= Solution
The <Bernstein polynomial> is
$$
B_n(f,x)=\sum_{j=0}^n f(j/n)\binom nj x^j(1-x)^{n-j}.
$$
It has degree at most $n$, since cancellation can lower the degree. For $0\le x\le1$, the <Bernstein basis> weights are nonnegative and their sum is $(x+(1-x))^n=1$ by the <binomial theorem>. Consequently
$$
|B_n(f,x)|\le\sum_{j=0}^n |f(j/n)|\binom nj x^j(1-x)^{n-j}\le\|f\|_\infty.
$$
Thus the <Bernstein polynomial> defines a <positive linear operator on continuous functions> satisfying \b[$\|B_n f\|_\infty\le\|f\|_\infty$]. Since $B_n1=1$, its <operator norm> in the <supremum norm> is exactly one.
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