This is the question:
(a) Given a function $f(x)=e^{-(a+1)x}$. Find the polynomial of degree $2$ , $p_{2}(x)$ such that (with $w(x)=1$):
$$||f-p_{2}||_{L^{2}(0,1)}=\min_{q \in P_{2}(0,1)}||f-q||_{L^{2}(0,1)}.$$
(b) Find the polynomial of degree $3$, $q_{3}(x)$ in $[-1,1]$
$$||q_{3}(x)||_{L^{\infty}(-1,1)}\leq ||x^3 -(\alpha + \beta)x -2||_{L^{\infty}(-1,1)}$$
I solved (a). I want some help solving (b). Someone from the community told me it has to do with the Chebyshev polynomials and is probably right about that. But i can't find a formal way to do (b).
Proof:
$$r(x)=x^{n+1}-q_{n}(x),~~~q_{n}(x) \in P_{n}[a,b]$$
We have that
$$P_{n+1}^1=\{r \in P_{n+1}[a,b]/r(x)=x^{n+1}-q_{n}(x),~q_{n}(x) \in P_{n}[a,b]\}$$
$$\min_{r \in P_{n+1}^1}||r||_{\infty}=\min_{q_{n} \in P_{n}[a,b]}||x^{n+1}-q_{n}(x)||_{\infty}=||x^{n+1}-(x^{n+1}-2^{-n}T_{n+1})||_{\infty}=\frac{1}{2^{n}}||T_{n+1}||_{\infty}$$
So here i proved that from all the monic polynomials of degree $n+1$ the $ 2^{-n}T_{n+1}(x)$ has the smallest $L^{\infty}$-norm on [-1,1].