Editing Bernstein polynomial (section)

=== Elementary proof ===

The probabilistic proof can also be rephrased in an elementary way, using the underlying probabilistic ideas but proceeding by direct verification:<ref>{{harvnb|Lorentz|1953|pages=5–6}}</ref><ref name="Beals 2004"/><ref>{{harvnb|Goldberg|1964}}</ref><ref>{{harvnb|Akhiezer|1956}}</ref><ref>{{harvnb|Burkill|1959}}</ref>

The following identities can be verified:

# <math> \sum_k {n \choose k} x^k (1-x)^{n-k} = 1</math> ("probability")
# <math> \sum_k {k\over n} {n \choose k} x^k (1-x)^{n-k} = x</math> ("mean")
# <math> \sum_k \left( x -{k\over n}\right)^2 {n \choose k} x^k (1-x)^{n-k} = {x(1-x)\over n}. </math> ("variance")

In fact, by the binomial theorem

<math display="block">(1+t)^n = \sum_k {n \choose k} t^k,</math>

and this equation can be applied twice to <math>t\frac{d}{dt}</math>. The identities (1), (2), and (3) follow easily using the substitution <math>t = x/ (1 - x)</math>.

Within these three identities, use the above basis polynomial notation

:<math> b_{k,n}(x) = {n\choose k} x^k (1-x)^{n-k},</math>

and let

:<math> f_n(x) = \sum_k f(k/n)\, b_{k,n}(x).</math>

Thus, by identity (1)

:<math>f_n(x) - f(x) =  \sum_k [f(k/n) - f(x)] \,b_{k,n}(x), </math>

so that

:<math>|f_n(x) - f(x)| \le   \sum_k |f(k/n) - f(x)| \, b_{k,n}(x).</math>

Since ''f'' is uniformly continuous, given <math>\varepsilon > 0</math>, there is a <math>\delta > 0</math> such that <math>|f(a) - f(b)| < \varepsilon</math> whenever
<math>|a-b| < \delta</math>.  Moreover, by continuity, <math>M= \sup |f| < \infty</math>. But then

:<math> |f_n(x) - f(x)| \le \sum_{|x -{k\over n}|< \delta} |f(k/n) - f(x)|\, b_{k,n}(x) + \sum_{|x -{k\over n}|\ge \delta} |f(k/n) - f(x)|\, b_{k,n}(x) .</math>

The first sum is less than ε. On the other hand, by identity (3) above, and since <math>|x - k/n| \ge \delta</math>, the second sum is bounded by <math>2M</math> times

:<math>\sum_{|x - k/n|\ge \delta} b_{k,n}(x) \le  \sum_k \delta^{-2} \left(x -{k\over n}\right)^2 b_{k,n}(x)  = \delta^{-2} {x(1-x)\over n} <  {1\over4} \delta^{-2} n^{-1}.</math>

:([[Chebyshev's inequality]])

It follows that the polynomials ''f''<sub>''n''</sub> tend to ''f'' uniformly.