 Solution

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.