Müntz–Szász theorem

Remarks

The idea of the proof: one first works in $L^2[0, 1]$. There the distance from $x^m$ to the span of finitely many powers $x^{\lambda_k}$ can be computed exactly with Gram determinants, which here are Cauchy determinants. The distance is a product of factors $\abs{m - \lambda_k} / (m + \lambda_k + 1)$ times a constant, and this product tends to $0$ exactly when $\sum 1/\lambda_k$ diverges. A further argument passes from $L^2$ to uniform approximation of the monomials $x^m$, and then the Weierstrass theorem finishes the proof.

Example: the exponents $\lambda_n = n$ give the ordinary polynomials, and $\sum 1/n = \infty$, which recovers the Weierstrass approximation theorem on $[0, 1]$. The primes also work, since $\sum 1/p = \infty$, so combinations of $1$ and $x^p$ for primes $p$ are dense. The squares do not: $\sum 1/n^2 < \infty$, so the combinations of $1, x, x^4, x^9, \dots$ are not dense in $C[0, 1]$.

The constant function $1$ is needed in the span: every $x^{\lambda_n}$ vanishes at $0$, so without $1$ no function with $f(0) \ne 0$ could be approximated.

Going deeper

For a student

The theorem sharpens the Weierstrass theorem and shows how much freedom there is in choosing building blocks. It is a classic application of complex analysis and Hilbert space methods to approximation; a common proof uses the zeros of bounded analytic functions in a half-plane. When the series converges, the proof shows more: no power $x^\lambda$ with $\lambda > 0$ outside the set $\set{\lambda_n}$ lies in the closure of the span.