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!

  • 2
    $\begingroup$ The answer to your title is certainly "no": in $(\mathbb R^2, \| \,\cdot\,\|_\infty)$, the element $(0, 1)$ has no unique best approximation in $\mathrm{span}((1, 0))$. $\endgroup$ Oct 16, 2018 at 10:29
  • $\begingroup$ Yes, it has been studied extensively. If you take any textbook on "Approximation Theory", this (uniform approximation of a continuous function by polynomial of given degree) is likely to be the first example discussed in the book. $\endgroup$
    – GEdgar
    Oct 22, 2018 at 13:55
  • $\begingroup$ I have taken the liberty of editing the title to reflect the body of the post. $\endgroup$ Oct 22, 2018 at 13:56
  • $\begingroup$ @GEdgar, the problem is often discussed, but I have yet to find an explicit answer. E.g., these course notes -- simply chosen for coming up high on a Google search -- claim that a "not necessarily unique" best approximation exists, but provides neither an example where uniqueness is not possible, nor goes on later to prove uniqueness (as far as I can tell). $\endgroup$ Oct 22, 2018 at 14:03
  • $\begingroup$ Here dlmf.nist.gov/3.11 we find the claim that a minimax approximation exists and is unique. Here home.iitk.ac.in/~sghorai/TEACHING/MTH308/minimax.pdf we find proofs of existence and uniqueness. $\endgroup$
    – GEdgar
    Oct 22, 2018 at 14:27

2 Answers 2


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 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.


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