Multivariable Advanced Calculus

106 CONTINUOUS FUNCTIONSand so ( )3 m2(f − ∑ m( 2) ) i−1i=1 3 gi can play the role of f in the first step of the proof.Therefore, there exists g m+1 defined and continuous on all of R n such that its valuesare in [−1/3, 1/3] and∣ ) ( m 3m∑( ) ∣∣ i−1 2 ∣∣∣∣ ∣∣∣∣M ∣ f −g∣∣( i − g m+1 ≤23) 2 3 .Hence ∣ ∣∣∣∣∣ ∣∣∣∣(f −i=1m∑( ) i−1 2g i3)i=1( ) ∣∣ m 2 ∣∣∣∣ ∣∣∣∣M ( ) m+1 2− g m+1 ≤ .33It follows there exists a sequence, {g i } such that each has its values in [−1/3, 1/3] andfor every m 5.13 holds. Then let∞∑( ) i−1 2g (x) ≡g i (x) .3It followsi=1i=1∞∑( ) i−1 2 |g (x)| ≤g∣i (x)∣m 3 ∣ ≤ ∑( ) i−1 2 13 3 ≤ 1and ∣ ∣∣∣∣ ( ) i−1 2 ( ) i−1 2g i (x)∣3 ∣ ≤ 13 3so the Weierstrass M test applies and shows convergence is uniform. Therefore g mustbe continuous. The estimate 5.13 implies f = g on M. The following is the Tietze extension theorem.Theorem 5.7.12 Let M be a closed nonempty subset of R n and let f : M →[a, b] be continuous at every point of M. Then there exists a function, g continuous onall of R n which coincides with f on M such that g (R n ) ⊆ [a, b] .Proof: Let f 1 (x) = 1 + 2b−a (f (x) − b) . Then f 1 satisfies the conditions of Lemma5.7.11 and so there exists g 1 : R n → [−1, 1] such that g is continuous on R n and equalsf 1 on M. Let g (x) = (g 1 (x) − 1) ( )b−a + b. This works. 2With the Tietze extension theorem, here is a better version of the Weierstrass approximationtheorem.Theorem 5.7.13 Let K be a closed and bounded subset of R n and let f : K → Rbe continuous. Then there exists a sequence of polynomials {p m } such thati=1lim (sup {|f (x) − p m (x)| : x ∈ K}) = 0.m→∞In other words, the sequence of polynomials converges uniformly to f on K.Proof: By the Tietze extension theorem, there exists an extension of f to a continuousfunction g defined on all R n such that g = f on K. Now since K is bounded,there exist intervals, [a k , b k ] such thatn∏K ⊆ [a k , b k ] = Rk=1Then by the Weierstrass approximation theorem, Theorem 5.7.6 there exists a sequenceof polynomials {p m } converging uniformly to g on R. Therefore, this sequence of polynomialsconverges uniformly to g = f on K as well. This proves the theorem. By considering the real and imaginary parts of a function which has values in C onecan generalize the above theorem.