B. Multivariate Bernstein Polynomials

Here we give the proofs of the theorems in Section 7.4.

Theorem B..1   Let $ f: I:=[0,1]\times[0,1]\to\mathbb{R}$ be a continuous function. Then the two-dimensional Bernstein polynomials

$\displaystyle B_{f,n_1,n_2}(x_1,x_2):=\sum_{k_1=0}^{n_1}\sum_{k_2=0}^{n_2}
f \l...
... k_1} {n_2\choose k_2}
x_1^{k_1} (1-x_1)^{n_1-k_1} x_2^{k_2} (1-x_2)^{n_2-k_2}

converge pointwise to $ f$ for $ n_1,n_2\to\infty$.

Proof. Let $ (x_1,x_2)\in I$ be a fixed point. Because of Theorem 7.2 we have
$\displaystyle {\vert B_{f,n_1,n_2}(x_1,x_2)-f(x_1,x_2)\vert}$
  $\displaystyle \le$ $\displaystyle \vert B_{f,n_1,n_2}(x_1,x_2)-B_{f(.,x_2),n_1}(x_1)\vert + \vert B_{f(.,x_2),n_1}(x_1)-f(x_1,x_2)\vert$  
  $\displaystyle \le$ $\displaystyle 2\epsilon$  

for all $ n_1\ge N_1(\epsilon)$ and $ n_2\ge N_2(\epsilon)$. The second summand is smaller than $ \epsilon$ for $ n_1\ge N_1(\epsilon)$ because

$\displaystyle B_{f(.,x_2),n_1}(x_1) = \sum_{k_1=0}^{n_1}
f \left( {k_1\over n_1}, x_2 \right) {n_1\choose k_1} x_1^{k_1} (1-x_1)^{n_1-k_1}

is the Bernstein polynomial for $ f(.,y)$, and the first summand is smaller than $ \epsilon$ for $ n_2\ge N_2(\epsilon)$ because $ B_{f,n_1,n_2}(x,y)$ is the (one-dimensional) Bernstein polynomial for $ B_{f(.,y),n_1}(x)$. Q.E.D.

Definition B..2 (Multivariate Bernstein Polynomials)   Let $ n_1,\ldots,n_m\in\mathbb{N}$ and $ f$ be a function of $ m$ variables. The polynomials

$\displaystyle B_{f,n_1,\ldots,n_m}(x_1,\ldots,x_m):=
\sum_{{0\le k_j\le n_j}\at...
\prod_{j=1}^m \left( {n_j\choose k_j} x_j^{k_j} (1-x_j)^{n_j-k_j} \right)

are called the multivariate Bernstein polynomials of $ f$.

Theorem B..3 (Pointwise Convergence)   Let $ f: [0,1]^m\to\mathbb{R}$ be a continuous function. Then the multivariate Bernstein polynomials $ B_{f,n_1,\ldots,n_m}$ converge pointwise to $ f$ for $ n_1,\ldots,n_m\to\infty$.

Proof. By applying Theorem 7.2 to each summand in
$\displaystyle {\vert B_{f,n_1,\ldots,n_m}(x_1,\ldots,x_m)-f(x_1,\ldots,x_m)\vert}$
  $\displaystyle \le$ $\displaystyle \vert B_{f,n_1,\ldots,n_m}(x_1,\ldots,x_m) - B_{f(.,\ldots,.,x_m),n_1,\ldots,n_{m-1}}(x_1,\ldots,x_{m-1})\vert$  
    $\displaystyle {} + \vert B_{f(.,\ldots,.,x_m),n_1,\ldots,n_{m-1}}(x_1,\ldots,x_{m-1})
- B_{f(.,\ldots,.,x_{m-1},x_m),n_1,\ldots,n_{m-2}}(x_1,\ldots,x_{m-2})\vert$  
    $\displaystyle {} + \cdots + \vert B_{f(.,x_2,\ldots,x_m),n_1}(x_1) - f(x_1,\ldots,x_m)\vert$  

we see that given an $ \epsilon>0$ there are $ N_1(\epsilon)$,..., $ N_m(\epsilon)$ such that

$\displaystyle \vert B_{f,n_1,\ldots,n_m}(x_1,\ldots,x_m)-f(x_1,\ldots,x_m)\vert \le m\epsilon

for all $ n_i\ge\max(N_1(\epsilon),\ldots,N_m(\epsilon))$. Q.E.D.

Lemma B..4   For all $ x\in\mathbb{R}$

$\displaystyle \sum_{k=0}^{n} (k-nx)^2 {n\choose k} x^k (1-x)^{n-k} = nx(1-x).

For all $ x\in[0,1]$ we have $ x(1-x)\le 1/4$ and hence

$\displaystyle 0 \le \sum_{k=0}^{n} (k-nx)^2 {n\choose k} x^k (1-x)^{n-k} \le {n\over 4}.

Theorem B..5 (Uniform Convergence)   Let $ f: [0,1]^m\to\mathbb{R}$ be a continuous function. Then the multivariate Bernstein polynomials $ B_{f,n_1,\ldots,n_m}$ converge uniformly to $ f$ for $ n_1,\ldots,n_m\to\infty$.

Proof. We first note that because of the uniform continuity of $ f$ on  $ I:=[0,1]^m$ we have

$\displaystyle \forall \epsilon>0:\quad
\exists \delta>0:\quad
\forall x,x'\in I...
... x-x'\Vert<\delta \;\Longrightarrow\;\Vert f(x)-f(x')\Vert < {\epsilon\over2}.

Given an $ \epsilon>0$, we can find such a $ \delta$. In order to simplify notation we set

$\displaystyle b_j:={n_j\choose k_j} x_j^{k_j} (1-x_j)^{n_j-k_j}

and $ k:=\left( {k_1\over n_1}, \ldots, {k_m\over n_m} \right)$. $ x$ always lies in $ I$. We have to estimate

$\displaystyle B_{f,n_1,\ldots,n_m}(x) - f(x) =
\sum_{{0\le k_j\le n_j}\atop{j\in\{1,\ldots,m\}}}
(f(k) - f(x)) b_1\cdots b_m

and to that end we split the sum into two parts, namely

$\displaystyle S_1:= \sum\nolimits' (f(k) - f(x)) b_1\cdots b_m,

where $ \sum\nolimits'$ means summation over all $ k_j$ with $ {0\le k_j\le
n_j}$ (where $ j\in\{1,\ldots,m\}$) and $ \Vert k - x \Vert _2 \ge \delta$, and

$\displaystyle S_2:= \sum\nolimits'' (f(k) - f(x)) b_1\cdots b_m,

where $ \sum\nolimits''$ means summation over the remaining terms. For $ S_2$ we have

$\displaystyle \vert S_2\vert \le \sum\nolimits'' \vert f(k) - f(x)\vert b_1\cdo...
...{{0\le k_j\le n_j}\atop{j\in\{1,\ldots,m\}}} b_1\cdots b_m
= {\epsilon\over2}.

We will now estimate $ S_1$. In the sum $ S_1$ the inequality $ \Vert k - x \Vert _2 \ge \delta$ holds, i.e.,

$\displaystyle \left( {k_1\over n_1} - x_1 \right)^2 + \cdots +
\left( {k_m\over n_m} - x_m \right)^2
\ge \delta^2.

Hence at least one of the summands on the left hand side is greater equal $ \delta^2/m$. Without loss of generality we can assume this is the case for the first summand:

$\displaystyle 1 \le {m\over\delta^2}{(k_1-n_1 x)^2\over n_1^2}.

Thus, using Lemma B.4,
$\displaystyle \sum\nolimits' b_1\cdots b_m$ $\displaystyle \le$ $\displaystyle \sum\nolimits' {m\over\delta^2 n_1^2} (k_1-n_1 x)^2 b_1\cdots b_m$  
  $\displaystyle \le$ $\displaystyle {m\over\delta^2 n_1^2} \sum_{{0\le k_j\le n_j}\atop{j\in\{1,\ldots,m\}}} (k_1-n_1 x)^2 b_1\cdots b_m$  
  $\displaystyle =$ $\displaystyle {m\over\delta^2 n_1^2} \sum_{k_1=0}^{n_1} (k_1-n_1 x)^2 {n_1\choose k_1} x_1^{k_1} (1-x_1)^{n_1-k_1}$  
  $\displaystyle \le$ $\displaystyle {m\over\delta^2 n_1^2} {n\over4} = {m \over 4 \delta^2 n_1}.$  

We can now estimate $ S_1$. Since $ f$ is continuous on a compact set $ M:=\max_{x\in I} \vert f(x)\vert$ exists.
$\displaystyle \vert S_1\vert$ $\displaystyle \le$ $\displaystyle \sum\nolimits' \vert f(k) - f(x)\vert b_1\cdots b_m$  
  $\displaystyle \le$ $\displaystyle 2M \sum\nolimits' b_1\cdots b_m$  
  $\displaystyle \le$ $\displaystyle {2Mm\over4\delta^2 n_1} = {Mm\over2\delta^2 n_1}$  

For $ n_1$ large enough we have $ Mm / 2\delta^2n_1 < \epsilon/2$ and thus

$\displaystyle \vert B_{f,n_1,\ldots,n_m}(x) - f(x)\vert \le \vert S_1\vert+\vert S_2\vert <
{\epsilon\over2} + {\epsilon\over2} = \epsilon,

which completes the proof. Q.E.D.

A reformulation of this fact is the following corollary.

Corollary B..6   The set of all polynomials is dense in $ C([0,1]^m)$.

Theorem B..7 (Error Bound for Lipschitz Condition)   If $ f: I:=[0,1]^m\to\mathbb{R}$ is a continuous function satisfying the Lipschitz condition

$\displaystyle \Vert f(x)-f(y)\Vert _2 < L \Vert x-y\Vert _2

on $ I$, then the inequality

$\displaystyle \Vert B_{f,n_1,\ldots,n_m}(x) - f(x) \Vert _2 <
{L\over2} \biggl( \sum_{j=1}^m {1\over n_j} \biggr)^{1\over2}


Proof. Abbreviating notation we set $ k:=\left( {k_1\over n_1}, \ldots, {k_m\over n_m} \right)$. We will use the Lipschitz condition, Corollary A.7, and Lemma B.4.
$\displaystyle {\Vert B_{f,n_1,\ldots,n_m}(x) - f(x) \Vert _2^2}$
  $\displaystyle \le$ $\displaystyle \biggl( \sum_{k_1,\ldots,k_m} \Vert f(k) - f(x) \Vert _2 b_1\cdots b_m \biggr)^2$  
  $\displaystyle <$ $\displaystyle \left( L \sum \Vert k-x \Vert _2 b_1\cdots b_m \right)^2$  
  $\displaystyle \le$ $\displaystyle L^2 \left( \sum \Vert k-x \Vert _2^2 b_1\cdots b_m \right) \left( \sum b_1\cdots b_m \right)$  
  $\displaystyle =$ $\displaystyle L^2 \sum \left( \left({k_1\over n_1}-x_1\right)^2 + \cdots + \left({k_m\over n_m}-x_m\right)^2 \right) b_1\cdots b_m$  
  $\displaystyle =$ $\displaystyle L^2 \sum_{j=1}^m {x_j(1-x_j)\over n_j}$  
  $\displaystyle \le$ $\displaystyle L^2 \sum_{j=1}^m {1\over 4 n_j}$  

This completes the proof. Q.E.D.

Theorem B..8 (Asymptotic Formula)   Let $ f: I:=[0,1]^m\to\mathbb{R}$ be a $ C^2$ function and $ x\in I$, then

$\displaystyle \lim_{n\to\infty} n ( B_{f,n,\ldots,n}(x) - f(x) )
= \sum_{j=1}^m...
...rtial x_j^2}
\le {1\over8} \sum_{j=1}^m {\partial ^2 f(x)\over\partial x_j^2}.

Proof. We define the vector $ h$ through $ h_j:=k_j/n-x_j$, where the $ k_j$ are the integers over which we sum in $ B_{f,n,\ldots,n}$. Using Theorem A.14 we see
$\displaystyle f\left({k_1\over n},\ldots,{k_m\over n}\right)$ $\displaystyle =$ $\displaystyle f(x) + \sum_{j=1}^m \left( {k_j\over n}-x_j \right) {\partial f(x)\over\partial x_j}$  
    $\displaystyle {} + {1\over2} \sum_{i=1}^m\sum_{j=1}^m
\left( {k_i\over n}-x_i \...
...) {\partial ^2 f(x_0)\over \partial x_i \partial x_j}
+ \Vert h\Vert^2 \rho(h),$  

where $ \lim_{h\to0} \rho(h)=0$. Summing this equation like the sum in $ B_{f,n,\ldots,n}$ we obtain

$\displaystyle B_{f,n,\ldots,n}
= f(x) + {1\over2} \sum_{i=1}^m {x_j(1-x_j)\over...
...r \partial x_j^2}
+ \sum_{k_1,\ldots,k_m} \Vert h\Vert^2 \rho(h) b_1\cdots b_m

since many terms vanish or can be summed because of Lemma B.4. Noting $ \lim_{h\to0} \rho(h)=0$ we can apply the same technique as in the proof of Theorem B.5 for estimating the last sum in the last equation, i.e., splitting the sum into two parts for $ \Vert h\Vert\ge\delta$ and $ \Vert h\Vert<\delta$. Hence we see that for all $ \epsilon$ this sum is less equal $ \epsilon/n$ for all sufficiently large $ n$, which yields the claim. Q.E.D.

Clemens Heitzinger 2003-05-08