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Cauchy–Schwarz Estimates for String Mode Coefficients

Statement

Let $D \geq 1$ be finite, let $η = diag (- 1, + 1, \dots, + 1)$ , and extend its bilinear contraction to $C^{D}$ . For $u \in C^{D}$ , define $∣ u ∣_{E}^{2} : = \sum_{μ = 0}^{D - 1} ∣ u^{μ} ∣^{2}$ . For a sequence $a = (a_{m})_{m \in Z}$ in $C^{D}$ , define $∥ a ∥_{ℓ^{2}, E}^{2} : = \sum_{m \in Z} ∣ a_{m} ∣_{E}^{2}$ . If $a, b$ have finite such norms, then for every $n \in Z$ the convolution is absolutely convergent and

$\sum_{m \in Z} ∣ a_{m} \cdot b_{n - m} ∣ \leq ∥ a ∥_{ℓ^{2}, E} ∥ b ∥_{ℓ^{2}, E} .$

For a continuous periodic function $Q : R \to C^{D}$ , define $∥ Q ∥_{2, E}^{2} : = (2 π)^{- 1} \int_{- π}^{π} ∣ Q (s) ∣_{E}^{2} d s$ ; for a continuous scalar periodic function $r$ , define $∥ r ∥_{1} : = (2 π)^{- 1} \int_{- π}^{π} ∣ r (s) ∣ d s$ . Then for continuous periodic $A, B : R \to C^{D}$ ,

$∥ A \cdot B ∥_{1} \leq ∥ A ∥_{2, E} ∥ B ∥_{2, E} .$

Here $\cdot$ is always the Minkowski contraction, while the norms used in estimates are positive Euclidean norms.

Facts & Assumptions

Given: A finite target dimension $D$ , the mostly-plus diagonal matrix $η$ , square-summable vector sequences, and continuous $2 π$ -periodic vector functions.

[A1]

The sequence norms are nonnegative extended sums that are finite where the statement assumes square summability; integrals in the function norms are normalized Riemann integrals of continuous functions on the compact interval.

Proof

technique · prove the finite Cauchy–Schwarz bound by a nonnegative quadratic, then pass to monotone limits and integrate

1.1givenalgebra

For nonnegative finite lists $(x_{j})$ and $(y_{j})$ , if $\sum_{j} y_{j}^{2} = 0$ , then every $y_{j} = 0$ and $\sum_{j} x_{j} y_{j} = 0$ . Otherwise put $t = (\sum_{j} x_{j} y_{j}) / (\sum_{j} y_{j}^{2})$ . Expanding $0 \leq \sum_{j} (x_{j} - t y_{j})^{2}$ gives $(\sum_{j} x_{j} y_{j})^{2} \leq (\sum_{j} x_{j}^{2}) (\sum_{j} y_{j}^{2})$ . Applying this to $x_{μ} = ∣ u^{μ} ∣$ and $y_{μ} = ∣ v^{μ} ∣$ , and using $∣ η_{μ μ} ∣ = 1$ , gives $∣ u \cdot v ∣ \leq \sum_{μ} ∣ u^{μ} ∣ ∣ v^{μ} ∣ \leq ∣ u ∣_{E} ∣ v ∣_{E}$ .

2.1givenstep 1.1algebra

For any finite index set $J \subset Z$ , step 1.1 applied to $x_{m} = ∣ a_{m} ∣_{E}$ , $y_{m} = ∣ b_{n - m} ∣_{E}$ gives $\sum_{m \in J} ∣ a_{m} \cdot b_{n - m} ∣ \leq (\sum_{m \in J} ∣ a_{m} ∣_{E}^{2})^{1 / 2} (\sum_{m \in J} ∣ b_{n - m} ∣_{E}^{2})^{1 / 2} \leq ∥ a ∥_{ℓ^{2}, E} ∥ b ∥_{ℓ^{2}, E} .$ Taking $J = {- N, \dots, N}$ makes the left sides increase to the full nonnegative series; their common finite bound proves absolute convergence and the asserted inequality.

3.1A1step 1.1algebra∎

Put $a (s) = ∣ A (s) ∣_{E}$ and $b (s) = ∣ B (s) ∣_{E}$ . If $\int b^{2} = 0$ , continuity and nonnegativity imply $b = 0$ everywhere: a positive value would make $b$ bounded below by a positive constant on a nonempty interval and force a positive integral. Otherwise set $t = ((2 π)^{- 1} \int a b) / ((2 π)^{- 1} \int b^{2})$ . Expanding $0 \leq (2 π)^{- 1} \int_{- π}^{π} (a - t b)^{2} d s$ gives $((2 π)^{- 1} \int a b)^{2} \leq ∥ A ∥_{2, E}^{2} ∥ B ∥_{2, E}^{2}$ . By step 1.1 pointwise, $∣ A (s) \cdot B (s) ∣ \leq a (s) b (s)$ ; integrate and take nonnegative square roots to obtain $∥ A \cdot B ∥_{1} \leq ∥ A ∥_{2, E} ∥ B ∥_{2, E}$ .

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Cauchy–Schwarz Estimates for String Mode Coefficients

Statement

Facts & Assumptions

Proof

Dependency tree · 0 levels

Sources