This is a problem from Mathematical Analysis by Zorich, from the chapter on continuous functions.
Definition:
Let $P_n$ be a polynomial of degree $n$. For a function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.
I have already proved the following:
There exist a polynomial $P_0(x) = a$ of best approximation of degree 0.
If $Q_\lambda(x) = \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$
I'm stuck on proving the following:
If there exists a polynomial of best approximation of degree $n$, there also exists a polynomial of best approximation of degree $n+1$.
My intuition is to prove $\lambda_0 x^{n+1}+P_n$ is the $n+1$ best approximation for $f$. Where $\Delta(\lambda_0 x^{n+1}) = \min_{\lambda\in \mathbb{R}} \Delta(\lambda x^{n+1})$ and $\Delta(P_n) = E(f(x) – \lambda_0 x^{n+1})$.
I don't know if this approach is right. I'm stuck on how to proceed.
Best Answer
Determining the best $n$th degree approximation explicitely is a very difficult problem and cannot be done by a simple induction (see the chapter on Tschebyscheff approximation in your favorite textbook of numerical analysis). But existence is easy, it follows from general principles: You have to set up a certain continuous function $\Phi$ on a certain compact set $K$ in a high-dimensional space.