[Math] Intuition about the Bernstein polynomials proof of the Weierstrass approximation theorem

Question

The Weierstrass approximation theorem can be stated as follows:

Let $f\in C([a,b])$. There exists a sequence $(p_n)_{n\in \mathbb{N}}$ of polynomials in $[a,b]$ such that $(p_n)$ converges uniformly to $f$.

One approach to prove this theorem is to notice that we just need to prove this for $[a,b]=[0,1]$ and then consider the Bernstein polynomials:

$$B_n(f)(x)=\sum_{k=0}^n f\left(\dfrac{k}{n}\right)\binom{n}{k}x^k(1-x)^{n-k},$$

and then prove that $(B_n(f))$ converges uniformly to $f$. The proof then indeed shows this convergence.

My question here is regarding intuition. This is the kind of thing that I wonder how could anyone think of defining those polynomials so that they converge to $f$.

So: what is the intuition behind the Bernstein polynomials? Considering that we want to find a sequence of polynomials which converge uniformly to a continuous function, how could we ever think about defining those polynomials? Is there some intuition here?

Answer 1

The sum of the coefficients ${n\choose k}x^k(1-x)^{n-k}$ is equal $1.$ For $ x$ close to ${k\over n}$ the coefficient ${n\choose k}x^k(1-x)^{n-k}$ is the largest.