Multivariable Advanced Calculus

104 CONTINUOUS FUNCTIONSProof: Let g k : [0, 1] → [a k , b k ] be linear, one to one, and onto and letx = g (y) ≡ (g 1 (y 1 ) , g 2 (y 2 ) , · · · , g n (y n )) .Thus g : [0, 1] n → ∏ nk=1 [a k, b k ] is one to one, onto, and each component function islinear. Then f ◦ g is a continuous function defined on [0, 1] n . It follows from Lemma5.7.3 there exists a sequence of polynomials, {p m (y)} each defined on [0, 1] n whichconverges uniformly to f ◦ g on [0, 1] n . Therefore, { p m(g −1 (x) )} converges uniformlyto f (x) on R. Buty = (y 1 , · · · , y n ) = ( g −11 (x 1 ) , · · · , g −1n (x n ) )and each g −1kis linear. Therefore, { (p m g −1 (x) )} is a sequence of polynomials. As tothe partial derivatives, it was shown above thatNow the chain rule implies thatlim ∥Dp m − D (f ◦ g)∥ [0,1] n = 0min(m)→∞D ( p m ◦ g −1) (x) = Dp m(g −1 (x) ) Dg −1 (x)Therefore, the following convergences are uniform in x ∈ R.lim D ( p m ◦ g −1) (x)min(m)→∞= lim Dp (m g −1 (x) ) Dg −1 (x)min(m)→∞= limmin(m)→∞ D (f ◦ g) ( g −1 (x) ) Dg −1 (x)= limmin(m)→∞ Df ( g ( g −1 (x) )) Dg ( g −1 (x) ) Dg −1 (x)= Df (x)The claim about higher order derivatives is more technical but follows in the same way.There is a more general version of this theorem which is easy to get. It depends onthe Tietze extension theorem, a wonderful little result which is interesting for its ownsake.5.7.1 The Tietze Extension TheoremTo generalize the Weierstrass approximation theorem I will give a special case of theTietze extension theorem, a very useful result in topology. When this is done, it willbe possible to prove the Weierstrass approximation theorem for functions defined on aclosed and bounded subset of R n rather than a box.Lemma 5.7.7 Let S ⊆ R n be a nonempty subset. Definedist (x, S) ≡ inf {|x − y| : y ∈ S} .Then x → dist (x, S) is a continuous function satisfying the inequality,|dist (x, S) − dist (y, S)| ≤ |x − y| . (5.12)