Let f:[0,1]→R be continuous and let n be a positive integer. For g:[0,1]→R a continuous function, write ∥f−g∥L∞=supx∈[0,1]∣f(x)−g(x)∣.
(i) Let p be a polynomial of degree at most n with the property that there are (n+2) distinct points x1,x2,…,xn+2∈[0,1] with x1<x2<…<xn+2 such that
f(xj)−p(xj)=(−1)j∥f−p∥L∞
for each j=1,2,…,n+2. Prove that
∥f−p∥L∞⩽∥f−q∥L∞
for every polynomial q of degree at most n.
(ii) Prove that there exists a polynomial p of degree at most n such that
∥f−p∥L∞⩽∥f−q∥L∞
for every polynomial q of degree at most n.
[If you deduce this from a more general result about abstract normed spaces, you must prove that result.]
(iii) Let Y={y1,y2,…,yn+2} be any set of (n+2) distinct points in [0,1].
(a) For j=1,2,…,n+2, let
rj(x)=k=1,k=j∏n+2yj−ykx−yk
t(x)=∑j=1n+2f(yj)rj(x) and r(x)=∑j=1n+2(−1)jrj(x). Explain why there is a unique number λ∈R such that the degree of the polynomial t−λr is at mostn.
(b) Let ∥f−g∥L∞(Y)=supx∈Y∣f(x)−g(x)∣. Deduce from part (a) that there exists a polynomial p of degree at most n such that
∥f−p∥L∞(Y)⩽∥f−q∥L∞(Y)
for every polynomial q of degree at most n.