-zero $\alpha_k$. Additionally note that if we define $\overline{z}_{n+1}:=\arg\max_{z\in[0,1]}{h_{n+1}(z)}$, $$\left|h_{n+1}(\overline{z}_{n+1})-\frac{1}{2}h_n(\overline{z}_{n+1})\right|<\frac{1}{4}h_n(\overline{z}_{n+1}),$$ so $\sup_{z\in[0,1]}{h_{n+1}(z)} = h_{n+1}(\overline{z}_{n+1}) < \frac{3}{4} h_n(\overline{z}_{n+1}) \le \frac{3}{4} \sup_{z\in[0,1]}{h_n(z)}$. Thus, everytime the procedure generates non-zero $\alpha_k$ it reduces the maximum approximation error by a factor of $\frac{3}{4}$. Thus $h_n$ converges uniformly to $0$, and $\sum_{k=1}^{\frac{n(n+1)}{2}}{\alpha_k z^{p_k} (1-z)^{q_k}}$ converges uniformly to $g$. Following the same proof as in CASE A, this gives uniform convergence of $\sum_{k=1}^\infty{\alpha_k f_k}$ to $f$ for some $f_k\in\hat{S}^B_1$. Hence $f_k\in\hat{S}^B_1$ is a countable conic generating set for $S_1^B$.
5. Conjectured lemma: If $\hat{S}_1^E$ is a countable conic generating set for $S_1^E$, then $$\left\{x\mapsto\prod_{i=1}^d{f_i(x_i)}\middle|f_1,\dots,f_d\in \hat{S}_1^E\right\}$$ is a countable conic generating set for $S_d^E$.
6. Alternative to 5. The $d>1$ could also be handled directly using the definition of the multivariate Lebesgue integral, and the properties of multivariate Bernstein polynomials (see e.g. Theorem 5 of this paper or the comments section here). Key would be the fact that an arbitrary function $f\in S_d^E$ may be decomposed as $f=\sum_{I\subseteq\{1,\dots,d\}}{f_I}$, where $f_{\{\}}(x)=f(0)\ge 0$, for all $x\in[0,1]^d$, and $f_I\ge 0$ for $I\ne\{\}$ is defined inductively by $f_I(x)=f(q_I(x))-\sum_{J\subset I}{f_J}(q_I(x))\ge 0$, for all $x\in[0,1]^d$, where for all $x\in[0,1]^d$ and $i\in\{1,\dots,d\}$, ${q_I(x)}_i$ equals $x_i$ if $i\in I$ or $0$ otherwise. (The proof that $\sum_{I\subseteq\{1,\dots,d\}}{f_I}=f$ is immediate from the definition of $f_{\{1,\dots,d\}}$.)
7. Given results 3, 4 and either 5 or 6 above, $\hat{S}_d^A$ and $\hat{S}_d^B$ are countable conic generating sets for $S_d^A$ and $S_d^B$, respectively.