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Suppose linear operators $T_n$ on a normed vector space satisfy $\Vert T_n\Vert \le C$ uniformly and $T_{n}p\to p$ on a dense subspace. The displayed triangle inequality for any element $p$ of that subspace first permits $p$ to approximate $f$, then lets $n$ increase, proving $T_{n}f\to f$ for every $f$. Applied to the Bernstein polynomial, the Weierstrass approximation theorem supplies the dense polynomial subspace and $C=1$.

The Weierstrass approximation theorem states that the polynomials are a dense subspace of $C[0,1]$ in the supremum norm. By linearity and (c), $B_{n}p\to p$ uniformly for every polynomial $p$. Given $f\in C[0,1]$ and $\epsilon >0$, choose a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. The operator norm estimate in (a) givesFirst let $n\to \infty$, then $\epsilon \downarrow 0$. Hence $\Vert B_{n}f-f{\Vert }_{\infty }\to 0$ for every continuous $f$. This is an instance of extension of approximation operators from a dense subspace.