Bernstein polynomials $\{b_{\nu, n}(x)\}_{\nu = 0, \dots, n}$ form a basis for any polynomial of degree $\leq n$ with real coefficients: i.e., $B_n(x) = \sum_{\nu=0}^n \beta_\nu b_{\nu,n}(x)$ where $\{\beta_{\nu} \in \mathbb{R}\}$.
I was wondering whether something can be said about the set of Bernstein coefficients $\{\beta_{\nu}\}$ that gives a positive polynomial $B_n(x)$ on the domain $x \in [0,1]$: $\{\{\beta_{\nu}\} : \sum_{\nu=0}^n \beta_\nu b_{\nu,n}(x) \geq 0, \forall x \in [0,1] \}$. (We know that the Bernstein polynomials $\{b_{\nu,n}(x)\}$ are positive for any $x \in [0,1]$, so I am looking for the more interesting case where at least one $\beta_{\nu} < 0$.)
In particular, I am looking for a way to compute this set of Bernstein coefficients (for an arbitrary value of $n$) in a way that is more efficient than a brute force grid search.
This is far from a complete answer. Just to obtain examples with negative coefficients I think the following works: Consider the identity $$ \sum_{\nu =0}^{2n} b_{\nu,2n}(x)=1. $$ Hence, for $\lambda > 0$ $$ \sum_{\nu =0,\nu \not=n}^{2n} b_{\nu,2n}(x)-\lambda b_{n,2n}(x)=1-(1+\lambda) b_{n,2n}(x). $$ The maximum of $b_{n,2n}$ (attained at $1/2$) is $(1/2)^{2n} {2n \choose n}$, and $(1/2)^{2n} {2n \choose n} \to 0$ $(n \to \infty)$. Thus, for $n$ sufficiently big $$ \sum_{\nu =0,\nu \not=n}^{2n} b_{\nu,2n}(x)-\lambda b_{n,2n}(x) > 0 \quad (x \in [0,1]). $$