Ex.7.23 Let P0=0,Pn+1(x)=Pn(x)+2x2−Pn2(x). Prove that Pn⇉[−1,1]∣x∣. (Stone-Weirestrass Th. can be proven without 1st prove Th.7.26) Hint: Use the identity ∣x∣−Pn+1=(∣x∣−Pn(x))(1−2∣x∣+Pn(x)) to prove that 0≤Pn(x)≤Pn+1(x)≤∣x∣(∣x∣<1) hence ∣x∣−Pn(x)≤∣x∣(1−2∣x∣)n<n+12 Proof From the recursive definition, we have ∣x∣−Pn+1(x)=∣x∣−Pn(x)−2x2−Pn2(x)=(∣x∣−Pn(x))(1−2∣x∣+Pn(x)).(⋆) For ∣x∣<1, since P0=0, by induction (⋆) implies 0≤Pn(x)≤Pn+1(x)≤∣x∣(∀n) and so (⋆)⟹0≤∣x∣−Pn+1(x)≤(∣x∣−Pn(x))(1−2∣x∣)≤⋯≤(∣x∣−Pn−k)(1−2∣x∣)k+1. Therefore 0≤∣x∣−Pn(x)≤∣x∣(1−2∣x∣)n≤n+12(∀n∈N∀x∈(−1,1)).□ The last inequality comes from φ(t)=t(1−2t)n,φ′(t)=(1−2t)n−1(1−2(n+1)t),maxφ([0,1])=φ(n+12)<n+12