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The Bernstein polynomial isIt has degree at most $n$, since cancellation can lower the degree. For $0\le x\le 1$, the Bernstein basis weights are nonnegative and their sum is $(x+(1-x){)}^n=1$ by the binomial theorem. ConsequentlyThus the Bernstein polynomial defines a positive linear operator on continuous functions satisfying $\Vert B_{n}f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$. Since $B_{n}1=1$, its operator norm in the supremum norm is exactly one.

The Bernstein polynomial is the positive weighted averageFor $0\le x\le 1$, the Bernstein basis weights are nonnegative and the binomial theorem gives ${\sum }_{k=0}^n{b}_{n,k}(x)=1$. Hence $|B_n(f,x)|\le {\sum }_k\Vert f{\Vert }_{\infty }{b}_{n,k}(x)=\Vert f{\Vert }_{\infty }$. In particular,This is the contraction property of the positive linear operator on continuous functions defined by the Bernstein polynomial.

Let $f(x)={\sum }_{j=0}^d{a}_j{x}^j$. For $n\ge \max (1,d)$, express $f$ in the Bernstein basis as $f={\sum }_{k=0}^n{\beta }_{k,n}{b}_{k,n}$, where $b_{k,n}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^k(1-x{)}^{n-k}$. Its precomputed coefficients arewith the ratio for $j=0$ interpreted as one. The identity follows from $x^j={\sum }_k[(k{)}_j/(n{)}_j]b_{k,n}$, obtained by the binomial theorem.

At this level solve the linear programThere are exactly $n+1$ inequalities. Equivalently, $f-\lambda$ must have nonnegative coefficients in the unnormalized basis; its coefficients are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)({\beta }_{k,n}-\lambda )$. Since the basis functions are nonnegative and sum to one, every feasible $\lambda$ satisfies $\lambda \le f(x)$ for all $x\in [0,1]$, so $v_n\le f_{\min }$.

Degree elevation of Bernstein coefficients gives, for $1\le k\le n$,with endpoint coefficients unchanged. These convex combinations imply $v_{n+1}\ge v_n$.

To cover every positive index even when $d>1$, put $\ell =a_0-{\sum }_{j=1}^d|a_j|$. For $1\le n<d$, use the one-inequality LP $\max \{\lambda :\lambda \le \ell \}$, giving $v_n=\ell$. This is a lower bound on $f$; moreover the explicit coefficient formula has all its factorial ratios in $[0,1]$, so ${\beta }_{k,d}\ge \ell$. Thus the transition to level $d$ is also monotone and every level respects the requested bound of at most $n+1$ inequalities. Constants are handled by the main formula at all positive indices.

Finally, for every $ϵ>0$, $f-(f_{\min }-ϵ)$ is strictly positive on the interval. Part (c) supplies a nonnegative-coefficient representation at some degree, and degree elevation preserves it at every higher degree. Hence eventually $v_n\ge f_{\min }-ϵ$. Combining with the upper bound proves the Bernstein linear programming hierarchy for polynomial minimization:

For nonnegative barycentric parameters, these total-degree Bernstein basis functions define a surface over a triangle. The triangular control net has $(n+1)(n+2)/2$ points. Its index constraint couples the three exponents, so it is not a tensor product of two independent univariate bases. The multinomial theorem gives nonnegativity and partition of unity.