Let $T_{n}$ be the $n$th Chebychev polynomial. Suppose that $\gamma_{i}>0$ for all $i$ and that $\sum_{i=1}^{\infty} \gamma_{i}$ converges. Explain why $f=\sum_{i=1}^{\infty} \gamma_{i} T_{3^{i}}$ is a well defined continuous function on $[-1,1]$.

Show that, if we take $P_{n}=\sum_{i=1}^{n} \gamma_{i} T_{3^{i}}$, we can find points $x_{k}$ with

$$-1 \leqslant x_{0}<x_{1}<\ldots<x_{3^{n+1}} \leqslant 1$$

such that $f\left(x_{k}\right)-P_{n}\left(x_{k}\right)=(-1)^{k+1} \sum_{i=n+1}^{\infty} \gamma_{i}$ for each $k=0,1, \ldots, 3^{n+1}$.

Suppose that $\delta_{n}$ is a decreasing sequence of positive numbers and that $\delta_{n} \rightarrow 0$ as $n \rightarrow \infty$. Stating clearly any theorem that you use, show that there exists a continuous function $f$ with

$$\sup _{t \in[-1,1]}|f(t)-P(t)| \geqslant \delta_{n}$$

for all polynomials $P$ of degree at most $n$ and all $n \geqslant 1$.