Multivariable Advanced Calculus

100 CONTINUOUS FUNCTIONSThus the expression is maximized when x = 1/2 and yields m/4 in this case. Thisproves the lemma. Now let f be a continuous function defined on [0, 1] . Let p n be the polynomialdefined byn∑( ( n kp n (x) ≡ f xk)n)k (1 − x) n−k . (5.7)k=0For f a continuous function defined on [0, 1] n and for x = (x 1 , · · · , x n ) ,consider thepolynomial,p m (x) ≡∑m 1k 1=0· · ·∑m nk n=0( )( ) ( )m1 m2 mn· · · x k11k 1 k 2 k (1 − x 1) m1−k1 x k22 (1 − x 2) m2−k2n· · · x knn(1 − x n ) m n−k nf(k1, · · · , k )n. (5.8)m 1 m nAlso define if I is a set in R n ||h|| I≡ sup {|h (x)| : x ∈ I} .Letmin (m) ≡ min {m 1 , · · · , m n } , max (m) ≡ max {m 1 , · · · , m n }Definition 5.7.2 Define p m converges uniformly to f on a set, I iflim ||p m − f|| I= 0.min(m)→∞To simplify the notation, let k = (k 1 , · · · , k n ) where each k i ∈ [0, m i ],(km ≡ k1, · · · , k )n,m 1 m nand let ( ) m≡kAlso define for k = (k 1 , · · · , k n )( )( ) ( )m1 m2 mn· · · .k 1 k 2 k nk ≤ m if 0 ≤ k i ≤ m i for each ix k (1 − x) m−k ≡ x k11 (1 − x 1) m1−k1 x k22 (1 − x 2) m2−k2 · · · x k nn (1 − x n ) mn−kn .Thus in terms of this notation,p m (x) = ∑ k≤m( )( mx k (1 − x) m−k kfk m)This is the n dimensional version of the Bernstein polynomials which is what results inthe case where n = 1.Lemma 5.7.3 For x ∈ [0, 1] n , f a continuous F valued function defined on [0, 1] n ,and p m given in 5.8, p m converges uniformly to f on [0, 1] n as m → ∞. More generally,one can have f a continuous function with values in an arbitrary real or complex normedlinear space. There is no change in the conclusions and proof. You just write ∥·∥ insteadof |·|.