= Solution
For a <continuous function> $f$ on $[0,1]$, its <Bernstein polynomial> is
$$
B_n(f,x)=\sum_{j=0}^n f(j/n)\binom njx^j(1-x)^{n-j}.
$$
The weights are nonnegative and, by the <binomial theorem>, sum to $(x+1-x)^n=1$. Thus $|B_n(f,x)|\le\sum_j|f(j/n)|\binom njx^j(1-x)^{n-j}\le\|f\|_\infty$ at every $x\in[0,1]$, including the endpoints. Taking the <supremum norm> gives
$$
\boxed{\|B_nf\|_\infty\le\|f\|_\infty.}
$$
In particular, $B_n$ is a <linear operator> of norm one, since it preserves the constant function one.
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