The $i^\text{th}$ degree-$n$ Bernstein basis polynomial is defined as $$ \begin{equation} b_i^n(x) = \binom{n}{i} (1-x)^{n-i} x^i. \end{equation} $$ The Bernstein basis polynomials have many interesting properties. For my application, I am interested that they satisfy $b_i^n(x) \ge 0$ on $x \in [0, 1]$ and $\sum_{i=0}^n b_i^n(x) = 1$.
For $j=0,\ldots,n$, define $f_j^n(x) \equiv \sum_{i=0}^n \alpha_{ij} b_i^n(x)$ such that $\sum_{i=0}^n \alpha_{ij} = 1$, an affine combination of the Bernstein basis polynomials. If $f_j^n(x)$ is a convex combination of the polynomials then it is likewise nonnegative on the unit interval. That said, it is clear that one may construct $f_j^n(x), j=0,\ldots,n$ that satisfy $f_j^n(x) \ge 0$ on $x \in [0, 1]$ and sum to $1$. As a trivial example, let $\alpha_{ij}$ be $1$ if $i = j$ and $0$ otherwise so that $f_i^n(x) = b_i^n(x)$.
I would like to prove that the only such sets of polynomials $f_j^n(x), j=0,\ldots,n$ that satisfy $f_j^n(x) \ge 0$ on $x \in [0, 1]$ and sum to $1$ are convex combinations. I attempted a proof by contradiction (e.g., suppose not and show that if any $\alpha_{ij} < 0$ then there exists a $j'$ and $\hat{x} \in [0, 1]$ such that $f_{j'}(\hat{x}) < 0$), but failed to arrive at a contradiction. Has this already been proved perhaps using other language/notation? If not, do any reasonable approaches come to mind? Thanks in advance.
Here is a counterexample: $$ f_0^3 = \frac{1}{2}\left(b_1^3+b_2^3\right) \\ f_1^3 = \frac{1}{2}\left(b_0^3-b_1^3+b_2^3+b_3^3\right) \\ f_2^3 = \frac{1}{2}\left(b_0^3+b_1^3-b_2^3+b_3^3\right) \\ f_3^3 = \frac{1}{2}\left(b_1^3+b_2^3\right) $$