Haar's uniqueness theorem

Remarks

The idea of the proof: best approximations always exist, since $V$ is finite-dimensional. If $V$ is Haar and $v_1, v_2$ are both best, then so is $(v_1 + v_2)/2$. Its error attains the norm only at points where $v_1 - v_2$ vanishes. For a Haar subspace the error of a best approximation attains its norm at $n + 1$ or more points, too many zeros for $v_1 - v_2 \ne 0$. Conversely, if a nonzero $v \in V$ vanishes at $n$ points, one builds a continuous $f$ that has $0$ and all small multiples of a suitable element of $V$ as best approximations.

Example: the polynomials of degree at most $n - 1$ on $[a, b]$ form a Haar subspace, so best uniform approximation of a continuous function by polynomials of a fixed degree is unique.

Non-example: on $[-1, 1]$, the span of $1$ and $x^2$ is not Haar, since $x^2 - 1$ has two zeros. Correspondingly $f(x) = x$ has many best approximations from it. For every $g = a + c x^2$, the error $x - g$ takes the values $1 - a - c$ and $-1 - a - c$ at $\pm 1$, which differ by $2$, so $\norm{x - g}_\infty \ge 1$. For $\abs{c} \le 1/2$, the error $h(x) = x - c(x^2 - 1)$ has $h' = 1 - 2cx \ge 0$ on $[-1, 1]$ and $h(\pm 1) = \pm 1$, so $\norm{h}_\infty = 1$. Hence every $c(x^2 - 1)$ with $\abs{c} \le 1/2$ is a best approximation of $x$.

Going deeper

For a student

The theorem explains the special role of polynomials and of Chebyshev systems in uniform approximation. For Haar subspaces the best approximation is characterized by alternation of the error at $n + 1$ points, which generalizes the equioscillation theorem. The Mairhuber–Curtis theorem shows that Haar subspaces of dimension at least $2$ do not exist on open sets in several variables. So best uniform approximation by multivariate polynomials of a fixed degree is not always unique.