This is the n dimensional version of the Bernstein polynomials which is what results in thecase where n = 1.
Lemma 3.11.2For x ∈
[0,1]
^{n},f a continuous F valued function defined on
[0,1]
^{n},and p_{m}given in 3.24, p_{m}converges uniformly to f on
[0,1]
^{n}as m → ∞. Moregenerally, one can have f a continuous function with values in an arbitrary real orcomplex normed linear space. There is no change in the conclusions and proof. You justwrite
∥⋅∥
instead of
|⋅|
.
Proof:The function f is uniformly continuous because it is continuous on a
sequentially compact set
[0,1]
^{n}. Therefore, there exists δ > 0 such that if
|x − y|
< δ,
then
|f (x)− f (y)| < ε.
Denote by G the set of k such that
(ki − mixi)
^{2}< η^{2}m^{2} for each i where η = δ∕
√--
n
. Note
this condition is equivalent to saying that for each i,
| |
||ki− xi||
mi
< η and
| |
||k-− x||< δ
|m |
A short computation shows that by the binomial theorem,
∑ (m )
xk (1− x)m−k = 1
k≤m k
and so for x ∈
[0,1]
^{n},
∑ (m ) k m− k|| ( k ) ||
|pm (x)− f (x)| ≤ k x (1− x) ||f m- − f (x)||
k≤m
∑ ( ) || ( ) ||
≤ m xk (1− x)m− k||f k- − f (x)||
k∈G k m
( ) | ( ) |
∑ m k m −k|| k- ||
+ k x (1 − x) |f m − f (x)| (3.25)
k∈GC
(3.25)
Now for k ∈ G it follows that for each i
||ki || δ
||m--− xi|| < √n- (3.26)
i
(3.26)
and so
| |
|f (km) − f (x)|
< ε because the above implies
| |
|km-− x|
< δ. Therefore, the first sum
on the right in 3.25 is no larger than
∑ ( ) ∑ ( )
m xk (1 − x)m−k ε ≤ m xk (1− x)m− kε = ε.
k∈G k k≤m k
m∑− 1-m-(m-−-1)!- k m −1−k( ( k+-1) ( k-))
= (m − 1− k)!k!x (1 − x ) f m − f m
k=0
m∑−1( ) ( (k+1) (k-))
= m − 1 xk(1− x)m −1− k f--m--−-f--m--
k=0 k 1∕m
By the mean value theorem,
( ) ( ) ( )
f-k+m1--− f-km--= f′(x ), x ∈ k-, k-+1
1∕m k,m k,m m m
Now the desired result follows as before from the uniform continuity of f^{′} on
[0,1]
. Let δ > 0
be such that if
|x − y| < δ, then |f′(x)− f′(y)| < ε
and let m be so large that 1∕m < δ∕2. Then if
|| k||
x − m
< δ∕2, it follows that
|x− xk,m |
< δ
and so
|| f (k+1)− f (k)||
|f ′(x)− f′(xk,m )| = ||f′(x)−----m-------m-|| < ε.
| 1∕m |
Now as before, letting M ≥
|f′(x)|
for all x,
′ ′ m∑−1( m − 1 ) k m−1−k ′ ′
|pm(x)− f (x)| ≤ k x (1− x) |f (xk,m) − f (x)|
k=0
∑ ( m − 1 ) m−1−k
≤ k xk (1− x) ε
{x:|x− km|<δ2}
m∑−1( ) 2
+M m − 1 4(k−-mx-)-xk(1− x)m−1−k
k=0 k m2 δ2
≤ ε+ 4M 1m --1--= ε + M -1--< 2ε
4 m2 δ2 m δ2
whenever m is large enough. Thus this proves uniform convergence. ■
Now consider the case where n ≥ 1. Applying the same manipulations to the sum which
corresponds to the i^{th} variable,
m1 mi−1 mn ( ) ( ) ( )
p (x ) ≡ ∑ ⋅⋅⋅ ∑ ⋅⋅⋅∑ m1 ⋅⋅⋅ mi − 1 ⋅⋅⋅ mn xk1(1− x )m1−k1 ⋅⋅⋅
mxi k=0 k =0 k =0 k1 ki kn 1 1
1 i n
ki mi−1−ki kn mn−kn
xi( (1 − xi) ⋅⋅⋅x)n (1(− xn) ⋅ )
f k1,⋅⋅⋅ ki+1,⋅⋅⋅ kn − f -k1,⋅⋅⋅-ki,⋅⋅⋅-kn-
---m1----mi-----mn------m1----mi----mn---
1∕mi
By the mean value theorem, the difference quotient is of the form
fxi (xk,m ),
the i^{th} component of x_{k,m} being between
kmi
i
and
kim+1
i
. Therefore, a repeat of the above
argument involving splitting the sum into two pieces, one for which k∕m is close to x, hence
close to x_{k,m} and one for which some k_{j}∕m_{j} is not close to x_{j} for some j yields the same
conclusion about uniform convergence on
[0,1]
^{n}. This has essentially proved the following
lemma.
Lemma 3.11.4Let f be in C^{k}
([0,1]n)
. Then there exists a sequence of polynomialsp_{m}
(x)
such that each partial derivative up to order k converges uniformly to thecorresponding partial derivative of f.