k m −k ( k1 kn )
⋅⋅⋅xnn(1− xn) n n f m-,⋅⋅⋅,m-- . (2.2)
1 n
(2.2)
Also define if I is a set in ℝ^{n}
||h||I ≡ sup{|h(x)| : x ∈ I}.
Let
min (m ) ≡ min{m1, ⋅⋅⋅,mn }, max (m ) ≡ max {m1,⋅⋅⋅,mn}
Definition B.1.1Define p_{m}converges uniformly to f on a set, I if
minl(imm)→∞ ||pm − f||I = 0.
To simplify the notation, let k =
(k1,⋅⋅⋅,kn)
where each k_{i}∈
[0,mi]
,
k- ( k1- -kn)
m ≡ m1 ,⋅⋅⋅,mn ,
and let
(m ) (m )(m ) (m )
≡ 1 2 ⋅⋅⋅ n .
k k1 k2 kn
Also define for k =
(k1,⋅⋅⋅,kn)
k ≤ m if 0 ≤ k ≤ m for each i
i i
k m− k k1 m1−k1 k2 m2−k2 kn mn−kn
x (1− x) ≡ x1 (1− x1) x2 (1− x2) ⋅⋅⋅xn (1− xn) .
Thus in terms of this notation,
∑ (m ) ( k)
pm (x) = xk(1− x)m −kf --
k≤m k m
This is the n dimensional version of the Bernstein polynomials which is what results in the case wheren = 1.
Lemma B.1.2For x ∈
[0,1]
^{n},f a continuous F valued function defined on
[0,1]
^{n}, and p_{m}givenin 2.2, p_{m}converges uniformly to f on
[0,1]
^{n}as m → ∞. More generally, one can have f acontinuous function with values in an arbitrary real or complex normed linear space. There is nochange in the conclusions and proof. You just write
∥⋅∥
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 ) k m− k
k x (1− x) = 1
k≤m
and so for x ∈
[0,1]
^{n},
∑ (m ) || ( k ) ||
|pm (x)− f (x)| ≤ xk(1− x)m− k||f -- − f (x)||
k≤m 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)| (2.3)
k∈GC
(2.3)
Now for k ∈ G it follows that for each i
| |
||ki-− x ||< √δ-- (2.4)
|mi i| n
(2.4)
and so
| (k) |
|f m- − f (x)|
< ε because the above implies
|k |
|m-− x|
< δ. Therefore, the first sum on the right in
2.3 is no larger than
( ) ( )
∑ m xk (1 − x)m−k ε ≤ ∑ m xk(1− x)m −kε = ε.
k∈G k k≤m k