Paper 3, Section II, F

Topics in Analysis
Part II, 2011

Let f:[0,1]Rf:[0,1] \rightarrow \mathbb{R} be continuous and let nn be a positive integer. For g:[0,1]Rg:[0,1] \rightarrow \mathbb{R} a continuous function, write fgL=supx[0,1]f(x)g(x)\|f-g\|_{L^{\infty}}=\sup _{x \in[0,1]}|f(x)-g(x)|.

(i) Let pp be a polynomial of degree at most nn with the property that there are (n+2)(n+2) distinct points x1,x2,,xn+2[0,1]x_{1}, x_{2}, \ldots, x_{n+2} \in[0,1] with x1<x2<<xn+2x_{1}<x_{2}<\ldots<x_{n+2} such that

f(xj)p(xj)=(1)jfpLf\left(x_{j}\right)-p\left(x_{j}\right)=(-1)^{j}\|f-p\|_{L^{\infty}}

for each j=1,2,,n+2j=1,2, \ldots, n+2. Prove that

fpLfqL\|f-p\|_{L^{\infty}} \leqslant\|f-q\|_{L^{\infty}}

for every polynomial qq of degree at most nn.

(ii) Prove that there exists a polynomial pp of degree at most nn such that

fpLfqL\|f-p\|_{L^{\infty}} \leqslant\|f-q\|_{L^{\infty}}

for every polynomial qq of degree at most nn.

[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}Y=\left\{y_{1}, y_{2}, \ldots, y_{n+2}\right\} be any set of (n+2)(n+2) distinct points in [0,1][0,1].

(a) For j=1,2,,n+2j=1,2, \ldots, n+2, let

rj(x)=k=1,kjn+2xykyjykr_{j}(x)=\prod_{k=1, k \neq j}^{n+2} \frac{x-y_{k}}{y_{j}-y_{k}}

t(x)=j=1n+2f(yj)rj(x)t(x)=\sum_{j=1}^{n+2} f\left(y_{j}\right) r_{j}(x) and r(x)=j=1n+2(1)jrj(x)r(x)=\sum_{j=1}^{n+2}(-1)^{j} r_{j}(x). Explain why there is a unique number λR\lambda \in \mathbb{R} such that the degree of the polynomial tλrt-\lambda r is at mostn\operatorname{most} n.

(b) Let fgL(Y)=supxYf(x)g(x)\|f-g\|_{L^{\infty}(Y)}=\sup _{x \in Y}|f(x)-g(x)|. Deduce from part (a) that there exists a polynomial pp of degree at most nn such that

fpL(Y)fqL(Y)\|f-p\|_{L^{\infty}(Y)} \leqslant\|f-q\|_{L^{\infty}(Y)}

for every polynomial qq of degree at most nn.