LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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Finite stationary Markov operators contract in weighted L2

Statement

Let $(X, P, π)$ be a finite reversible chain and let $f : X \to C$ . Define $∥ f ∥_{2, π}^{2} : = \sum_{x} π_{x} ∣ f_{x} ∣^{2}$ and $(P f)_{x} : = \sum_{y} P_{x y} f_{y}$ . Then

$∥ P f ∥_{2, π}^{2} \leq ∥ f ∥_{2, π}^{2},$

so $P$ is a contraction on the finite weighted inner-product space $L^{2} (X, π)$ . Detailed balance is not needed for this inequality beyond the stated stationary-law hypotheses.

Facts & Assumptions

Given: A finite reversible chain $(X, P, π)$ , where $π_{x} > 0$ , $\sum_{x} π_{x} = 1$ , $π P = π$ , and $P_{x y} \geq 0$ with $\sum_{y} P_{x y} = 1$ , and a function $f : X \to C$ .

[F1]

Finite reversible Markov chain with positive stationary law: $P$ is row-stochastic and $π P = π$ with every $π_{x} > 0$ .

Proof

1.1givenF1algebra

For each $x$ , set $m_{x} : = \sum_{y} P_{x y} f_{y}$ . Expanding the nonnegative sum gives $\sum_{y} P_{x y} ∣ f_{y} - m_{x} ∣^{2} = \sum_{y} P_{x y} ∣ f_{y} ∣^{2} - ∣ m_{x} ∣^{2} \geq 0$ , because the row weights sum to one. Thus $∣ m_{x} ∣^{2} \leq \sum_{y} P_{x y} ∣ f_{y} ∣^{2}$ .

2.1F1step 1.1algebra

Multiply the inequality in step 1.1 by $π_{x}$ and sum over $x$ . All sums are finite, so reindexing gives $\sum_{x} π_{x} ∣ m_{x} ∣^{2} \leq \sum_{y} (\sum_{x} π_{x} P_{x y}) ∣ f_{y} ∣^{2} = \sum_{y} π_{y} ∣ f_{y} ∣^{2}$ by stationarity.

3.1F1step 2.1algebra∎

The left and right sides in step 2.1 are $∥ P f ∥_{2, π}^{2}$ and $∥ f ∥_{2, π}^{2}$ by definition; taking nonnegative square roots proves the contraction claim. Positivity of every $π_{x}$ makes the weighted form positive definite.

Depends on

Used by

Absolute gap equals one minus the largest nonconstant eigenvalue modulus Theorem

Dependency tree · two levels

9 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

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 18 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, Chapter 8 (standard reference, not scraped)