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analysis.special_functions.bernstein

Bernstein approximations and Weierstrass' theorem #

We prove that the Bernstein approximations

 k : fin (n+1), f (k/n : ) * n.choose k * x^k * (1-x)^(n-k)

for a continuous function f : C([0,1], ℝ) converge uniformly to f as n tends to infinity.

Our proof follows [Richard Beals' Analysis, an introduction][beals-analysis], §7D. The original proof, due to Bernstein in 1912, is probabilistic, and relies on Bernoulli's theorem, which gives bounds for how quickly the observed frequencies in a Bernoulli trial approach the underlying probability.

The proof here does not directly rely on Bernoulli's theorem, but can also be given a probabilistic account.

(You don't need to think in these terms to follow the proof below: it's a giant calc block!)

This result proves Weierstrass' theorem that polynomials are dense in C([0,1], ℝ), although we defer an abstract statement of this until later.

The Bernstein polynomials, as continuous functions on [0,1].

Equations
@[simp]
theorem bernstein_apply (n ν : ) (x : unit_interval) :
(bernstein n ν) x = (((n.choose ν)) * x ^ ν) * (1 - x) ^ (n - ν)
theorem bernstein_nonneg {n ν : } {x : unit_interval} :
0 (bernstein n ν) x

We now give a slight reformulation of bernstein_polynomial.variance.

def bernstein.z {n : } (k : fin (n + 1)) :

Send k : fin (n+1) to the equally spaced points k/n in the unit interval.

Equations
theorem bernstein.probability (n : ) (x : unit_interval) :
∑ (k : fin (n + 1)), (bernstein n k) x = 1
theorem bernstein.variance {n : } (h : 0 < n) (x : unit_interval) :
∑ (k : fin (n + 1)), ((x - (bernstein.z k)) ^ 2) * (bernstein n k) x = (x) * (1 - x) / n

The n-th approximation of a continuous function on [0,1] by Bernstein polynomials, given by ∑ k, f (k/n) * bernstein n k x.

Equations

We now set up some of the basic machinery of the proof that the Bernstein approximations converge uniformly.

A key player is the set S f ε h n x, for some function f : C(I, ℝ), h : 0 < ε, n : ℕ and x : I.

This is the set of points k in fin (n+1) such that k/n is within δ of x, where δ is the modulus of uniform continuity for f, chosen so |f x - f y| < ε/2 when |x - y| < δ.

We show that if k ∉ S, then 1 ≤ δ^-2 * (x - k/n)^2.

@[simp]
def bernstein_approximation.δ (f : C(unit_interval, )) (ε : ) (h : 0 < ε) :

The modulus of (uniform) continuity for f, chosen so |f x - f y| < ε/2 when |x - y| < δ.

Equations
def bernstein_approximation.S (f : C(unit_interval, )) (ε : ) (h : 0 < ε) (n : ) (x : unit_interval) :
finset (fin (n + 1))

The set of points k so k/n is within δ of x.

Equations
theorem bernstein_approximation.lt_of_mem_S {f : C(unit_interval, )} {ε : } {h : 0 < ε} {n : } {x : unit_interval} {k : fin (n + 1)} (m : k bernstein_approximation.S f ε h n x) :
|f (bernstein.z k) - f x| < ε / 2

If k ∈ S, then f(k/n) is close to f x.

theorem bernstein_approximation.le_of_mem_S_compl {f : C(unit_interval, )} {ε : } {h : 0 < ε} {n : } {x : unit_interval} {k : fin (n + 1)} (m : k (bernstein_approximation.S f ε h n x)) :

If k ∉ S, then as δ ≤ |x - k/n|, we have the inequality 1 ≤ δ^-2 * (x - k/n)^2. This particular formulation will be helpful later.

The Bernstein approximations

 k : fin (n+1), f (k/n : ) * n.choose k * x^k * (1-x)^(n-k)

for a continuous function f : C([0,1], ℝ) converge uniformly to f as n tends to infinity.

This is the proof given in [Richard Beals' Analysis, an introduction][beals-analysis], §7D, and reproduced on wikipedia.