Does any continuous function on $[0,1]$ have a best $n$th degree polynomial approximation in the supremum norm?

Question

Recently I am stuck in a problem in approximation theory which actually is problem in functional analysis.

$C[0,1]$ is a normed vector space with $||\cdot ||_{\infty}$. $\Pi_n$ is a subspace which contains all the polynomials whose degree is no more than $n$. It is easy to conclude that $C[0,1]$ is a banach space and $\Pi_n$is a finite dimensional space.

$\forall f\in C[0,1]$, does there exist a unique $p\in\Pi_n$ such that$||f-p||_{\infty}=\inf_{g\in\Pi_n}||f-g||_{\infty}$?

As far as I know, a element in normed vector space have a best approximation in a finite dimensional subspace. Moreover, if the subspace is strictly convex, the best approximation is unique.

Therefore, $\forall f\in C[0,1]$, there exist $p\in\Pi_n$ such that$||f-p||_{\infty}=\inf_{g\in\Pi_n}||f-g||_{\infty}$. However, we cannnot guarantee that $p$ is unique because $\Pi_n$ is not strictly convex.(for example,$1,t\in \Pi_n,||\frac{1}{2} 1+\frac{1}{2} t||_{\infty}=1$, not less than $1$.)

All I want to know is that whether or not the best approximation element is unique. If it is indeed unique, could we conclude that by functional analysis?

Thanks for reading! Detailed comments or proofs are appreciated!

Answer 1

There is a unique best approximation in $\Pi_n$. This is a classical result in approximation theory.

Let $f\in C([0,1])$ with $\mathrm{dist}(f,\Pi_n)=d$. The key observation is that for every best approximation $p\in\Pi_n$ there exist $n+2$ points where $f(x)-p(x)=\pm d$ with alternating signs. This result is known as the Chebyshev Equioscillation Theorem. The proof is a bit lengthy, see Theorem 1.2 in Theory of uniform approximation of functions by polynomials by Dzjadyk and Ševčuk.

Answer 2

answer to the original version of the question
We get unique best approximation in subspaces only when the norm is strictly convex. Unfortunately, the norm $\|\cdot\|_\infty$ in your problem is not strictly convex. So unique approximation fails for some finite-dimensional subspaces of $C[0,1]$. Whether it holds for your special subspaces $\Pi_n$ will require additional argument.