Prove that the Bernstein operator preserves function degree. (If $f$ has degree $m$, then $\mathcal{B}(f)$ has degree $m$)

Question

This is a copy of the problem presented here, however, I did not understand the problem due to the manipulations being done without much motivation. I therefore restate the problem here.

Problem:

Given a function $f: [0, 1] \to \mathbb{R}$ of degree $m$ we define the Bernstein approximation of order $n$ to $f$ as the function $$g(x) = \sum_{i=0}^{n} f\left(\frac{i}{n}\right) B_i^n(x).$$

The goal is to show that if $f$ has degree $m \leq n$, then $g$ has degree $m$.

Preliminaries:

The $i$'th Bernstein polynomial of degree $n$ is defined as $$ B_i^n(x) = {n \choose i}x^i(1 - x)^{n-i}. $$

A Bézier curve $p(t)$ is a function on the form $$ p(t) = \sum_{i=0}^n c_i B_i^n(u) $$ where $u = (t - a) / (b - a)$ and $t \in [a, b]$. The $r$'th derivative of p with respect to $t$ is $$ p^{(r)}(t) = \frac{n!}{(n - r)!(b - a)^r}\sum_{i=0}^{n-r}\Delta^{r}c_iB_i^{n-r}(u) $$ where $\Delta^r c_i = \Delta^{r-1}c_{i+1} - \Delta^{r-1}c_i$ denotes the $r$'th forward difference of $c_i$.

What I have tried:

1. My initial idea for how to solve this is by looking at the derivatives of $g$ with respect to $t$. If I can show that $g$ differentiated $m$ times is a constant, and that $g$ differentiated $m + 1$ times is zero, then $g$ has degree $m$. This corresponds to showing that the $m$'th forward difference of $c_i$ is non-zero, while the $j$'th forward difference of $c_i$ is zero for $j > m$.
2. Notice that the Bernstein approximation is a linear operator. We can therefore just consider the case where $f(x) = x^m$. This is what the linked answer above does, but it does it without motivation for the various manipulations, which leaves me a bit baffled. The trick here is to manipulate the expression to end up with something where you have coefficients $D_i$ in front of the $x^i$ terms, and proving the claim by showing that $D_i$ is zero for $i > m$.

Any nudges in the right direction is appreciated.

Thanks in advance!

Answer 1

As you observe, it is enough to consider the case $f(x) = x^m$ for $m\le n$. Here's a probabilistic approach. Let $S_n$ be a random variable with the binomial distribution with parameters $n$ and $x$: $$ P[S_n=k]={n\choose k}x^k(1-x)^{n-k},\qquad k=0,1,\ldots,n. $$ Then $$ g(x) = E[f(S_n/n)], $$ Pulling out the factor $n^{-m}$, it's enough to show that $$ E[S_n^m] $$ is a polynomial in $x$ of degree $\le m$, provided $0\le m\le n$. Now think of coin tossing: $S_n=X_1+X_2+\cdots+X_n$ where the $X_k$ are independent and $P[X_k=1]=1-P[X_k=0]=x$. Expand $S_n^m$ using the multinomial theorem $$ S_n^m=(X_1+\cdots+X_n)^m=\sum_{k_1+k_2+\cdots+k_n=m}{m\choose k_1,k_2,\dots,k_n}X_1^{k_1}X_2^{k_2}\cdots X_n^{k_n}. $$ Use this expansion and independence to find an expression for $E[S_n^m]$. It will be evident that this expression is a polynomial in $x$ of degree $\le m$.