Müntz–Szász Theorem | Real Analysis

Approximate

\sqrt{x}

by least squares and watch the theorem decide: the dense basis

\{1, x, \ldots, x^N\}

(

\lambda_n = n

\sum \frac{1}{n} = \infty

) shrinks its error

0.1003 \to 0.0708 \to 0.0544

, while the sparse basis

\{1, x, x^4, \ldots, x^{N^2}\}

(

\sum \frac{1}{n^2} = \frac{\pi^2}{6} < \infty

) stays stuck near

0.16

Examples & Exercises

Prove $\sum_{n \ge 1} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty$ , and deduce from the necessity direction of the Müntz–Szász theorem that $\operatorname{span}\{1, x, x^4, x^9, \ldots\}$ is not dense in $C[0,1]$ . The proof of the necessity direction itself is marked as an idea-level sketch.

\frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \sum_{k=1}^{\infty} \frac{b_k^2}{2}, \qquad b_k = \frac{2(-1)^{k+1}}{k}

\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty \;\Rightarrow\; \operatorname{span}\{1, x^{n^2}\} \text{ not dense}

N = 6

: the sparse-basis error

\approx 0.1676

stays stuck near

0.16

— the proof below is why this stall was inevitable.

Show proof

First compute $\sum 1/n^2 = \pi^2/6$ (complete proof). Take $f(x) = x$ as an odd function on $(-\pi, \pi)$ ; integration by parts gives its sine coefficients $b_k = \frac{1}{\pi}\int_{-\pi}^{\pi} x \sin(kx)\,dx = \frac{2(-1)^{k+1}}{k}$ . Parseval's identity says $\frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \sum_{k=1}^{\infty} \frac{b_k^2}{2}$ ; the left side is $\frac{2\pi^2}{3}$ , so $\frac{2\pi^2}{3} = 2\sum_{k=1}^{\infty} \frac{1}{k^2}$ , i.e. $\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} \approx 1.6449 < \infty$ .
Now quote the necessity direction of the Müntz–Szász theorem: if $\sum \frac{1}{\lambda_n} < \infty$ then $\operatorname{span}\{1, x^{\lambda_1}, x^{\lambda_2}, \ldots\}$ is not dense in $C[0,1]$ . [Sketch marker: the full proof of this direction needs complex analysis — convergence of $\sum 1/\lambda_n$ makes a Blaschke-type infinite product converge, producing a nonzero analytic function orthogonal (in a weighted $L^2$ inner product) to every $x^{\lambda_n}$ ; Hahn–Banach then separates a continuous function at fixed positive distance from the span. We use only the conclusion here.] For $\lambda_n = n^2$ we have $\sum \frac{1}{\lambda_n} = \frac{\pi^2}{6} < \infty$ , so $\operatorname{span}\{1, x, x^4, x^9, \ldots\}$ is not dense in $C[0,1]$ — which is why the sparse basis in the scene stays stuck at error $\approx 0.16$ .

Prove that every uniform limit of even polynomials on $[-1, 1]$ is an even function (complete proof), and conclude that $\operatorname{span}\{1, x^2, x^4, \ldots\}$ is not dense in $C[-1, 1]$ — a warm-up done directly, without quoting Müntz–Szász.

f(x) = \lim_{n \to \infty} p_n(x) = \lim_{n \to \infty} p_n(-x) = f(-x)

g(x) = x \in C[-1,1] \text{ is not even} \;\Rightarrow\; \operatorname{span}\{x^{2k}\} \text{ not dense}

N = 6

: the dense basis closes in on

\sqrt{x}

to error

\approx 0.0708

— with all polynomials available (unlike even degrees only), every continuous function is reachable.

Show proof

Let $p_n$ be even polynomials with $p_n \to f$ uniformly on $[-1, 1]$ . Uniform convergence implies pointwise convergence, so $f(x) = \lim_{n \to \infty} p_n(x)$ for each $x \in [-1, 1]$ . Each $p_n$ is even, so $p_n(x) = p_n(-x)$ , and hence $f(x) = \lim_{n \to \infty} p_n(x) = \lim_{n \to \infty} p_n(-x) = f(-x)$ . So $f$ is even — a uniform limit of even polynomials is always even. (End of proof.)
$g(x) = x$ is continuous on $[-1, 1]$ but not even. If $\operatorname{span}\{1, x^2, x^4, \ldots\}$ were dense in $C[-1, 1]$ , some sequence of even polynomials would converge uniformly to $g$ , and by what we just proved the limit $g$ would have to be even — a contradiction. Hence $\operatorname{span}\{1, x^2, x^4, \ldots\}$ is not dense in $C[-1, 1]$ . (Note: this is a $C[-1,1]$ statement, distinct from the scene's sparse basis $\{1, x, x^4, \ldots\}$ on $C[0,1]$ — the non-density of the latter is the job of the Müntz–Szász necessity direction in the first problem.)

For $\lambda_n = n$ , show $\sum \frac{1}{n} = \infty$ and prove directly — via the Weierstrass approximation theorem, without quoting the sufficiency direction of Müntz–Szász — that $\operatorname{span}\{1, x, x^2, \ldots\}$ is dense in $C[0,1]$ . Concretely, check $x^2 \in \operatorname{span}\{1, x, \ldots, x^N\}$ for $N \ge 2$ , and read off the numbers with which the scene's dense basis closes in on $\sqrt{x}$ .

H_n = \sum_{k=1}^{n} \frac{1}{k} > \int_1^{n+1} \frac{dx}{x} = \ln(n+1) \to \infty

\operatorname{span}\{1, x, x^2, \ldots\} = \text{all polynomials}

\text{all polynomials} \xrightarrow{\text{Weierstrass}} \text{dense in } C[0,1]

N = 8

: dense-basis error

\approx 0.0544

— down from

0.1003

N = 4

and

0.0708

N = 6

. A basis with

\sum \frac{1}{n} = \infty

never stalls.

Show proof

The harmonic series diverges: for $H_n = \sum_{k=1}^{n} \frac{1}{k}$ , each term satisfies $\frac{1}{k} \ge \int_{k}^{k+1} \frac{dx}{x}$ , so $H_n \ge \int_{1}^{n+1} \frac{dx}{x} = \ln(n+1) \to \infty$ . Hence for $\lambda_n = n$ we have $\sum \frac{1}{\lambda_n} = \infty$ . (Numerical witness: $H_{1000} \approx 7.4855 > \ln(1001) \approx 6.9088$ .)
For $\lambda_n = n$ , $\operatorname{span}\{1, x, x^2, \ldots\}$ is exactly the set of all (real-coefficient) polynomials — $x^k$ is itself the $k$ -th basis element, so every polynomial of degree at most $d$ is a linear combination of the first $d+1$ basis elements. By the Weierstrass approximation theorem, polynomials are uniformly dense in $C[0,1]$ , so this span is dense too. No need to quote the sufficiency direction of Müntz–Szász at all — Weierstrass handles this case directly.
Concrete check: for $N \ge 2$ , $x^2$ is itself a basis element, so it trivially lies in $\operatorname{span}\{1, x, \ldots, x^N\}$ — exact membership, not approximation. The scene shows this density working in practice even on a function with a cusp like $\sqrt{x}$ : the dense basis's least-squares error shrinks $0.1003 \to 0.0708 \to 0.0544$ for $N = 4, 6, 8$ . (End of proof.)