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Chebyshev extremal nodes converge to the arcsine equilibrium measure

Statement

Assume the Axiom of Choice. For $n \geq 1$ let $x_{k, n} : = \cos \frac{k π}{n}$ for $0 \leq k \leq n$ and let $ν_{n} : = \frac{1}{n + 1} \sum_{k = 0}^{n} δ_{x_{k, n}}$ . Then $ν_{n}$ converges weakly, as $n \to \infty$ , to the arcsine equilibrium measure $d x / (π \sqrt{1 - x^{2}})$ of $[- 1, 1]$ , and the points $x_{0, n}, \dots, x_{n, n}$ are exactly the points of $[- 1, 1]$ at which the Chebyshev polynomial $T_{n}$ attains its extreme values $\pm 1$ , alternately signed.

Facts & Assumptions

Given: the nodes $x_{k, n} = \cos \frac{k π}{n}$ , the empirical measures $ν_{n}$ , the Chebyshev polynomials $T_{n}$ of Chebyshev polynomials of the first and second kinds by their three-term recurrences, and the Axiom of Choice.

[F1]

The multiple-angle identity $T_{n} (\cos θ) = \cos (n θ)$ holds for all real $θ$ ( $T_{n} (\cos θ) = \cos (n θ)$ and $U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ)$ for every $n \in N$ ), and $∣ \cos u ∣ \leq 1$ for real $u$ , with $\cos (k π) = (- 1)^{k}$ (Parity and the Pythagorean identity for sine and cosine).

[F2]

The arcsine measure $μ$ of $[- 1, 1]$ is the unique equilibrium measure of $[- 1, 1]$ , and for every continuous $f$ on $[- 1, 1]$ one has $\int f d μ = \frac{1}{π} \int_{0}^{π} f (\cos θ) d θ$ (Arcsine equilibrium measure and capacity of a segment).

[F3]

Dirac measures are probability measures, finite nonnegative weighted sums of measures are measures, and $ν_{n}$ is therefore a Borel probability measure on $[- 1, 1]$ (The Dirac set function at a point, A Dirac set function is a probability measure, Nonnegative scalar multiples and countable weighted sums of measures are measures).

[F4]

Weak convergence $ν_{n} \Rightarrow μ$ means $\int f d ν_{n} \to \int f d μ$ for every bounded continuous real $f$ (Weak convergence of borel probability measures).

[F5]

A continuous real function on the closed bounded interval $[0, π]$ is Riemann integrable, and its uniform-mesh Riemann sums converge to the integral: $\frac{1}{m} \sum_{j = 0}^{m - 1} g (\frac{j π}{m}) \to \frac{1}{π} \int_{0}^{π} g (θ) d θ$ (apply Every continuous function on a closed nondegenerate rectangle in $R^{m}$ is Riemann integrable in dimension $1$ , then The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ ). Its Riemann integral equals its Lebesgue integral by A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, whose Countable Choice hypothesis is supplied by the assumed Axiom of Choice through AC implies DC implies countable choice.

Verification

technique · direct

1.1F3givenalgebra

For each $n \geq 1$ the points $x_{k, n} = \cos \frac{k π}{n}$ lie in $[- 1, 1]$ ; since $\cos$ is strictly decreasing on $[0, π]$ and $\frac{k π}{n}$ are strictly increasing for $k = 0, \dots, n$ , the $n + 1$ points are pairwise distinct, and by [F3] each $ν_{n}$ is a Borel probability measure on $[- 1, 1]$ .

1.2F1givenalgebra

By the multiple-angle identity [F1], $T_{n} (x_{k, n}) = \cos (k π) = (- 1)^{k}$ ; moreover every $x \in [- 1, 1]$ is $x = \cos θ$ for some $θ \in [0, π]$ , so $∣ T_{n} (x) ∣ = ∣ \cos (n θ) ∣ \leq 1$ for every $x \in [- 1, 1]$ , with equality exactly at the points where $∣ \cos n θ ∣ = 1$ , that is, where $n θ$ is an integer multiple of $π$ .

1.3F5given

Weak convergence. Let $f$ be a continuous real function on $[- 1, 1]$ and put $g (θ) : = f (\cos θ)$ ; then $g$ is continuous on $[0, π]$ , so [F5] applied with $m = n$ gives $\frac{1}{n} \sum_{k = 0}^{n - 1} g (\frac{k π}{n}) \to \frac{1}{π} \int_{0}^{π} g (θ) d θ$ .

2.1step 1.2

Consequently the points $x_{k, n}$ are precisely the points of $[- 1, 1]$ at which $T_{n}$ attains $\pm 1$ , with alternating signs, which is the second assertion.

2.2step 1.3algebra

The empirical sum $\frac{1}{n + 1} \sum_{k = 0}^{n} g (\frac{k π}{n})$ differs from $\frac{1}{n} \sum_{k = 0}^{n - 1} g (\frac{k π}{n})$ by at most $\frac{2}{n} \sup_{[0, π]} ∣ g ∣$ , which tends to $0$ , so it has the same limit $\frac{1}{π} \int_{0}^{π} g d θ$ .

3.1step 2.2F2F3algebra

By the definition of $ν_{n}$ and [F3], $\int f d ν_{n} = \frac{1}{n + 1} \sum_{k = 0}^{n} f (x_{k, n}) = \frac{1}{n + 1} \sum_{k = 0}^{n} g (\frac{k π}{n})$ , and by [F2] the limit $\frac{1}{π} \int_{0}^{π} g d θ$ equals $\int f d μ$ ; hence $\int f d ν_{n} \to \int f d μ$ for every continuous $f$ on $[- 1, 1]$ .

4.1step 2.1step 3.1F2F4given∎

Since $[- 1, 1]$ is compact, every continuous real $f$ on it is bounded; step 3.1 therefore gives convergence of integrals for every bounded continuous test function on the metric space $[- 1, 1]$ . By [F4] this is $ν_{n} \Rightarrow μ$ , which together with step 2.1 proves both assertions.

Remarks

Why the weights are $\frac{1}{n + 1}$ . The extremal points of $T_{n}$ are the $n + 1$ points $\cos \frac{k π}{n}$ ; the Riemann sum of step 1.3 is naturally indexed by $k = 0, \dots, n - 1$ , and step 2.2 records that replacing $n$ weights by $n + 1$ weights and adjoining the endpoint $θ = π$ changes the average by $O (1 / n)$ only. The endpoint contribution vanishes in the limit and does not affect the weak limit.

The limit is the equilibrium measure. The identification of the limit with the arcsine measure is exactly the equilibrium computation of Arcsine equilibrium measure and capacity of a segment; this example supplies the discrete approximation of that measure by Chebyshev nodes.

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Sources

E. B. Saff, Logarithmic Potential Theory with Applications to Approximation Theory, §1 (standard reference, not scraped)
B. Khoruzhenko, LTCC Potential Theory notes, §5 (standard reference, not scraped)