Bernstein control values bound a piecewise polynomial

Fix a cell and $t\in [0,1]$. Both $t$ and $1-t$ are nonnegative, so their natural powers are nonnegative by induction using the recursive definition [F10] and closure of nonnegative reals under multiplication. By [F3]–[F4], $\left(\genfrac{}{}{0px}{}{n}{k}\right)$ is a natural that is either zero or positive; [F5] sends its zero case to zero, and [F6] sends its positive case to a positive real. Thus every $B_{k,n}(t)\ge 0$.

Apply [F2] with $x=t$ and $y=1-t$. Since $t+(1-t)=1$, [F10] and induction give $1^n=1$, so the displayed Bernstein basis terms sum to one.

By [F8]–[F9], let $c_{\min }$ and $c_{\max }$ be the minimum and maximum of the finite control list. For each $k$, [F1], step 1.1, and [F7] give $c_{\min }{B}_{k,n}(t)\le c_{i,k}{B}_{k,n}(t)\le c_{\max }{B}_{k,n}(t)$: the control differences and basis weights are nonnegative, so their products are nonnegative. Summing with [F7] and using step 1.2 gives $c_{\min }\le p_i(x)\le c_{\max }$. Therefore nonnegative controls imply $p_i(x)\ge 0$. Each $p_i$ is a polynomial in $x$ after substituting the affine local coordinate, so it is continuous on its cell by [F11]. For $n_i\ge 1$, [F3], [F5], [F10], step 1.2, and [F12] give $B_{0,n_i}(0)=B_{n_i,n_i}(1)=1$ and all other endpoint basis values zero; for $n_i=0$ the sole basis term is constantly one. Thus at a shared node $p_i(x_{i+1})=c_{i,n_i}=c_{i+1,0}=p_{i+1}(x_{i+1})$. Given $\epsilon >0$, the minimum of the two positive continuity radii for the adjacent cell restrictions then gives continuity of the piecewise function at that node; at cell interiors continuity follows from [F11], and at the outer endpoints it follows from the single adjacent cell restriction. Applying the bound cellwise gives nonnegativity throughout the partition.