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Finitely many polynomial moments control the uniform distance on bounded Lipschitz profiles

Statement

Fix $I = [a, b]$ with $a \leq b$ and let $Σ_{I}$ be the set of real functions $σ$ with support in $I$ satisfying $∣ σ (x) - σ (y) ∣ \leq ∣ x - y ∣$ for all $x, y$ . Then for every $ε > 0$ there exist $K \in N$ and $δ > 0$ such that every $σ \in Σ_{I}$ with $∣ \int_{R} σ (x) x^{k} d x ∣ \leq δ (k = 0, 1, \dots, K)$ satisfies $\sup_{x \in R} ∣ σ (x) ∣ \leq ε$ . Consequently, if $(σ_{n}) \subseteq Σ_{I}$ and $\int_{R} σ_{n} (x) x^{k} d x \to 0$ for every $k \geq 0$ , then $σ_{n} \to 0$ uniformly on $R$ ; equivalently, on $Σ_{I}$ the topology of all polynomial moments coincides with the topology of uniform convergence.

Facts & Assumptions

Given: reals $a \leq b$ , the set $Σ_{I}$ of real functions $σ$ vanishing outside $I$ with $∣ σ (x) - σ (y) ∣ \leq ∣ x - y ∣$ for all $x, y$ , and a real $ε > 0$ . For $σ \in Σ_{I}$ the integral $\int_{R} σ (x) x^{k} d x$ of the Statement is read as $\int_{a}^{b} σ (x) x^{k} d x$ (Riemann-Darboux, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ) when $a < b$ , and as $0$ when $a = b$ ; the convention $x^{0} = 1$ is used.

[F1]

A function with $∣ g (x) - g (y) ∣ \leq ∣ x - y ∣$ for all reals $x, y$ is continuous on $R$ : at every point and every real $η > 0$ , $δ : = η$ witnesses continuity (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[F2]

For $a \leq b$ , every continuous real function on $[a, b]$ is a uniform limit of polynomials (Polynomials are uniformly dense in $C ([a, b], R)$ for every closed interval).

[F3]

For $a < b$ , every continuous real function on $[a, b]$ is bounded and Riemann integrable, so its Darboux integral exists (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[F4]

For $a < b$ , if $f, g$ are integrable on $[a, b]$ then so are $f + g$ and $λ f$ for real $λ$ , with $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ); if $f \leq g$ pointwise on $[a, b]$ then $\int_{a}^{b} f \leq \int_{a}^{b} g$ , and if $m \leq f \leq M$ then $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[F5]

Absolute value and its basic inequalities: $∣ u ∣ = u$ for $u \geq 0$ and $∣ u ∣ = - u$ for $u < 0$ (Absolute value in an ordered field); for every real $c > 0$ and real $x$ , $∣ x ∣ \leq c$ if and only if $- c \leq x \leq c$ (Basic properties of the absolute value); and $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ for all reals $x, y$ (The triangle inequality).

[F6]

A sequence $(g_{n})$ of real functions converges uniformly to $0$ on $R$ when for every real $η > 0$ there is $N$ with $∣ g_{n} (x) ∣ < η$ for all $n \geq N$ and all $x$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Proof

technique · direct

1.1givenF1algebra

Boundedness of the profiles: let $σ \in Σ_{I}$ . If $a = b$ , then for every real $r > 0$ the point $y : = a + r$ lies outside $I$ , so $σ (y) = 0$ and $∣ σ (a) ∣ \leq ∣ a - y ∣ = r$ ; hence $∣ σ (a) ∣ = 0$ and $σ \equiv 0$ . If $a < b$ , the Lipschitz bound together with $σ (y) = 0$ for $y < a$ gives $∣ σ (a) ∣ = ∣ σ (a) - σ (y) ∣ \leq a - y$ for every $y < a$ , hence $∣ σ (a) ∣ \leq 0$ , and symmetrically $σ (b) = 0$ ; then for $x \in I$ one has $∣ σ (x) ∣ \leq ∣ x - a ∣$ and $∣ σ (x) ∣ \leq ∣ b - x ∣$ , so $∣ σ (x) ∣ \leq (b - a) / 2$ . Put $C : = \max {1, (b - a) / 2}$ , so $\sup_{R} ∣ σ ∣ \leq C$ for every $σ \in Σ_{I}$ , and $σ$ is continuous on $R$ by [F1]. For the rest of the proof assume $a < b$ ; the case $a = b$ is finished below.

1.2givenF1F3F4F5constructalgebra

The comparison bump: fix $x \in I$ and a real $η > 0$ , and put $u : = \max (a, x - η / 2)$ , $v : = \min (b, x + η / 2)$ , so $u < v$ because $a < b$ and $x \in I$ . Let $c : = (u + v) / 2$ and define $F (y) : = \max {0, \frac{2}{v - u} (1 - \frac{2 ∣ y - c ∣}{v - u})}$ for real $y$ . Then $F \geq 0$ is continuous, its support is $[u, v] \subseteq I \cap [x - η / 2, x + η / 2]$ , and $\int_{R} F = 1$ , the graph of $F$ being a triangle of height $2 (v - u)^{- 1}$ and base $v - u$ . Since every $y$ in the support of $F$ satisfies $∣ y - x ∣ \leq η / 2$ , $σ (y) \geq σ (x) - ∣ y - x ∣ \geq σ (x) - η / 2$ ; hence if $σ (x) > η$ then $σ F \geq (η / 2) F$ pointwise on $I$ and, as $σ F$ and $F$ are continuous there, [F4] and [F3] give $\int_{R} σ F \geq (σ (x) - η / 2) \int_{R} F = σ (x) - η / 2 > η / 2$ , while if $σ (x) < - η$ then symmetrically $\int_{R} σ F \leq σ (x) + η / 2 < - η / 2$ . In either case $∣ σ (x) ∣ > η$ implies $∣ \int_{R} σ F ∣ > η / 2$ , so ${σ \in Σ_{I} : ∣ \int_{R} σ F ∣ \leq η / 2} \subseteq V (x, η) : = {σ \in Σ_{I} : ∣ σ (x) ∣ \leq η}$ .

1.3givenF1F3F4F5algebra

Integral triangle inequality and polynomial bounds: let $f$ be continuous on $[a, b]$ . Then $f$ and $∣ f ∣$ are continuous and integrable by [F3]. Applying [F4] to the two pointwise chains $- ∣ f ∣ \leq f \leq ∣ f ∣$ and $- f \leq ∣ f ∣$ gives $\int f \leq \int ∣ f ∣$ and $- \int f = \int (- f) \leq \int ∣ f ∣$ ; by [F5], $∣ \int f ∣ \leq \int ∣ f ∣$ . Now let $P = \sum_{k = 0}^{K} a_{k} x^{k}$ be a real polynomial with $K \geq 0$ , put $S : = 1 + \sum_{k = 0}^{K} ∣ a_{k} ∣ \geq 1$ , let $η > 0$ and set $δ_{0} : = η / (4 S)$ . If $σ \in Σ_{I}$ satisfies $∣ \int_{R} σ x^{k} ∣ \leq δ_{0}$ for $k = 0, \dots, K$ , then $σ P$ is continuous on $[a, b]$ , [F4] gives $\int_{R} σ P = \sum_{k = 0}^{K} a_{k} \int_{R} σ x^{k}$ , and the integral triangle inequality just proved together with [F5] yields $∣ \int_{R} σ P ∣ \leq \sum_{k = 0}^{K} ∣ a_{k} ∣ ∣ \int_{R} σ x^{k} ∣ \leq \sum_{k = 0}^{K} ∣ a_{k} ∣ δ_{0} \leq η / 4$ .

1.4givenchoosealgebra

A finite mesh: for every real $η > 0$ there are finitely many points $a = x_{1} < x_{2} < \dots < x_{m} = b$ with $x_{i + 1} - x_{i} \leq η$ for all $i < m$ ; one may take $m : = ⌈ (b - a) / η ⌉ + 1$ and split $[a, b]$ into equal parts. Every $x \in I$ then satisfies $∣ x - x_{i} ∣ \leq η$ for at least one mesh point $x_{i}$ .

2.1givenF1F2F3F4F5step 1.1step 1.2step 1.3algebra

Polynomial replacement of the bump: keep the notation of step 1.2 and put $θ : = η / (4 C (b - a)) > 0$ with $C$ from step 1.1. By [F2] choose a polynomial $P$ with $\sup_{y \in I} ∣ F (y) - P (y) ∣ \leq θ$ . The functions $σ F$ , $σ P$ and $σ (F - P)$ are continuous on $[a, b]$ by [F1], hence integrable by [F3], and all three vanish outside $I$ ; therefore, by step 1.3 and [F4], $∣ \int_{R} σ (F - P) ∣ \leq \int_{R} ∣ σ (F - P) ∣ \leq C (b - a) \sup_{I} ∣ F - P ∣ \leq η / 4$ , and if $∣ \int_{R} σ P ∣ \leq η / 4$ , then [F5] gives $∣ \int_{R} σ F ∣ \leq ∣ \int_{R} σ P ∣ + ∣ \int_{R} σ (F - P) ∣ \leq η / 2$ . Combined with step 1.2, ${σ \in Σ_{I} : ∣ \int_{R} σ P ∣ \leq η / 4} \subseteq V (x, η)$ .

2.2givenstep 1.4algebra

From mesh values to the supremum: let $η > 0$ , let $a = x_{1} < \dots < x_{m} = b$ be a mesh as in step 1.4, and let $σ \in Σ_{I}$ satisfy $∣ σ (x_{i}) ∣ \leq η$ for all $i$ . For $x \in I$ choose $i$ with $∣ x - x_{i} ∣ \leq η$ ; then $∣ σ (x) ∣ \leq ∣ σ (x_{i}) ∣ + ∣ x - x_{i} ∣ \leq 2 η$ , while $∣ σ (x) ∣ = 0 \leq 2 η$ for $x \notin I$ . Hence $\sup_{R} ∣ σ ∣ \leq 2 η$ .

3.1givenstep 1.1step 1.3step 1.4step 2.1step 2.2algebra

First claim: given $ε > 0$ , apply steps 2.1 and 1.3 with $η : = ε / 2$ at each mesh point $x_{1}, \dots, x_{m}$ of step 1.4: for each $i$ this produces a polynomial $P_{i} = \sum_{k = 0}^{K_{i}} a_{i, k} x^{k}$ such that ${σ \in Σ_{I} : ∣ \int_{R} σ P_{i} ∣ \leq ε / 8} \subseteq V (x_{i}, ε / 2)$ , and a threshold $δ_{i} : = \frac{ε / 8}{1 + \sum_{k = 0}^{K_{i}} ∣ a_{i, k} ∣} > 0$ such that the moments up to $K_{i}$ being at most $δ_{i}$ force $∣ \int_{R} σ P_{i} ∣ \leq ε / 8$ . Put $K : = \max_{i} K_{i}$ and $δ : = \min_{i} δ_{i} > 0$ ; both depend only on $I$ and $ε$ . Let $σ \in Σ_{I}$ satisfy $∣ \int_{R} σ x^{k} ∣ \leq δ$ for $k = 0, \dots, K$ . For each $i$ the moments up to $K_{i} \leq K$ are at most $δ \leq δ_{i}$ , so $∣ \int_{R} σ P_{i} ∣ \leq ε / 8$ and hence $∣ σ (x_{i}) ∣ \leq ε / 2$ ; step 2.2 with $η = ε / 2$ gives $\sup_{R} ∣ σ ∣ \leq ε$ . In the case $a = b$ every $σ \in Σ_{I}$ is $σ \equiv 0$ by step 1.1, so any $K$ and $δ$ work.

4.1givenF3F4F6step 1.1step 1.3step 3.1algebra∎

Consequence and topology: for a sequence with every moment tending to zero, apply step 3.1 with tolerance $ε / 2$ ; the finitely many moment conditions hold eventually, giving $\sup ∣ σ_{n} ∣ \leq ε / 2 < ε$ , hence uniform convergence by [F6]. To compare the topologies at an arbitrary $τ \in Σ_{I}$ , put $g : = (σ - τ) / 2 \in Σ_{I}$ . Given $ε > 0$ , step 3.1 at tolerance $ε / 4$ supplies $K, δ > 0$ ; if $∣ \int (σ - τ) x^{k} ∣ < 2 δ$ for $k \leq K$ , then $\sup ∣ σ - τ ∣ = 2 \sup ∣ g ∣ \leq ε / 2 < ε$ . Thus a finite intersection of moment neighborhoods of $τ$ lies in each uniform neighborhood. Conversely, for every $k$ , continuity and steps 1.3 and [F4] give $∣ \int (σ - τ) x^{k} ∣ \leq (b - a) \max {1, ∣ a ∣, ∣ b ∣}^{k} \sup ∣ σ - τ ∣$ , so each moment functional is continuous for the uniform topology. These two neighborhood containments prove equality of the topologies; when $a = b$ the space is the singleton zero profile by step 1.1.

Depends on

Used by

Plancherel Young diagrams converge to the limit shape Theorem

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45 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Vladimir Ivanov and Grigori Olshanski, Kerov's central limit theorem for the Plancherel measure on Young diagrams, arXiv:math/0304010; survey-paper version in Symmetric Functions 2001, NATO Science Series II 74 (2002), 93-151 (standard reference, not scraped)